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Unit 3: Trigonometric functions

Degrees and radians.

  • Intro to radians (Opens a modal)
  • Radians & degrees (Opens a modal)
  • Degrees to radians (Opens a modal)
  • Radians to degrees (Opens a modal)
  • Radian angles & quadrants (Opens a modal)
  • Radians & degrees Get 3 of 4 questions to level up!

Unit circle

  • Unit circle (Opens a modal)
  • The trig functions & right triangle trig ratios (Opens a modal)
  • Trig values of π/4 (Opens a modal)
  • Trig unit circle review (Opens a modal)
  • Plotting angles on unit circle Get 3 of 4 questions to level up!
  • Sign of trigonometric functions Get 3 of 4 questions to level up!
  • Trig values of special angles Get 3 of 4 questions to level up!

Trigonometric functions

  • Graph of y=sin(x) (Opens a modal)
  • Graph of y=tan(x) (Opens a modal)
  • Intersection points of y=sin(x) and y=cos(x) (Opens a modal)
  • Find value of other trigonometric functions from given trigonometric function Get 3 of 4 questions to level up!

Trigonometric identities: Symmetry

  • Sine & cosine identities: symmetry (Opens a modal)
  • Tangent identities: symmetry (Opens a modal)
  • Sine & cosine identities: periodicity (Opens a modal)
  • Tangent identities: periodicity (Opens a modal)

Trigonometric identities: Sum and difference

  • Trig angle addition identities (Opens a modal)
  • Using the cosine angle addition identity (Opens a modal)
  • Using the cosine double-angle identity (Opens a modal)
  • Proof of the sine angle addition identity (Opens a modal)
  • Proof of the cosine angle addition identity (Opens a modal)
  • Trig challenge problem: cosine of angle-sum (Opens a modal)
  • Trigonometric functions of sum and difference of angles Get 3 of 4 questions to level up!
  • Sum and difference of trigonometric functions Get 3 of 4 questions to level up!
  • Evaluate trigonometric expressions (intermediate) Get 3 of 4 questions to level up!

Trigonometric equations

  • Proof of the Pythagorean trig identity (Opens a modal)
  • Using the Pythagorean trig identity (Opens a modal)
  • Solving sinusoidal equations of the form sin(x)=d (Opens a modal)
  • Solving cos(θ)=1 and cos(θ)=-1 (Opens a modal)
  • Use the Pythagorean identity Get 3 of 4 questions to level up!
  • Principal solutions of trigonometric equation Get 3 of 4 questions to level up!
  • General solution of trigonometric equation Get 3 of 4 questions to level up!

Solutions to select NCERT problems

  • Select problems from exercise 3.3 (Opens a modal)
  • Select problems from miscellaneous exercise (Opens a modal)

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NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions

  • NCERT Solutions

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NCERT Solutions for Class 11 Maths Chapter 3 provides 100% accurate and comprehensive answers to all questions from NCERT textbooks. All students aspiring to excel in their entrance exams should refer to these study guides for more profound knowledge and better grades in this subject. These solutions have been drafted by teaching experts following the latest CBSE guidelines.

The other chapters in class 11 maths ncert solutions are very important and available in PDF format. You can also download physics class 11 and chemistry class 11 solutions on vedantu site.

Trigonometric Functions Chapter at a Glance - Class 11 NCERT Solutions

$1^a-\left(\frac{1}{360}\right)^2$ of a revolution

$1^{\circ}-60^{\prime}, 1^{\prime}-60^{\circ}$

1 Revolution $-2 \pi$ radians

$1^2=\frac{\pi}{180}$ radian

If in a circle of radius $r$, an arc of length $l$ subtends an angle of $\theta$ radians, then $\theta^*=\frac{l}{r}$

Trigonometric identities

(i) $\sin ^2 \theta+\cos ^2 \theta=1$

(ii) $\sec ^2 \theta-1+\tan ^2 \theta$

(iii) $\operatorname{coses}^2 \theta=1+\cot ^2 \theta$

(i) $\sin (-x)=-\sin x$

(ii) $\cos (-x)=\cos x$

(iii) $\tan (-x)=-\tan x$

(iv) $\operatorname{cosec}(-x)=-\operatorname{cosec} x$

(v) $\sec (-x)-\sec x$

(vi) $\cot (-x)=-\cot x$

(i) $\sin \left(\frac{\pi}{2} \pm \theta\right)=\cos \theta$

(ii) $\cos \left(\frac{\pi}{2} \pm \theta\right)=\mp \sin \theta$

(iii) $\tan \left(\frac{\pi}{2} \pm \theta\right)=\mp \cot \theta$

(iv) $\cot \left(\frac{\pi}{2} \pm \theta\right)=\mp \tan \theta$

(v) $\sec \left(\frac{\pi}{2} \pm \theta\right)=\mp \operatorname{cosec} \theta$

(vi) $\operatorname{cosec}\left(\frac{\pi}{2} \pm \theta\right)=\sec \theta$

(i) $\sin (\pi \pm \theta)= \pm \sin \theta$

(ii) $\cos (\pi \pm \theta)=-\cos \theta$

(iii) $\tan (\pi \pm \theta)=\mp \tan \theta$

(iv) $\cot (\pi \pm \theta)=\mp \cot \theta$

(v) $\sec (\pi \pm \theta)=-\sec \theta$

(vi) $\operatorname{coses}(\pi \pm \theta)=\mp \operatorname{cosec} \theta$

(i) $\sin \left(\frac{3 \pi}{2} \pm \theta\right)=-\cos \theta$

(ii) $\cos \left(\frac{3 \pi}{2} \pm \theta\right)=\sin \theta$

(iii) $\tan \left(\frac{3 \pi}{2} \pm \theta\right)=-\cot \theta$

(iv) $\cot \left(\frac{3 \pi}{2} \pm \theta\right)=\tan \theta$

(v) $\sec \left(\frac{3 \pi}{2} \pm \theta\right)-\operatorname{cosec} \theta$

(vi) $\operatorname{cosec}\left(\frac{3 \pi}{2} \pm \theta\right)=-\sec \theta$

(i) $\sin (2 \pi \pm \theta)=\mp \sin \theta$

(ii) $\cos (2 \pi \pm \theta)=\cos \theta$

(iii) $\tan (2 \pi \pm \theta)=\mp \tan \theta$

(iv) $\cot (2 \pi \pm \theta)=\mp \cot \theta$

(v) $\sec (2 \pi \pm \theta)=\sec \theta$

(vi) $\operatorname{cosec}(2 \pi \pm \theta)-\mp \operatorname{cosec} \theta$

Trigonometric functions of sum and difference of angles:

$\sin (x \pm y)=\sin x \cos y \pm \cos x \sin y$

$\cos (x \pm y)=\cos x \cos y \mp \sin x \sin y$

If nome of the angles $x, y$ and $(x \pm y)$ is an odd multiple of $\frac{\pi}{2}$, then $\tan (x \pm y)=\frac{\tan x \pm \tan y}{17 \tan x \tan y}$

If nome of the angles $x, y$ and $(x \pm y)$ is a multiple of $\pi$, then $\cot (x \pm y)-\frac{\cot x \cot y \mp 1}{\cot y \pm \cot x}$

Multiple and Submultiple Angle formula:

$\cos 2 x=\cos ^2 x-\sin ^2 x$

$$ \begin{aligned} & -2 \cos ^2 x-1-1-2 \sin ^2 x \\ & =\frac{1-\tan ^2 x}{1+\tan ^2 x} \end{aligned} $$

$\sin 2 x=2 \sin x \cdot \cos x=\frac{2 \tan x}{1+\tan ^2 x}$

$\tan 2 x=\frac{2 \tan x}{1-\tan ^2 x}$

$\sin 3 x=3 \sin x-4 \sin ^3 x$

$\cos 3 x-4 \cos ^3 x-3 \cos x$

$\tan 3 x=\frac{3 \tan x-\tan ^3 x}{1-3 \tan ^2 x}$

Trigonometric Transformation formulae:

$\sin C+\sin D$ $$ =2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right) $$ $\sin C-\sin D$ $$ =2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right) $$ $\cos C+\cos D$ $$ 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right) $$ $\cos C-\cos D$ $$ =-2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right) $$ - $2 \sin A \cos B$ $$ -\sin (A+B)+\sin (A-B) $$ - $2 \cos A \sin B$ $$ -\sin (A+B)-\sin (A-B) $$ - $2 \cos A \cos B$ $$ =\cos (A+B)+\cos (A-B) $$ - $2 \sin A \sin B$ $$ -\cos (A-B)-\cos (A+B) $$

(i) $\sin \theta=\sin \alpha$ $$ \Rightarrow \theta=n \pi+(-1)^n \alpha, n \in Z $$

(ii)$\cos \theta=\cos x$ $$ \Rightarrow \theta=2 n x \pm \alpha, n \in Z $$

(iii)$\tan \theta=\tan x$ $$ \Rightarrow \theta=n \pi+\alpha, n \in Z $$ $\sin ^2 \theta=\sin ^2 \alpha$

(iv) $\left.\cos ^2 \theta=\cos ^2 \alpha\right\} \theta=n \pi \pm a, n \in Z$ $\tan ^2 x=\tan ^2 \alpha$

Exercises under NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Exercise 3.1: In this exercise, students are introduced to trigonometric ratios of acute angles and their applications in solving problems related to heights and distances. The exercise covers basic concepts such as the definition of trigonometric ratios, Pythagoras theorem, and the concept of complementary angles.

Exercise 3.2: This exercise focuses on the evaluation of trigonometric ratios of special angles such as 0, 30, 45, 60, and 90 degrees. The exercise also includes the derivation of trigonometric ratios of 30, 45, and 60 degrees using the concept of the unit circle.

Exercise 3.3: This exercise covers the trigonometric ratios of angles between 0 and 90 degrees. The exercise includes problems on finding the values of trigonometric ratios using various techniques such as the use of Pythagoras theorem, the double-angle formula, and the half-angle formula.

Exercise 3.4: In this exercise, students learn about the trigonometric ratios of complementary angles and the concept of co-functions. The exercise covers problems on finding the values of trigonometric ratios using the co-function identity.

Miscellaneous Exercise: This exercise includes a variety of problems that cover different topics such as the use of trigonometric ratios in solving real-life problems, the use of trigonometric identities to simplify expressions, and the solution of trigonometric equations. This exercise provides an opportunity for students to apply the concepts learned in the chapter to solve more complex problems.

Access NCERT Solution for Class 11 Maths Chapter 3 - Trigonometric Functions

Exercise 3.1

1. Find the radian measures corresponding to the following degree measures:

(i) $\text{2}{{\text{5}}^{\text{o}}}$

Ans: We know that $\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }$ radian

Therefore  ${{1}^{\circ }}=\dfrac{\pi }{180}$ radian 

hence, 

$\text{2}{{\text{5}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ 25}$ radian  

      $\text{=}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{36}}$radian

(ii) $\text{-4}{{\text{7}}^{\text{o}}}\text{30 }\!\!'\!\!\text{ }$

Ans: Here we have,

  $\text{-4}{{\text{7}}^{\text{o}}}\text{30 }\!\!'\!\!\text{ =-47}{{\dfrac{\text{1}}{\text{2}}}^{\text{o}}}$   

           $\text{=-}\dfrac{\text{95}}{\text{2}}$ degree

Since we know that, $\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }$ radian

Therefore  ${{\text{1}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}$ radian 

$\text{-}\dfrac{\text{95}}{\text{2}}$ degree$\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ }\left( \dfrac{\text{-95}}{\text{2}} \right)$ radian

                  $\text{=}\left( \dfrac{\text{-19}}{\text{36 }\!\!\times\!\!\text{ 2}} \right)\text{ }\!\!\pi\!\!\text{ }$  radian

                  $\text{=}\dfrac{\text{-19}}{\text{72}}\text{ }\!\!\pi\!\!\text{ }$radian

$\text{-4}{{\text{7}}^{\text{o}}}\text{30 }\!\!'\!\!\text{ =-}\dfrac{\text{19}}{\text{72}}\text{ }\!\!\pi\!\!\text{ }$ radian

(iii) $\text{24}{{\text{0}}^{\text{o}}}$

Ans: We know that,

$\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }$ radian

$\text{24}{{\text{0}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ 240}$ radian

       $\text{=}\dfrac{\text{4}}{\text{3}}\text{ }\!\!\pi\!\!\text{ }$radian

(iv) $\text{52}{{\text{0}}^{\text{o}}}$

$\text{52}{{\text{0}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ 520}$ radian

       $\text{=}\dfrac{\text{26 }\!\!\pi\!\!\text{ }}{\text{9}}$radian

2. Find the degree measures corresponding to the following radian measures

(Use$\text{ }\!\!\pi\!\!\text{ =}\dfrac{\text{22}}{\text{7}}$ ) 

(i) $\dfrac{\text{11}}{\text{16}}$

 $\text{ }\!\!\pi\!\!\text{ }$ radian$\text{=18}{{\text{0}}^{\text{o}}}$

Therefore  $\text{1 radian =}{{\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}}^{\text{o}}}$ 

$\dfrac{\text{11}}{\text{16}}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{11}}{\text{16}}$ degree

              $\text{=}\dfrac{\text{45 }\!\!\times\!\!\text{ 11}}{\text{ }\!\!\pi\!\!\text{  }\!\!\times\!\!\text{ 4}}$degree

              \[\text{=}\dfrac{\text{45 }\!\!\times\!\!\text{ 11 }\!\!\times\!\!\text{ 7}}{\text{22 }\!\!\times\!\!\text{ 4}}\] degree

              $\text{=}\dfrac{\text{315}}{\text{8}}$degree

Further computing,

$\dfrac{\text{11}}{\text{16}}$ radian$\text{=39}\dfrac{\text{3}}{\text{8}}$ degree

                $\text{=3}{{\text{9}}^{\text{o}}}\text{+}\dfrac{\text{3 }\!\!\times\!\!\text{ 60}}{\text{8}}$ minutes

Since ${{\text{1}}^{\text{o}}}\text{=60 }\!\!'\!\!\text{ }$

    $\dfrac{\text{11}}{\text{16}}$ radian $\text{=3}{{\text{9}}^{\text{o}}}\text{+22 }\!\!'\!\!\text{ +}\dfrac{\text{1}}{\text{2}}$minutes

Since  $\text{1 }\!\!'\!\!\text{ =60''}$ 

$\dfrac{\text{11}}{\text{16}}$ radian$\text{ }\!\!~\!\!\text{ =3}{{\text{9}}^{\text{o}}}\text{22 }\!\!'\!\!\text{ 30''}$

(ii) $\text{-4}$

$\text{ }\!\!\pi\!\!\text{ }$ radian$\text{=18}{{\text{0}}^{\text{o}}}$

$\text{-4}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\left( \text{-4} \right)$ degree

                $\text{=}\dfrac{\text{180 }\!\!\times\!\!\text{ 7}\left( \text{-4} \right)}{\text{22}}$degree

                $\text{=}\dfrac{\text{-2520}}{\text{11}}$ degree

                $\text{=-229}\dfrac{\text{1}}{\text{11}}$degree

$\text{-4}$ radian$\text{=-22}{{\text{9}}^{\text{o}}}\text{+}\dfrac{\text{1 }\!\!\times\!\!\text{ 60}}{\text{11}}$ minutes                        

               $\text{=-22}{{\text{9}}^{\text{o}}}\text{+5 }\!\!'\!\!\text{ +}\dfrac{\text{5}}{\text{11}}$ minutes

Since $\text{1 }\!\!'\!\!\text{ =60 }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$

$\text{-4}$ radian$\text{=-22}{{\text{9}}^{\text{o}}}\text{5 }\!\!'\!\!\text{ 27''}$

(iii) $\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}$

$\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}$ degree

                 $\text{=30}{{\text{0}}^{\text{o}}}$

(iv)$\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

 $\pi $ radian$\text{=18}{{\text{0}}^{\text{o}}}$

$\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

                 $\text{=21}{{\text{0}}^{\text{o}}}$

3. A wheel makes $\text{360}$ revolutions in one minute. Through how many radians does it turn in one second?

Ans: Number of revolutions the wheel makes in $\text{1}$ minute$\text{=360}$

 Number of revolutions the wheel make in $\text{1}$ second$\text{=}\dfrac{\text{360}}{\text{60}}$

                                                                                   $\text{=6}$

In one complete revolution, the wheel turns an angle of \[\text{2 }\!\!\pi\!\!\text{ }\] radian.

Hence, it will turn an angle of $\text{6 }\!\!\times\!\!\text{ 2 }\!\!\pi\!\!\text{ =12 }\!\!\pi\!\!\text{ }$ radian, in $\text{6}$ complete revolutions.

Therefore, the wheel turns an angle of $\text{12 }\!\!\pi\!\!\text{ }$ radian in one second.

4. Find the degree measure of the angle subtended at the centre of a circle of radius $\text{100}$cm by an arc of length  $\text{22}$  cm.

 in a circle of radius $\text{r}$ unit, if  an angle $\text{ }\!\!\theta\!\!\text{ }$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}$                                ……(1)

Therefore, 

Substituting $\text{r=100cm}$ ,\[\text{l=22cm}\] in the formula (1) , we have,

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{22}}{\text{100}}$ radian

Since $\text{1 radian=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}$ 

 $\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{22}}{\text{100}}$ degree

     $\text{=}\dfrac{\text{180 }\!\!\times\!\!\text{ 7 }\!\!\times\!\!\text{ 22}}{\text{22 }\!\!\times\!\!\text{ 100}}$degree

     $\text{=}\dfrac{\text{63}}{\text{5}}$ degree

    $\text{=12}\dfrac{\text{3}}{\text{5}}$degree

Since  ${{\text{1}}^{\text{o}}}\text{=60 }\!\!'\!\!\text{ }$ , we have,

$\text{ }\!\!\theta\!\!\text{ =1}{{\text{2}}^{\text{o}}}\text{36 }\!\!'\!\!\text{ }$

Hence , the required angle is $\text{1}{{\text{2}}^{\text{o}}}\text{36 }\!\!'\!\!\text{ }$.

5. In a circle of diameter $\text{40}$ cm, the length of a chord is $\text{20}$ cm. Find the length of minor arc of the chord.

Ans: Given that, diameter of the circle$=40$ cm

 Hence Radius $\left( r \right)$ of the circle$=\dfrac{40}{2}cm$

                                                   $=20cm$

Let $\text{AB}$ be a chord  of the circle whose length is $20$ cm.

(Image will be uploaded soon)

             

In $\Delta \text{OAB,}$

$\text{OA=OB}$ 

     $=$ Radius of the circle

     $\text{=20cm}$

Now also, $\text{AB=20cm}$

Therefore, $\Delta \text{OAB}$ is an equilateral triangle.

\[\therefore \text{ }\!\!\theta\!\!\text{ =6}{{\text{0}}^{\text{o}}}\]

     $\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ radian

We know that,

Substituting $\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$  in the formula (1),

       $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\dfrac{\text{arc AB}}{\text{20}}$

$\text{arc AB=}\dfrac{\text{20 }\!\!\pi\!\!\text{ }}{\text{3}}\text{cm}$

Therefore, the length of the minor arc of the chord is $\dfrac{\text{20 }\!\!\pi\!\!\text{ }}{\text{3}}\text{cm}$ .

6. If in two circles, arcs of the same length subtend angles $\text{6}{{\text{0}}^{\text{o}}}$ and $\text{7}{{\text{5}}^{\text{o}}}$ at the centre, find the ratio of their radii.

Ans: Let the radii of the two circles be ${{\text{r}}_{\text{1}}}$ and ${{\text{r}}_{\text{2}}}$ . Let an arc of length ${{\text{l}}_{\text{1}}}$ subtends an angle of $\text{6}{{\text{0}}^{\text{o}}}$ at the centre of the circle of radius ${{\text{r}}_{\text{1}}}$ , whereas let an arc of length ${{\text{l}}_{\text{2}}}$ subtends an angle of $\text{7}{{\text{5}}^{\text{o}}}$ at the centre of the circle of radius ${{\text{r}}_{\text{2}}}$ .

Now, we have,

$\text{6}{{\text{0}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ radian  and 

${{75}^{\circ }}=\dfrac{5\pi }{12}$ radian

\[\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}\]  

$\text{l=r }\!\!\theta\!\!\text{ }$          

Hence we obtain,                   

$\text{l=}\dfrac{{{\text{r}}_{\text{1}}}\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ 

and  $\text{l=}\dfrac{{{\text{r}}_{\text{2}}}\text{5 }\!\!\pi\!\!\text{ }}{\text{12}}$

according to the ${{\text{l}}_{\text{1}}}\text{=}{{\text{l}}_{\text{2}}}$ 

thus we have,

$\dfrac{{{\text{r}}_{\text{1}}}\text{ }\!\!\pi\!\!\text{ }}{\text{3 }\!\!\pi\!\!\text{ }}\text{=}\dfrac{{{\text{r}}_{\text{2}}}\text{5 }\!\!\pi\!\!\text{ }}{\text{12}}$

  ${{\text{r}}_{\text{1}}}\text{=}\dfrac{{{\text{r}}_{\text{2}}}\text{5}}{\text{4}}$

 $\dfrac{{{\text{r}}_{\text{1}}}}{{{\text{r}}_{\text{2}}}}\text{=}\dfrac{\text{5}}{\text{4}}$

Hence , the ratio of the radii is $\text{5:4}$ .

7. Find the angle in radian through which a pendulum swings if its length is $\text{75}$ cm and the tip describes an arc of length.

(i)  $\text{10}$ cm

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}$  

 Given that $\text{r=75cm}$

And here, $\text{l=10cm}$

Hence substituting the values in the formula,

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{10}}{\text{75}}$ radian

  $\text{=}\dfrac{\text{2}}{\text{15}}$radian

(ii) $\text{15}$ cm

And here, $\text{l=15cm}$  

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{15}}{\text{75}}$ radian

       $\text{=}\dfrac{\text{1}}{\text{5}}$radian

(iii) $\text{21}$ cm

And here, $\text{l=21cm}$

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{21}}{\text{75}}$ radian

       $\text{=}\dfrac{\text{7}}{\text{25}}$radian

Exercise 3.2

1. Find the values of the other five trigonometric functions if  $\text{cos x=-}\dfrac{\text{1}}{\text{2}}$ , $x$ lies in the third quadrant.

Ans: Here given that,  $\text{cos x=-}\dfrac{\text{1}}{\text{2}}$

Therefore we have,

$\text{sec x=}\dfrac{\text{1}}{\text{cos x}}$

          $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{1}}{\text{2}} \right)}$

          $\text{=-2}$

Now we know that,$\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{si}{{\text{n}}^{\text{2}}}\text{x=1-co}{{\text{s}}^{\text{2}}}\text{x}$

Substituting  $\text{cos x=-}\dfrac{\text{1}}{\text{2}}$ in the formula, we obtain,

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}{{\left( \text{-}\dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}$

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}\dfrac{\text{1}}{\text{4}}$

            $\text{=}\dfrac{\text{3}}{\text{4}}$

   $\text{sin x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{3}}}{\text{2}}$

Since $\text{x}$ lies in the ${{\text{3}}^{\text{rd}}}$quadrant, the value of $\sin x$ will be negative.

$\text{sin x=-}\dfrac{\sqrt{\text{3}}}{\text{2}}$

Therefore, $\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}$ 

                            $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)}$ 

                            $\text{=-}\dfrac{\text{2}}{\sqrt{\text{3}}}$ 

 $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ 

        $\text{=}\dfrac{\left( \text{-}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)}{\left( \text{-}\dfrac{\text{1}}{\text{2}} \right)}$

        $\text{=}\sqrt{\text{3}}$

$\text{cot x=}\dfrac{\text{1}}{\text{tan x}}$

       $\text{=}\dfrac{\text{1}}{\sqrt{\text{3}}}$

2. Find the values of other five trigonometric functions if $\text{sin  x=}\dfrac{\text{3}}{\text{5}}$  , $\text{x}$ lies in second quadrant.

Here given that,  $\text{sin x=}\dfrac{\text{3}}{\text{5}}$

$\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}$

          $=\dfrac{1}{\left( \dfrac{3}{5} \right)}$

           $=\dfrac{5}{3}$

Now we know that , ${{\sin }^{2}}x+{{\cos }^{2}}x=1$

Therefore we have, $\text{co}{{\text{s}}^{\text{2}}}\text{x=1-si}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\sin x=\dfrac{3}{5}$  in the formula, we obtain,

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}{{\left( \dfrac{\text{3}}{\text{5}} \right)}^{\text{2}}}$

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}\dfrac{\text{9}}{\text{25}}$

         $\text{=}\dfrac{\text{16}}{\text{25}}$ 

   $\text{cos x= }\!\!\pm\!\!\text{ }\dfrac{\text{4}}{\text{5}}$

Since $x$ lies in the ${{2}^{nd}}$quadrant, the value of $\cos x$ will be negative.

$\text{cos x=-}\dfrac{\text{4}}{\text{5}}$

Therefore, $sec x=\dfrac{1}{\cos x}$ 

                          $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{4}}{\text{5}} \right)}$ 

                          $\text{=-}\dfrac{\text{5}}{\text{4}}$ 

 $\tan x=\dfrac{\sin  x}{\cos x}$ 

        $\text{=}\dfrac{\left( \dfrac{\text{3}}{\text{5}} \right)}{\left( \text{-}\dfrac{\text{4}}{\text{5}} \right)}$

        $\text{=-}\dfrac{\text{3}}{\text{4}}$

$\cot x=\dfrac{1}{\tan x}$

        $\text{=-}\dfrac{\text{4}}{\text{3}}$

3. Find the values of other five trigonometric functions if   $\text{cot x=}\dfrac{\text{3}}{\text{4}}$ , $\text{x}$ lies in third quadrant.

Ans: Here given that,  $\cot x=\dfrac{3}{4}$

$\tan x=\dfrac{1}{\cot x}$

       $=\dfrac{1}{\left( \dfrac{3}{4} \right)}$

       $=\dfrac{4}{3}$

Now we know that , \[\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}\]

Therefore we have, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\text{tan x=}\dfrac{\text{4}}{\text{3}}$  in the formula, we obtain,

${{\sec }^{2}}x=1+{{\left( \dfrac{4}{3} \right)}^{2}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\dfrac{\text{16}}{\text{9}}$

             $=\dfrac{25}{9}$ 

   $\text{sec x= }\!\!\pm\!\!\text{ }\dfrac{\text{5}}{\text{3}}$

Since $x$ lies in the ${{3}^{rd}}$quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\dfrac{\text{5}}{\text{3}}$

Therefore, $\cos  x=\dfrac{1}{\sec x}$ 

                         $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{5}}{\text{3}} \right)}$ 

                          $\text{=-}\dfrac{\text{3}}{\text{5}}$ 

Now  , $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

 Hence we have, $\text{sin x=}\dfrac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{3}}{\text{5}} \right)$   

                                    \[\text{=}\left( \text{-}\dfrac{\text{4}}{\text{5}} \right)\] 

 And 

          $\text{=-}\dfrac{\text{5}}{\text{4}}$

4. Find the values of other five trigonometric functions if $\text{sec  x=}\dfrac{\text{13}}{\text{5}}$  , $\text{x}$ lies in fourth quadrant.

Ans: Here given that,  $\sec x=\dfrac{13}{5}$

$\cos x=\dfrac{1}{\sec x}$

        $=\dfrac{1}{\left( \dfrac{13}{5} \right)}$

        $=\dfrac{5}{13}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{ta}{{\text{n}}^{\text{2}}}\text{x=se}{{\text{c}}^{\text{2}}}\text{x-1}$

Substituting  \[\text{sec x=}\dfrac{\text{13}}{\text{5}}\]  in the formula, we obtain,

\[\text{ta}{{\text{n}}^{\text{2}}}\text{x=}{{\left( \dfrac{\text{13}}{\text{5}} \right)}^{\text{2}}}\text{-1}\]

$\text{ta}{{\text{n}}^{\text{2}}}\text{x=}\dfrac{\text{169}}{\text{25}}\text{-1}$

         $\text{=}\dfrac{\text{144}}{\text{25}}$ 

   \[\text{tanx= }\!\!\pm\!\!\text{ }\dfrac{\text{12}}{\text{5}}\]

Since $x$ lies in the ${{4}^{th}}$ quadrant, the value of $\tan x$ will be negative.

$\text{tan x=-}\dfrac{\text{12}}{\text{5}}$

Therefore, \[\text{cot x=}\dfrac{\text{1}}{\text{tan x}}\] 

                          $\text{=-}\dfrac{\text{5}}{\text{12}}$ 

 Hence we have, $\text{sin x=}\dfrac{\text{5}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{12}}{\text{5}} \right)$   

                                  $\text{=}\left( \text{-}\dfrac{\text{12}}{\text{13}} \right)$ 

            $\text{=-}\dfrac{\text{13}}{\text{12}}$

5. Find the values of other five trigonometric functions if  $\text{tan x=-}\dfrac{\text{5}}{\text{12}}$ , $\text{x}$ lies in second quadrant.

Ans: Here given that,  $\text{tan x=-}\dfrac{\text{5}}{\text{12}}$

       $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{5}}{\text{12}} \right)}$

       $\text{=-}\dfrac{\text{12}}{\text{5}}$

Substituting  $\text{tan x=-}\dfrac{\text{5}}{\text{12}}$  in the formula, we obtain,

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}{{\left( \text{-}\dfrac{\text{5}}{\text{12}} \right)}^{\text{2}}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\dfrac{\text{25}}{\text{144}}$

             $=\dfrac{169}{144}$ 

   $\text{sec x= }\!\!\pm\!\!\text{ }\dfrac{\text{13}}{\text{12}}$

Since $x$ lies in the ${{2}^{nd}}$ quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\dfrac{\text{13}}{\text{12}}$

Therefore, $\text{cos x=}\dfrac{\text{1}}{\text{sec x}}$ 

                          $\text{=-}\dfrac{\text{12}}{\text{13}}$ 

 Hence we have, $\text{sin x=}\left( \text{-}\dfrac{\text{5}}{\text{12}} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{12}}{\text{13}} \right)$   

                                    $=\left( \dfrac{5}{13} \right)$ 

           $\text{=}\dfrac{\text{13}}{\text{5}}$

6. Find the value of the trigonometric function $\text{sin76}{{\text{5}}^{\text{o}}}$ .

Ans: We know that the values of $\sin x$ repeat after an interval of $2\pi $ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{sin76}{{\text{5}}^{\text{o}}}\text{=sin}\left( \text{2 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}}\text{+4}{{\text{5}}^{\text{o}}} \right)$

            $\text{=sin4}{{\text{5}}^{\text{o}}}$

            $\text{=}\dfrac{\text{1}}{\sqrt{\text{2}}}\text{.}$

7. Find the value of the trigonometric function $\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)$  

Ans: We  know that the values of $\text{cosec x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

           $\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+4 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}} \right)$

                               $\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+144}{{\text{0}}^{\text{o}}} \right)$

                               $\text{=cosec3}{{\text{0}}^{\text{o}}}$

                               $=2$ 

8. Find the value of the trigonometric function  $\text{tan}\dfrac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}$ .

Ans: We know that the values of $\text{tan x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or \[{{180}^{\circ }}\].

$\text{tan}\dfrac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}\text{=tan6}\dfrac{\text{1}}{\text{3}}\text{ }\!\!\pi\!\!\text{ }$

            $\text{=tan}\left( \text{6 }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

            \[\text{=tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

            $\text{=}\sqrt{\text{3}}$      

9. Find the value of the trigonometric function $\text{sin}\left( \text{-}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}} \right)$ 

Ans: We know that the values of $\text{sin x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

$\text{sin}\left( \text{-}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=sin}\left( \text{-}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}}\text{+2 }\!\!\times\!\!\text{ 2 }\!\!\pi\!\!\text{ } \right)$ 

                   $\text{=sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

                    $=\dfrac{\sqrt{3}}{2}$

10. Find the value of the trigonometric function $\text{cot}\left( \text{-}\dfrac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

Ans: We  know that the values of $\text{cot x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or \[{{180}^{\circ }}\].

$\text{cot}\left( \text{-}\dfrac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{=cot}\left( \text{-}\dfrac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+4 }\!\!\pi\!\!\text{ } \right)$ 

                   $\text{=cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

              $=1$

Exercise 3.3

1. Prove that $\text{si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=-}\dfrac{\text{1}}{\text{2}}$

Ans: Substituting the values of  \[\text{sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\] on left hand side,

$\text{si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}\text{+}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}\text{-}{{\left( \text{1} \right)}^{\text{2}}}$

                                  \[\text{=}\dfrac{\text{1}}{\text{4}}\text{+}\dfrac{\text{1}}{\text{4}}\text{-1}\]

                                  $=-\dfrac{1}{2}$

                                  $=$ R.H.S.

Hence proved.

2. Prove that $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\dfrac{\text{3}}{\text{2}}$

Ans: Substituting the values of  $\text{sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ on left hand side,

L.H.S.$\text{=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

           $\text{=2}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}\text{+cose}{{\text{c}}^{\text{2}}}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right){{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}$

           $\text{=2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{4}}\text{+}{{\left( \text{-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)}^{\text{2}}}\left( \dfrac{\text{1}}{\text{4}} \right)$

          $\text{=}\dfrac{\text{1}}{\text{2}}\text{+}{{\left( \text{-2} \right)}^{\text{2}}}\left( \dfrac{\text{1}}{\text{4}} \right)$

Since $\text{cosec x}$ repeat its value after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ , 

we have, $\text{cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$  

L.H.S  $=\dfrac{1}{2}+\dfrac{4}{4}$

           $=\dfrac{3}{2}$ 

          $=$ R.H.S.

3. Prove that $\text{co}{{\text{t}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{+3ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=6}$

Ans: Substituting the values of  $\text{cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{,tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$ on left hand side,

L.H.S.$\text{=co}{{\text{t}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{+3ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

           $\text{=}{{\left( \sqrt{\text{3}} \right)}^{\text{2}}}\text{+cosec}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{+3}{{\left( \dfrac{\text{1}}{\sqrt{\text{3}}} \right)}^{\text{2}}}$

         $\text{=3+cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+3 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{3}}$

we have, $\text{cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$  

L.H.S $=3+2+1$ 

          $=1$ 

4. Prove that $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2se}{{\text{c}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=10}$

Substituting the values of  $\text{sin}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{,cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{,sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ on left hand side,

L.H.S.$\text{=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2se}{{\text{c}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

           $\text{=2}{{\left\{ \text{sin}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right\}}^{\text{2}}}\text{+2}{{\left( \dfrac{\text{1}}{\sqrt{\text{2}}} \right)}^{\text{2}}}\text{+2}{{\left( \text{2} \right)}^{\text{2}}}$

           $\text{=2}{{\left\{ \text{sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right\}}^{\text{2}}}\text{+2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}\text{+8}$

Since $\text{sin x}$ repeat its value after an interval of \[\text{2 }\!\!\pi\!\!\text{ }\] , 

we have, \[\text{sin}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{=sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]  

 L.H.S  $=1+1+8$ 

            $=10$

            $=$ R.H.S.

5. Find the value of :

(i) $\text{sin7}{{\text{5}}^{\text{o}}}$

Ans: We have,

$\text{sin7}{{\text{5}}^{\text{o}}}\text{=sin(4}{{\text{5}}^{\text{o}}}\text{+3}{{\text{0}}^{\text{o}}}\text{)}$

          $\text{=sin4}{{\text{5}}^{\text{o}}}\text{cos3}{{\text{0}}^{\text{o}}}\text{+cos4}{{\text{5}}^{\text{o}}}\text{sin3}{{\text{0}}^{\text{o}}}$

Since we know that, $\text{sin}\left( \text{x+y} \right)\text{=sin x cos y+cos x sin y}$ 

$Sin 75^o = \dfrac{1}{\sqrt 2}\times \dfrac{\sqrt 3}{2} + \dfrac{1}{\sqrt 2}\times\dfrac{1}{2}$

$Sin 75^o = \dfrac{{\sqrt 3}+1}{2\sqrt 2}$

(ii) $\text{tan1}{{\text{5}}^{\text{o}}}$

\[\text{tan1}{{\text{5}}^{\text{o}}}\text{=tan}\left( \text{4}{{\text{5}}^{\text{o}}}\text{-3}{{\text{0}}^{\text{o}}} \right)\]

        $\text{=}\dfrac{\text{tan4}{{\text{5}}^{\text{o}}}\text{-tan3}{{\text{0}}^{\text{o}}}}{\text{1+tan4}{{\text{5}}^{\text{o}}}\text{tan3}{{\text{0}}^{\text{o}}}}$

Since we know, $\text{tan}\left( \text{x-y} \right)\text{=}\dfrac{\text{tan x-tan y}}{\text{1+tan x tan y}}$

$\text{tan1}{{\text{5}}^{\text{o}}}\text{=}\dfrac{\text{1-}\dfrac{\text{1}}{\sqrt{\text{3}}}}{\text{1+1}\left( \dfrac{\text{1}}{\sqrt{\text{3}}} \right)}$

           $\text{=}\dfrac{\dfrac{\sqrt{\text{3}}\text{-1}}{\sqrt{\text{3}}}}{\dfrac{\sqrt{\text{3}}\text{+1}}{\sqrt{\text{3}}}}$

         $\text{=}\dfrac{\sqrt{\text{3}}\text{-1}}{\sqrt{\text{3}}\text{+1}}$

         $\text{=}\dfrac{{{\left( \sqrt{\text{3}}\text{-1} \right)}^{\text{2}}}}{\left( \sqrt{\text{3}}\text{+1} \right)\left( \sqrt{\text{3}}\text{-1} \right)}$ 

Further computing we have,

$\text{tan1}{{\text{5}}^{\text{o}}}\text{=}\dfrac{\text{3+1-2}\sqrt{\text{3}}}{{{\left( \sqrt{\text{3}} \right)}^{\text{2}}}\text{-}{{\left( \text{1} \right)}^{\text{2}}}}$ 

           \[\text{=}\dfrac{\text{4-2}\sqrt{\text{3}}}{\text{3-1}}\]

           \[\text{=2-}\sqrt{\text{3}}\]

6. Prove that $\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{-sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{=sin}\left( \text{x+y} \right)$

Ans: We know that, $\text{cos}\left( \text{x+y} \right)\text{=cos xcos y-sin xsin y}$ 

$\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{-sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{=cos}\left[ \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x+}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right]$

$\text{=cos}\left[ \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-}\left( \text{x+y} \right) \right]$

$\text{=sin}\left( \text{x+y} \right)$ 

L.H.S  $=$ R.H.S.

Hence  proved.

7. Prove that $\dfrac{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)}{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}\text{=}{{\left( \dfrac{\text{1+tanx}}{\text{1-tanx}} \right)}^{\text{2}}}$

Ans: We  know that ,$\text{tan}\left( \text{A+B} \right)\text{=}\dfrac{\text{tan A+tan B}}{\text{1-tan Atan B}}$

and $\text{tan}\left( \text{A-B} \right)\text{=}\dfrac{\text{tan A-tan B}}{\text{1+tan Atan B}}$

L.H.S.$\text{=}\dfrac{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)}{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}$

Using the above formula,

          $\text{L}\text{.H}\text{.S=}\dfrac{\left( \dfrac{\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+tanx}}{\text{1-tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{tanx}} \right)}{\dfrac{\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-tanx}}{\text{1+tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{tanx}}}$

                    $\text{=}\dfrac{\left( \dfrac{\text{1+tan x}}{\text{1-tan x}} \right)}{\left( \dfrac{\text{1-tan x}}{\text{1+tan x}} \right)}$      [ substituting $\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=1}$ ]

                    $\text{=}{{\left( \dfrac{\text{1+tan x}}{\text{1-tan x}} \right)}^{\text{2}}}$

                    $=$ R.H.S.

8.  Prove that  $\dfrac{\text{cos}\left( \text{ }\!\!\pi\!\!\text{ +x} \right)\text{cos}\left( \text{-x} \right)}{\text{sin}\left( \text{ }\!\!\pi\!\!\text{ -x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+x} \right)}\text{=co}{{\text{t}}^{\text{2}}}\text{x}$

Ans: Observe that $\text{cos x}$ repeats same value after an interval $\text{2 }\!\!\pi\!\!\text{ }$ 

and $\text{sin x}$ repeats same value after an interval  $\text{2 }\!\!\pi\!\!\text{ }$.

L.H.S.$\text{=}\dfrac{\text{cos}\left( \text{ }\!\!\pi\!\!\text{ +x} \right)\text{cos}\left( \text{-x} \right)}{\text{sin}\left( \text{ }\!\!\pi\!\!\text{ -x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+x} \right)}$

           $\text{=}\dfrac{\left[ \text{-cos x} \right]\left[ \text{cos x} \right]}{\left( \text{sin x} \right)\left( \text{-sin x} \right)}$

           $\text{=}\dfrac{\text{-co}{{\text{s}}^{\text{2}}}\text{x}}{\text{-si}{{\text{n}}^{\text{2}}}\text{x}}$

           $\text{=co}{{\text{t}}^{\text{2}}}\text{x}$

9. Prove that,

$\text{Cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{2}}\text{+x} \right)\text{Cos}\left( \text{2 }\!\!\pi\!\!\text{ +x} \right)\left[ \text{cot}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{2}}\text{-x} \right)\text{+cot}\left( \text{2 }\!\!\pi\!\!\text{ +x} \right) \right]\text{=1}$

Ans: We know that $\text{cot x}$  repeats same value after an interval $2\pi $ .

L.H.S.$=Cos\left( \dfrac{3\pi }{2}+x \right)Cos\left( 2\pi +x \right)\left[ cot\left( \dfrac{3\pi }{2}-x \right)+cot\left( 2\pi +x \right) \right]$

           $\text{=sin x cos x}\left[ \text{tan x+cot x} \right]$

Substituting $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ and

$\text{cot x=}\dfrac{\text{cos x}}{\text{sin x}}$ ,

$\text{L}\text{.H}\text{.S=sin xcos x}\left( \dfrac{\text{sin x}}{\text{cos x}}\text{+}\dfrac{\text{cos x}}{\text{sin x}} \right)$

         $\text{=}\left( \text{sin x cos x} \right)\left[ \dfrac{\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x}}{\text{sin x cos x}} \right]$

         $\text{=1}$ 

        $\text{=}$ R.H.S.

10. Prove that $\text{sin}\left( \text{n+1} \right)\text{xsin}\left( \text{n+2} \right)\text{x+cos (n+1)x cos (n+2)x=cos x}$

Ans: We know that , $\text{cos}\left( \text{x-y} \right)\text{=cosxcosy+sinxsiny}$ 

L.H.S.$\text{=sin}\left( \text{n+1} \right)\text{xsin}\left( \text{n+2} \right)\text{x+cos (n+1)x cos (n+2)x}$

           $\text{=cos}\left[ \left( \text{n+1} \right)\text{x-}\left( \text{n+2} \right)\text{x} \right]$ 

           $\text{=cos}\left( \text{-x} \right)$ 

           $\text{=cosx}$ 

           $=$  R.H.S.

11. Prove that $\text{cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{-cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{=-}\sqrt{\text{2}}\text{sinx}$

Ans: We  know that , $\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\therefore $ L.H.S.$\text{=cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{-cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)$

               $\text{=-2sin}\left\{ \dfrac{\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{+}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}{\text{2}} \right\}\text{.sin}\left\{ \dfrac{\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{-}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}{\text{2}} \right\}$

               $\text{=-2sin}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{sin x}$ 

Since $\text{sin x}$ repeats the same value after an interval $\text{2 }\!\!\pi\!\!\text{ }$ , 

we have, $\text{sin}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{=sin}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

therefore, 

$\text{L}\text{.H}\text{.S=-2sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{sin x}$ 

           $\text{=-2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\sqrt{\text{2}}}\text{ }\!\!\times\!\!\text{ sinx}$

           $\text{=-}\sqrt{\text{2}}\text{sin x}$

           $\text{=}$ R.H.S.

12. Prove that $\text{si}{{\text{n}}^{\text{2}}}\text{6x-si}{{\text{n}}^{\text{2}}}\text{4x=sin 2x sin 10x}$

Ans: We know that,$\text{sinA+sinB=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And $\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\therefore $ L.H.S.$\text{=si}{{\text{n}}^{\text{2}}}\text{6x-si}{{\text{n}}^{\text{2}}}\text{4xa}$

               $\text{=}\left( \text{sin 6x+sin 4x} \right)\left( \text{sin 6x-sin 4x} \right)$

               $\text{=}\left[ \text{2sin}\left( \dfrac{\text{6x+4x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{6x-4x}}{\text{2}} \right) \right]\left[ \text{2cos}\left( \dfrac{\text{6x+4x}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{6x-4x}}{\text{2}} \right) \right]$

               $\text{=}\left( \text{2sin 5x cos x} \right)\left( \text{2cos 5x sin x} \right)$

Now we know that, $\text{sin 2x=2sin x cos x}$ ,

$\text{L}\text{.H}\text{.S=}\left( \text{2sin 5x cos 5x} \right)\left( \text{2sin x cos x} \right)$ 

          $\text{=sin 10x sin 2x}$

          $\text{=}$ R.H.S.

13. Prove that $\text{co}{{\text{s}}^{\text{2}}}\text{2x-co}{{\text{s}}^{\text{2}}}\text{6x=sin 4x sin 8x}$

Ans: We  know that,

$\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And $\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

 L.H.S.$\text{=co}{{\text{s}}^{\text{2}}}\text{2x-co}{{\text{s}}^{\text{2}}}\text{6x}$

            $\text{=}\left( \text{cos 2x+cos 6x} \right)\left( \text{cos 2x-6x} \right)$

            $\text{=}\left[ \text{2cos}\left( \dfrac{\text{2x+6x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{2x-6x}}{\text{2}} \right) \right]\left[ \text{-2sin}\left( \dfrac{\text{2x+6x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{2x-6x}}{\text{2}} \right) \right]$

Further computing, we have,

$\text{L}\text{.H}\text{.S=}\left[ \text{2cos 4x cos}\left( \text{-2x} \right) \right]\left[ \text{-2sin 4xsin}\left( \text{-2x} \right) \right]$

         $\text{=}\left[ \text{2cos  4x cos 2x} \right]\left[ \text{-2sin 4x}\left( \text{-sin 2x} \right) \right]$

         $\text{=}\left( \text{2sin 4x cos 4x} \right)\left( \text{2sin 2xcos 2x} \right)$

Now we know that, $\text{sin 2x=2sin x cos x}$ 

Therefore we have,                

$\text{L}\text{.H}\text{.S=sin 8x sin 4x}$

         $\text{=}$  R.H.S.

.Hence proved.

14. Prove that $\text{sin 2x+2sin 4x+sin6=4co}{{\text{s}}^{\text{2}}}\text{xsin 4x}$

Ans: We know that, \[\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)\]

L.H.S.$\text{=sin 2x+2sin 4x+sin 6x}$

.           $\text{=}\left[ \text{sin 2x+sin 6x} \right]\text{+2sin 4x}$

           $\text{=}\left[ \text{2sin}\left( \dfrac{\text{2x+6x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{2x-6x}}{\text{2}} \right) \right]\text{+2sin4x}$

           $\text{=2sin 4xcos}\left( \text{-2x} \right)\text{+2sin 4x}$

We have, $\text{L}\text{.H}\text{.S=2sin 4x cos 2x+2sin 4x}$ 

                         $\text{=2sin 4x}\left( \text{cos 2x+1} \right)$ 

Now we know that, $\text{cos 2x+1=2co}{{\text{s}}^{\text{2}}}\text{x}$ 

$\text{L}\text{.H}\text{.S=2sin 4x}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)$ 

         $\text{=4co}{{\text{s}}^{\text{2}}}\text{xsin 4x}$

 = R.H.S.

15. Prove that $\text{cot 4x}\left( \text{sin 5x+sin 3x} \right)\text{=cot x}\left( \text{sin 5x-sin 3x} \right)$

Ans: We know that, $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

L.H.S.$\text{=cot 4x}\left( \text{sin 5x+sin 3x} \right)$

          $\text{=}\dfrac{\text{cot 4x}}{\text{sin 4x}}\left[ \text{2sin}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right) \right]$

          $\text{=}\left( \dfrac{\text{cos 4x}}{\text{sin 4x}} \right)\left[ \text{2sin 4x cos x} \right]$

          $\text{=2cos 4x cos x}$

Now also ,we know that, $\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

 R.H.S.$\text{=cot x}\left( \text{sin 5x-sin 3x} \right)$

            $\text{=}\dfrac{\text{cos x}}{\text{sin x}}\left[ \text{2cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{5x-3x}}{\text{2}} \right) \right]$

            $\text{=}\dfrac{\text{cos x}}{\text{sin x}}\left[ \text{2cos 4x sin x} \right]$

            $\text{=2cos 4x cos x}$

Therefore , we can conclude that,

L.H.S.=R.H.S.

16. Prove that $\dfrac{\text{cos 9x-cos 5x}}{\text{sin 17x-sin 3x}}\text{=-}\dfrac{\text{sin 2x}}{\text{cos 10x}}$

$\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And  $\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

 L.H.S.$\text{=}\dfrac{\text{cos 9x-cos 5x}}{\text{sin 17x-sin 3x}}$

            \[\text{=}\dfrac{\text{-2sin}\left( \dfrac{\text{9x+5x}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{9x-5x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{17x+3x}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{17x-3x}}{\text{2}} \right)}\]                    (following the formula)

           

            $\text{=}\dfrac{\text{-2sin 7x}\text{.sin 2x}}{\text{2cos 10x}\text{.sin 7x}}$

            $\text{=-}\dfrac{\text{sin 2x}}{\text{cos 10x}}$ 

17. Prove that:$\dfrac{\text{sin 5x+sin 3x}}{\text{cos 5x+cos 3x}}\text{=tan 4x}$

We  know that

$\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)\text{,}$

Now , L.H.S.$\text{=}\dfrac{\text{sin 5x+sin 3x}}{\text{cos 5x+cos 3x}}$

                      $\text{=}\dfrac{\text{2sin}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}$                  (using the formula)

                      $\text{=}\dfrac{\text{2sin}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}$

                      $\text{=}\dfrac{\text{2sin 4x cos x}}{\text{2cos 4x cos x}}$

$\text{L}\text{.H}\text{.S=tan 4x}$ 

18. Prove that \[\dfrac{\text{sin x-sin y}}{\text{cos x+cos y}}\text{=tan}\dfrac{\text{x-y}}{\text{2}}\]

$\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)\text{,}$

.$\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

  L.H.S.\[\text{=}\dfrac{\text{sin x-sin y}}{\text{cosx+cosy}}\]

               $\text{=}\dfrac{\text{2cos}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{x-y}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)}$

               $\text{=}\dfrac{\text{sin}\left( \dfrac{\text{x-y}}{\text{2}} \right)}{\text{cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)}$

              $\text{=tan}\left( \dfrac{\text{x-y}}{\text{2}} \right)$

Therefore $\text{L}\text{.H}\text{.S=R}\text{.H}\text{.S}$ 

19. Prove that $\dfrac{\text{sin x+sin 3x}}{\text{cos x+cos 3x}}\text{=tan 2x}$

Ans: We  know that

$\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{,}$

.\[\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)\]

Now , L.H.S.$\text{=}\dfrac{\text{sinx+sin3x}}{\text{cos x+cos 3x}}$

                      $\text{=}\dfrac{\text{2sin}\left( \dfrac{\text{x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-3x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-3x}}{\text{2}} \right)}$                   (using the formula)

                      $\text{=}\dfrac{\text{sin 2x}}{\text{cos 2x}}$

                      $\text{=tan 2x}$

Therefore  L.H.S$=$ R.H.S.

20. Prove that $\dfrac{\text{sin x-sin 3x}}{\text{si}{{\text{n}}^{\text{2}}}\text{x-co}{{\text{s}}^{\text{2}}}\text{x}}\text{=2sin x}$

$\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And \[\text{co}{{\text{s}}^{\text{2}}}\text{A-si}{{\text{n}}^{\text{2}}}\text{A=cos 2A}\]

 L.H.S.$\text{=}\dfrac{\text{sin x-sin 3x}}{\text{si}{{\text{n}}^{\text{2}}}\text{x-co}{{\text{s}}^{\text{2}}}\text{x}}$

       \[\text{=}\dfrac{\text{2cos}\left( \dfrac{\text{x+3x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{x-3x}}{\text{2}} \right)}{\text{-cos2x}}\]

            $\text{=}\dfrac{\text{2cos2xsin}\left( \text{-x} \right)}{\text{-cos 2x}}$

            $\text{=-2 }\!\!\times\!\!\text{ }\left( \text{-sinx} \right)$

Therefore , we have,

$\text{L}\text{.H}\text{.S=2sin x}$

        $=$ R.H.S.

21. Prove that $\dfrac{\text{cos 4x+cos 3x+cos 2x}}{\text{sin 4x+sin 3x+sin 2x}}\text{=cot 3x}$

And, $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

Now, L.H.S.$\text{=}\dfrac{\text{cos 4x+cos 3x+cos 2x}}{\text{sin 4x+sin 3x+sin 2x}}$

                     \[\text{=}\dfrac{\left( \text{cos 4x+cos 2x} \right)\text{+cos 3x}}{\left( \text{sin4x+sin2x} \right)\text{+sin 3x}}\]

                     \[\text{=}\dfrac{\text{2cos}\left( \dfrac{\text{4x+2x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{4x-2x}}{\text{2}} \right)\text{+cos3x}}{\text{2sin}\left( \dfrac{\text{4x+2x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{4x-2x}}{\text{2}} \right)\text{+sin 3x}}\]   (using the formulas)

                      \[\text{=}\dfrac{\text{2cos 3x cos x+cos 3x}}{\text{2sin 3x cos x+sin 3x}}\]

Further computing, we obtain,

L.H.S$\text{=}\dfrac{\text{cos 3x}\left( \text{2cos x+1} \right)}{\text{sin 3x}\left( \text{2cos x+1} \right)}$ 

          \[\text{=cot 3x}\] 

22. Prove that \[\text{cot x cot 2x-cot 2x cot 3x-cot 3x cot x=1}\]

We know that, \[\text{cot}\left( \text{A+B} \right)\text{=}\dfrac{\text{cotAcotB-1}}{\text{cot A+cot B}}\]

Now , L.H.S.$\text{=cot xcot 2x-cot 2x cot 3x-cot 3x cot x}$

                      \[\text{=cot x cot 2x-cot 3x}\left( \text{cot 2x+cot x} \right)\]

                      \[\text{=cot x cot 2x-cot}\left( \text{2x+x} \right)\left( \text{cot 2x+cot x} \right)\]

                      \[\text{=cot x cot 2x-}\left[ \dfrac{\text{cot 2x cot x-1}}{\text{cot x+cot 2x}} \right]\left( \text{cot 2x+cot x} \right)\]

Further computing we obtain,

$\text{L}\text{.H}\text{.S=cot x cot 2x-}\left( \text{cot 2x cot x-1} \right)$ 

         \[\text{=1}\]

         $\text{=}$ R.H.S.

23. Prove that $\text{tan 4x=}\dfrac{\text{4tan x}\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}{\text{1-6ta}{{\text{n}}^{\text{2}}}\text{x+ta}{{\text{n}}^{\text{4}}}\text{x}}$

Ans: We  know that $\text{tan 2A=}\dfrac{\text{2tan A}}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{A}}$

L.H.S.$\text{=tan 4x}$

           $\text{=tan2}\left( \text{2x} \right)$

           \[\text{=}\dfrac{\text{2tan 2x}}{\text{1-ta}{{\text{n}}^{\text{2}}}\left( \text{2x} \right)}\][using the formula]

           $\text{=}\dfrac{\left( \dfrac{\text{4tan x}}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{x}} \right)}{\left[ \text{1-}\dfrac{\text{4ta}{{\text{n}}^{\text{2}}}\text{x}}{{{\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}^{\text{2}}}} \right]}$

Further  computing, we obtain,

L.H.S $\text{=}\dfrac{\left( \dfrac{\text{4tan x}}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{x}} \right)}{\left[ \dfrac{{{\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}^{\text{2}}}\text{4ta}{{\text{n}}^{\text{2}}}\text{x}}{{{\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}^{\text{2}}}} \right]}$$$$$

          $\text{=}\dfrac{\text{4tan x}\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}{\text{1+ta}{{\text{n}}^{\text{4}}}\text{x-2ta}{{\text{n}}^{\text{2}}}\text{x-4ta}{{\text{n}}^{\text{2}}}\text{x}}$

          $\text{=}\dfrac{\text{4tan x}\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}{\text{1-6ta}{{\text{n}}^{\text{2}}}\text{x+ta}{{\text{n}}^{\text{4}}}\text{x}}$ 

24. Prove that $\text{cos 4x=1-8si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}$

Ans: We know that, $\text{cos 2x=1-2si}{{\text{n}}^{\text{2}}}\text{x}$ 

And $\text{sin 2x=2sin x cos x}$ 

L.H.S.\[\text{=cos 4x}\]

           $\text{=cos 2}\left( \text{2x} \right)$

           $\text{=1-2si}{{\text{n}}^{\text{2}}}\text{2x}$

           $\text{=1-2}{{\left( \text{2sin x cos x} \right)}^{\text{2}}}$

Further computing we get,

L.H.S$\text{=1-8si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}$ 

          $=$R.H.S.

25. Prove that $\text{cos 6x=32xco}{{\text{s}}^{\text{6}}}\text{x-48co}{{\text{s}}^{\text{4}}}\text{x+18co}{{\text{s}}^{\text{2}}}\text{x-1}$

Ans: We know that, $\text{cos 3A=4co}{{\text{s}}^{\text{3}}}\text{A-3cosA}$

and  $\text{cos 2x=1-2si}{{\text{n}}^{\text{2}}}\text{x}$

L.H.S.$\text{=cos 6x}$

           $\text{=cos 3}\left( \text{2x} \right)$

           \[\text{=4co}{{\text{s}}^{\text{3}}}\text{2x-3cos 2x}\]

           \[\text{=4}\left[ {{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x-1} \right)}^{\text{3}}}\text{-3}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x-1} \right) \right]\]

L.H.S$\text{=4}\left[ {{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)}^{\text{3}}}\text{-}{{\left( \text{1} \right)}^{\text{3}}}\text{-3}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right) \right]\text{-6co}{{\text{s}}^{\text{2}}}\text{x+3}$

         $\text{=4}\left[ {{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)}^{\text{3}}}\text{-}{{\left( \text{1} \right)}^{\text{3}}}\text{-3}{{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)}^{\text{2}}}\text{+3}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right) \right]\text{-6co}{{\text{s}}^{\text{2}}}\text{x+3}$

         $\text{=4}\left[ \text{8co}{{\text{s}}^{\text{6}}}\text{x-1-12co}{{\text{s}}^{\text{4}}}\text{x+6co}{{\text{s}}^{\text{2}}}\text{x} \right]\text{-6co}{{\text{s}}^{\text{2}}}\text{x+3}$

         $\text{=32co}{{\text{s}}^{\text{6}}}\text{x-48co}{{\text{s}}^{\text{4}}}\text{x+18co}{{\text{s}}^{\text{2}}}\text{x-1}$

L.H.S $=$ R.H.S.

Exercise 3.4

1. Find the principal and general solutions of the \[\text{tan x=}\sqrt{\text{3}}\].

Ans: Here given that,

\[\text{tan x=}\sqrt{\text{3}}\]

We know that $\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\sqrt{\text{3}}$

and $\text{tan}\left( \dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=tan}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\sqrt{\text{3}}$

Therefore, the principal solutions are\[\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\] and \[\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}}\] .

Now, \[\text{tan x=tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

Which implies,

$\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where \[\text{n}\in \text{Z}\]

Therefore, the general solution is \[\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\], where $\text{n}\in \text{Z}$ .

2. Find the principal and general solutions of the equation $\text{secx=2}$

Ans: Here it is given that,

$\text{sec x=2}$

Now we know that

$\text{sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=2}$ and 

$\text{sec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}\text{=sec}\left( \text{2 }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

          $\text{=sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

          $\text{=2}$ 

Therefore, the principal solutions are$\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ and \[\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}\].

Now, $\text{sec x=sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

and we know , 

$\sec x=\dfrac{1}{\cos  x}$

\[\text{cos x=cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

$\text{x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where $\text{n}\in \text{Z}$ .

Therefore, the general solution is $\text{x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where $n\in Z$ .

3. Find the principal and general solutions of the equation $\text{cot x=-}\sqrt{\text{3}}$

\[\text{cot x=-}\sqrt{\text{3}}\]

Now we know that $\text{cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=}\sqrt{\text{3}}$

\[\text{cot}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\]

                 \[\text{=-}\sqrt{\text{3}}\]

and  $\text{cot}\left( \text{2 }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

                          \[\text{=-}\sqrt{\text{3}}\]

$\text{cot}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-}\sqrt{\text{3}}$ 

and $\text{cot}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-}\sqrt{\text{3}}$

Therefore, the principal solutions are $\text{x=}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ and $\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}$.

.Now, $\text{cot x=cot}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

And we know \[\text{cot x=}\dfrac{\text{1}}{\text{tan x}}\]

$\text{tan x=tan}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ , where $\text{n}\in \text{Z}$

Therefore, the general solution is $\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ , where $\text{n}\in \text{Z}$ .

4. Find the general solution of $\text{cosec x=-2}$

$\text{cosec x=-2}$

$\text{cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=2}$

$\text{cosec}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{a}$ 

                     $\text{=-2}$

and \[\text{cosec}\left( \text{2 }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\]

                                        \[\text{=-2}\]

therefore we have,

\[\text{cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-2}\]and $\text{cosec}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-2}$

Hence , the principal solutions are$\text{x=}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\,$ and $\text{ }\dfrac{11\pi }{6}$.

Now, $\text{cosec x=cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

And we know, \[\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}\]

$\text{sin x=sin}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$  

,where $\text{n}\in \text{Z}$.

Therefore, the general solution is $\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{ }$ ,where $\text{n}\in \text{Z}$.

5. Find the general solution of the equation $\text{cos 4x=cos 2x}$

Ans: Here it is given that, $\text{cos 4x=cos 2x}$

$\text{cos 4x-cos 2x=0}$

Now we know that, $\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$ 

$\text{-2sin}\left( \dfrac{\text{4x+2x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{4x-2x}}{\text{2}} \right)\text{=0}$

                            $\text{sin 3x sin x=0}$

Hence we have, $\text{sin 3x=0}\,\,$

Or, $\text{ sin x=0}$

Therefore, $\text{3x=n }\!\!\pi\!\!\text{ }$

 or    $\text{x=n }\!\!\pi\!\!\text{ }$    ,where $\text{ n}\in \text{Z}$

 therefore, \[\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}\]     

 or  $\text{x=n }\!\!\pi\!\!\text{ }$  ,where $\text{ n}\in \text{Z}$.

6. Find the general solution of the equation $\text{cos 3x+cos x-cos 2x=0}$.

$\text{cos 3x+cos x-cos 2x=0}$

Now we know that, $\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

Therefore  $\text{cos 3x+cos x-cos 2x=0}$ implies

$\text{2cos}\left( \dfrac{\text{3x+x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{3x-x}}{\text{2}} \right)\text{-cos 2x=0}$

                     $\text{2cos 2x cos x-cos 2x=0}$

                        $\text{cos 2x}\left( \text{2cos x-1} \right)\text{=0}$

Hence we have, 

 Either $\text{cos 2x=0}$

Or $\text{cos x=}\dfrac{\text{1}}{\text{2}}$ 

Which in turn implies that,

Either $\text{2x=}\left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\,$

Or,  $\text{cos x=cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$    , where  $\text{n}\in \text{Z}$

Either  $\text{x=}\left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{4}\,\,$

Or,  $\text{x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$  ,where  \[\text{n}\in \text{Z}\].

7. Find the general solution of the equation \[\text{sin 2x+cos x=0}\] .

\[\text{sin 2x cos x=0}\]

\[\text{2sin x cos x+cos x=0}\]

     \[\text{cos x(2sin x+1)= }\!\!~\!\!\text{ 0}\]

Either $\text{cos x=0}$ 

Or, $\text{sin x=-}\dfrac{\text{1}}{\text{2}}$  

Hence we have,

 \[\text{x= }\!\!~\!\!\text{ (2n+1)}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\]   , where $\text{n}\in \text{Z}$ .

Either 

Or,  \[\text{sin x=-}\dfrac{\text{1}}{\text{2}}\]

             $\text{=-sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$ 

             $\text{=sin}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$ 

             $\text{=sin}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$ 

Which implies

$\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$  , where  $\text{n}\in \text{Z}$

Therefore, the general solution is \[\left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\,\] or  \[\text{n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{,n}\in \text{Z}\].

8. Find the general solution of the equation \[\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}\]

Ans: Here given that , \[\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}\]

Now we know that, $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$ 

      \[\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}\]  implies

           \[\text{1+ta}{{\text{n}}^{\text{2}}}\text{2x=1-tan 2x}\]

   \[\text{ta}{{\text{n}}^{\text{2}}}\text{2x+tan 2x=0}\]

 \[\text{tan 2x(tan 2x+1)= }\!\!~\!\!\text{ 0}\]

Hence  either $\text{tan 2x=0}$ 

Or, $\text{tan 2x=-1}$ 

Which implies  either  \[\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\]  , where $\text{n}\in \text{Z}$ ,

               $\text{=-tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

               $\text{=tan}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$

                $\text{=tan}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$

Which in turn implies that,  

$\text{2x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{,}$ where  $\text{n}\in \text{Z}$

i.e,  $\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\text{+}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{8}}\text{,}$ where $\text{n}\in \text{Z}$.

Therefore, the general solution is $\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\,\,$ or  $\,\,\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\text{+}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{8}}\text{,n}\in \text{Z}$.

9. Find the general solution of the equation \[\text{sin x+sin 3x+sin 5x=0}\]

Here given that ,\[\text{sin x+sin 3x+sin 5x=0}\]

Now we know that,  $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

Therefore ,

                             \[\text{sin x+sin 3x+sin 5x=0}\]

                          \[\left( \text{sin x+sin 3x} \right)\text{+sin 5x=0}\] 

     $\left[ \text{2sin}\left( \dfrac{\text{x+5x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-5x}}{\text{2}} \right) \right]\text{+sin 3x=0}\,$

                       \[\text{2sin 3x cos (-2x)+sin 3x= }\!\!~\!\!\text{ 0}\]

Simplifying we get,

\[\text{2sin 3xcos 2x+sin 3x=0}\]

     \[\text{sin 3x(2cos 2x+1)= }\!\!~\!\!\text{ 0}\]

Hence either $\text{sin 3x=0}$ 

Or, $\text{cos 2x=-}\dfrac{\text{1}}{\text{2}}$ 

Which implies  $\text{3x=n }\!\!\pi\!\!\text{ }$ , where $\text{n}\in \text{Z}$ 

Or,   $\text{cos 2x=-}\dfrac{\text{1}}{\text{2}}$

                  $\text{=-cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ 

                  $\text{=cos}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$ 

                     $\text{=cos}\dfrac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$ 

i.e., either $\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}$   , where $\text{n}\in \text{Z}$ 

or,  $\text{2x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$   ,where $\text{n}\in \text{Z}$ .

Therefore, the general solution is $\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}\,$ or $\text{n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{,n}\in \text{Z}$.

Miscellaneous Exercise

1. Prove that: $\text{2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}\text{=0}$

Ans: We know that $\text{cos x+cos y=2cos}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)$

Now L.H.S.$\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}$

                    $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+2cos}\left( \dfrac{\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right)\text{cos}\left( \dfrac{\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{-}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right)$       (using the formula)

                    $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+2cos}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\left( \dfrac{\text{- }\!\!\pi\!\!\text{ }}{\text{13}} \right)$

                    $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+2cos}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}} \right)$

Simplifying, 

$\text{L}\text{.H}\text{.S=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\left[ \text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}} \right]$

         $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\left[ \text{2cos}\left( \dfrac{\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right)\text{cos}\dfrac{\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{-}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right]$

          $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\left[ \text{2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{26}} \right]$

Substituting $\text{cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{=0}$ , we get,

$\text{L}\text{.H}\text{.S=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{ }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{26}}$ 

           $\text{=0}$

           $=\text{R}\text{.H}\text{.S}$

2. Prove that: $\left( \text{sin 3x+sin x} \right)\text{sin x+}\left( \text{cos 3x-cos x} \right)\text{cos x=0}$

We know that, $\text{sin x+sin y=2sin}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)$ 

And  $\text{cos x-cos y=-2sin}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{x-y}}{\text{2}} \right)$ 

L.H.S.$\text{=}\left( \text{sin 3x+sin x} \right)\text{sin x+}\left( \text{cos 3x-cos x} \right)\text{cos x}$

           $\text{=sin 3x sin x+si}{{\text{n}}^{\text{2}}}\text{x+cos 3x cos x-co}{{\text{s}}^{\text{2}}}\text{x}$    (using the formula)

           $\text{=cos 3x cos x+sin 3x sin x-}\left( \text{co}{{\text{s}}^{\text{2}}}\text{x-si}{{\text{n}}^{\text{2}}}\text{x} \right)$

  Simplifying  we get,       

$\text{L}\text{.H}\text{.S=cos}\left( \text{3x-x} \right)\text{-cos 2x}\,$

         $\text{=cos 2x-cos 2x}$

         $\text{=0}$

         $=\text{R}\text{.H}\text{.S}\text{.}$

3. Prove that: ${{\left( \text{cos x+cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}\text{=4co}{{\text{s}}^{\text{2}}}\dfrac{\text{x+y}}{\text{2}}$

Ans: We know that, $\text{cos}\left( \text{x+y} \right)\text{=cos x cos y-sin xsin y}$

and  L.H.S$\text{=}{{\left( \text{cos x+cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}$

                  \begin{align} & \text{=co}{{\text{s}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{y+2cos x cos y+si}{{\text{n}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{y-2sin x sin y} \\ &  \\ \end{align}

                 $\text{=}\left( \text{co}{{\text{s}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{x} \right)\text{+}\left( \text{co}{{\text{s}}^{\text{2}}}\text{y+si}{{\text{n}}^{\text{2}}}\text{y} \right)\text{+2}\left( \text{cos x cos y-sin x sin y} \right)$

Simplifying and using the formula,

L.H.S$\text{=1+1+2cos}\left( \text{x+y} \right)$ 

         $\text{=2}\left[ \text{1+cos}\left( \text{x+y} \right) \right]$ 

         $\text{=2}\left[ \text{1+2co}{{\text{s}}^{\text{2}}}\dfrac{\left( \text{x+y} \right)}{\text{2}}\text{-1} \right]$ 

 [since $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\left( \text{x+y} \right)}{\text{2}}\text{-1=cos}\left( \text{x+y} \right)$ ]

          $\text{=4co}{{\text{s}}^{\text{2}}}\left( \text{x+y} \right)$ 

Therefore  L.H.S$=$ R.H.S

4. Prove that:   ${{\left( \text{cos x-cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}\text{=4si}{{\text{n}}^{\text{2}}}\dfrac{\text{x-y}}{\text{2}}$ 

Ans: We know that, $\text{cos}\left( \text{x-y} \right)\text{=cos x cos y+sin x sin y}$ 

L.H.S.$\text{=}{{\left( \text{cos x-cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}$

           $\text{=co}{{\text{s}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{y-2cos x cos y+si}{{\text{n}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{y-2sin x sin y}$

           \[\text{=}\left( \text{co}{{\text{s}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{x} \right)\text{+}\left( \text{co}{{\text{s}}^{\text{2}}}\text{y+si}{{\text{n}}^{\text{2}}}\text{y} \right)\text{-2}\left[ \text{cos x cos y+sin x sin y} \right]\]

Simplifying and using the formula  we get,

         L.H.S $\text{=1+1-2}\left[ \text{cos}\left( \text{x-y} \right) \right]\,\,$

                   $\text{=2}\left[ \text{1-cos}\left( \text{x-y} \right) \right]$

                   $\text{=2}\left[ \text{1-}\left\{ \text{1-2si}{{\text{n}}^{\text{2}}}\left( \dfrac{\text{x-y}}{\text{2}} \right) \right\} \right]\,$ 

[since  $\text{1-2si}{{\text{n}}^{\text{2}}}\dfrac{\left( \text{x-y} \right)}{\text{2}}\text{=cos}\left( \text{x-y} \right)$ ]

                   $\text{=4si}{{\text{n}}^{\text{2}}}\left( \dfrac{\text{x-y}}{\text{2}} \right)$ 

5. Prove that: $\text{sin x+sin 3x+sin 5x+sin 7x=4cos xcos 2xsin 4x}$

Ans: We  know  that $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\text{L}\text{.H}\text{.S}\text{. =sin x+sin 3x+sin 5x+sin 7x}$

            \[\text{=}\left( \text{sin x+sin 5x} \right)\text{+}\left( \text{sin 3x+sin 7x} \right)\]      

Using the formula and simplifying,

          $\text{=2sin}\left( \dfrac{\text{x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{x-5x}}{\text{2}} \right)\text{+2sin}\left( \dfrac{\text{3x+7x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{3x-7x}}{\text{2}} \right)$       

          \[\text{=2cos 2x}\left[ \text{sin 3x+sin 5x} \right]\]

          \[\text{=2cos 2x}\left[ \text{2sin}\left( \dfrac{\text{3x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{3x-5x}}{\text{2}} \right) \right]\] 

          \[\text{=2cos 2x}\left[ \text{2sin 4x}\text{.cos}\left( \text{-x} \right) \right]\] 

\[\text{L}\text{.H}\text{.S=4cos 2x sin 4x cos x}\] 

           \[=\text{R}\text{.H}\text{.S}\]

6. Prove that: $\dfrac{\left( \text{sin 7x+sin 5x} \right)\text{+}\left( \text{sin 9x+sin 3x} \right)}{\left( \text{cos 7x+cos 5x} \right)\text{+}\left( \text{cos 9x+cos 3x} \right)}\text{=tan 6x}$

Ans: We  known that,

$\text{sinA+sinB=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And $\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\text{L}\text{.H}\text{.S}\text{. =}\dfrac{\left( \text{sin 7x+sin 5x} \right)\text{+}\left( \text{sin9x+sin3x} \right)}{\left( \text{cos 7x+cos 5x} \right)\text{+}\left( \text{cos9x+cos3x} \right)}$

          $\text{=}\dfrac{\left[ \text{2sin}\left( \dfrac{\text{7x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{7x-5x}}{\text{2}} \right) \right]\text{+}\left[ \text{2sin}\left( \dfrac{\text{9x+3x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{9x-3x}}{\text{2}} \right) \right]}{\left[ \text{2cos}\left( \dfrac{\text{7x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{7x-5x}}{\text{2}} \right) \right]\text{+}\left[ \text{2cos}\left( \dfrac{\text{9x+3x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{9x-3x}}{\text{2}} \right) \right]}$

          $\text{=}\dfrac{\left[ \text{2sin 6x}\text{.cos x} \right]\text{+}\left[ \text{2sin 6x}\text{.cos 3x} \right]}{\left[ \text{2cos 6x}\text{.cos x} \right]\text{+}\left[ \text{2cos 6x}\text{.cos 6x} \right]}$

          $\text{=}\dfrac{\text{2sin 6x}\left[ \text{cos x+cos 3x} \right]}{\text{2cos 6x}\left[ \text{cos x+cos 3x} \right]}$

           $\text{=tan 6x}$

Therefore L.H.S$=$ R.H.S

7. Prove that: 

$\text{sin 3x+sin 2x-sin x=4sin xcos}\dfrac{\text{x}}{\text{2}}\text{cos}\dfrac{\text{3x}}{\text{2}}$

$\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And  $\text{sin A-sin B=2sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)$ 

$\text{L}\text{.H}\text{.S}\text{.=sin3x+sin2x-sinx}$

          $\text{=sin 3x+}\left[ \text{2cos}\left( \dfrac{\text{2x+x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{2x-x}}{\text{2}} \right) \right]\,$

          $\text{=sin 3x+}\left[ \text{2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{x}}{\text{2}} \right) \right]$

Since we know that, $\text{sin 2x=2sin xcos x}$ 

$\text{L}\text{.H}\text{.S=2sin}\dfrac{\text{3x}}{\text{2}}\text{.cos}\dfrac{\text{3x}}{\text{2}}\text{+2cos}\dfrac{\text{3x}}{\text{2}}\text{sin}\dfrac{\text{x}}{\text{2}}\,\,\,\,\,\,$

         $\text{=2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\left[ \text{sin}\left( \dfrac{\text{3x}}{\text{2}} \right)\text{+sin}\left( \dfrac{\text{x}}{\text{2}} \right) \right]$

         \[\text{=2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\left[ \text{2sin}\left\{ \dfrac{\left( \dfrac{\text{3x}}{\text{2}} \right)\text{+}\left( \dfrac{\text{x}}{\text{2}} \right)}{\text{2}} \right\}\text{cos}\left\{ \dfrac{\left( \dfrac{\text{3x}}{\text{2}} \right)\text{-}\left( \dfrac{\text{x}}{\text{2}} \right)}{\text{2}} \right\} \right]\]

          $\text{=2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\text{.2sin xcos}\left( \dfrac{\text{x}}{\text{2}} \right)$

Therefore 

 $\text{L}\text{.H}\text{.S=4sin xcos}\left( \dfrac{\text{x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{3x}}{\text{2}} \right)$

          $\text{=R}\text{.H}\text{.S}$ 

8.Find $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ ,if $\text{tanx=-}\dfrac{\text{4}}{\text{3}}$ , $\text{x}$ lies in 2nd quadrant.   

Ans: Here, $\text{x}$ is in 2nd quadrant.

$\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{x }\!\!\pi\!\!\text{ }$

i.e,  $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}<\dfrac{\text{x}}{\text{2}}<\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$

hence $\dfrac{\text{x}}{\text{2}}$ lies in 1st quadrant.

Therefore, \[\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\,\,\] and $\text{tan}\dfrac{\text{x}}{\text{2}}$ are positive.

 Given that $\text{tan x=-}\dfrac{\text{4}}{\text{3}}$

We know that, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

         $\text{=1+}{{\left( \text{-}\dfrac{\text{4}}{\text{3}} \right)}^{\text{2}}}$ 

         \[\text{=}\dfrac{\text{25}}{\text{9}}\]

As \[\text{x}\] is in 2nd quadrant, $\text{sec x}$ is negative.

Therefore , $\text{secx=-}\dfrac{\text{5}}{\text{3}}$ 

Then $\text{cos x=-}\dfrac{\text{3}}{\text{5}}$ 

Now we know that, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=cos x+1}$ 

Computing we get, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{2}}{\text{5}}$ 

Hence \[\text{cos}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{1}}{\sqrt{\text{5}}}\] 

Now we know that, $\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$ 

Therefore substituting $\text{cos}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{1}}{\sqrt{\text{5}}}$ and computing ,

$\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{2}}{\sqrt{\text{5}}}$ 

$\text{tan}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{sin}\dfrac{\text{x}}{\text{2}}}{\text{cos}\dfrac{\text{x}}{\text{2}}}$ 

          $=2$ 

Thus, the respective values of$\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\text{,tan}\dfrac{\text{x}}{\text{2}}\,$

are $\,\dfrac{2\sqrt{5}}{5},\dfrac{\sqrt{5}}{5},\,\,2$ .

9.Find $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ ,if $\cos x\text{=-}\dfrac{1}{\text{3}}$ , $\text{x}$ lies in 3rd quadrant.   

Ans: Here, $\text{x}$ is in 3rd quadrant.

$\text{  }\!\!\pi\!\!\text{ x}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{2}}$

i.e,  $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{}\dfrac{\text{x}}{\text{2}}\text{}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$

hence $\dfrac{\text{x}}{\text{2}}$ lies in 2nd quadrant.

Therefore, $\text{cos}\dfrac{\text{x}}{\text{2}}\,\,\,$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ are negative and $\text{sin}\dfrac{\text{x}}{\text{2}}$ is positive.

 Given that $\text{cos x=-}\dfrac{1}{\text{3}}$

Now we know that, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=cosx+1}$ 

Computing we get, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{2}}{\text{3}}$ 

Hence $\text{cos}\dfrac{\text{x}}{\text{2}}\text{=-}\dfrac{\text{1}}{\sqrt{\text{3}}}$ 

Therefore substituting $\text{cos}\dfrac{\text{x}}{\text{2}}\text{=-}\dfrac{\text{1}}{\sqrt{\text{3}}}$ and computing ,

$\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{2}}{\text{3}}}$ 

          $\text{=-}\sqrt{\text{2}}$ 

Thus, the respective values of $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\text{,tan}\dfrac{\text{x}}{\text{2}}\,$

are  $\,\sqrt{\dfrac{\text{2}}{\text{3}}}\text{,-}\dfrac{\text{1}}{\sqrt{\text{3}}}\text{,}\,\text{-}\,\sqrt{\text{2}}$.

10.Find$\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ ,if $\text{sin x=}\dfrac{1}{4}$ , $\text{x}$ lies in 2nd quadrant.   

Ans: Here, $\text{x}$ lies in 2nd quadrant.

hence  $\dfrac{\text{x}}{\text{2}}$ lies in 1st quadrant.

Therefore, $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\,\,$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ are positive.

 Given that $\text{sin x=}\dfrac{\text{1}}{\text{4}}$

Therefore substituting $\text{sin x=}\dfrac{\text{1}}{\text{4}}$ and computing ,

$\text{cos x=-}\dfrac{\sqrt{\text{15}}}{\text{4}}$ 

since $\text{x}$ lies in 2nd quadrant,  $\text{cos x}$ is negative.

Now we know that, $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=1-cos x}$ 

Computing we get, $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=1+}\dfrac{\sqrt{\text{15}}}{\text{4}}$ 

Hence $\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{4+}\sqrt{\text{15}}}{\text{8}}}$ 

Therefore substituting $\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{4+}\sqrt{\text{15}}}{\text{8}}}$ and computing ,

$\text{cos}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{4-}\sqrt{\text{15}}}{\text{8}}}$ 

          $\text{=}\dfrac{\sqrt{\text{4+}\sqrt{\text{15}}}}{\sqrt{\text{4-}\sqrt{\text{15}}}}$ 

       \[\text{=4+}\sqrt{\text{15}}\] 

are  $\sqrt{\dfrac{\text{4+}\sqrt{\text{15}}}{\text{8}}}\text{,}\sqrt{\dfrac{\text{4-}\sqrt{\text{15}}}{\text{8}}}\text{,4+}\sqrt{\text{15}}$ .

NCERT Solutions for CBSE Class 11 Maths Chapter 3 Exercise 3.2 Trigonometric Functions

Ncert solutions for class 11 maths chapter 3 – free pdf download.

The Mathematical jargon in Trigonometry Class 11 is undoubtedly a complex affair for most students to grasp. But at the same time, it is an indispensable topic in 11 th standard syllabus. Stronghold on this topic constitutes a very basic understanding of geometric calculations and is crucial for solving multiple numerical problems in other subjects besides Mathematics.

NCERT Solutions for Class 11 Maths Chapter 3 are formulated by experienced and highly learned tutors focusing specifically on the problematic areas students tend to encounter in this chapter. Numerical sums based on the application of trigonometric formulas, graphs of trigonometric functions, conversion of angles from radian to degree and vice versa, and general solutions have been explained in great detail and elaborate steps.

Additionally, these study guides come with topic-wise exercises and probable questions as well, because increased practice will only help students improve their problem-solving speed and accuracy. In order to avail these suggestions offline, you can click on NCERT Solutions for Class 11 Maths Chapter 3 PDF Download option on Vedantu’s website.

NCERT Solutions for Class 11 Maths Chapter 3 Subtopics

Before wandering into the conceptual details of Trigonometric Functions Class 11, you can have a look at the summary of all topics discussed in this chapter.

Section 1: Introduction

In this part, students are introduced to fundamental trigonometric functions and their application. You will be required to perform calculations of distances based on these ratios.

Section 2: Angles

Class 11 Maths Trigonometry comes with four subsections under this topic. Students will learn about measurement of angles in different units, radians and degrees, and relations between them. Solutions to numerical sums involving the conversion of angle measurements from one form to another have been discussed in detail in NCERT Solutions for Class 11 Maths Chapter 3.

Section 3: Trigonometric Functions

This section deals with two topics. First one teaches the signs and symbols of generalised trigonometric functions in all four quadrants of a graph. In the next subsection, students will be acquainted with the domain and range of the different functions, that is, how the value of a function increases or decreases for an increasing angle value. You can find tables and detailed explanations on the working of these concepts in Trigonometric Functions Class 11 NCERT Solutions.

Section 4: Trigonometric Functions of Sum and Difference of Two Angles

The concluding section of Chapter 3 Maths Class 11 PDF focuses on the derivation of formulas and expressions based on trigonometric functions of the sums and differences between two angles. These formulas have been explained with the help of examples. Students must have a clear understanding of these as they make up an important part of geometric evaluations and are applied in a wide range of calculations.

Important Points Covered in Class 11 Maths Chapter 3 - Trigonometric Functions

The term “Angle” refers to a measurement of rotation of a ray about its fixed end point. The initial position of the ray before starting the rotation is called the initial side of the angle and the final position of the ray after rotation is called the terminal side of the angle.

In the Sexagesimal system, the angles are measured in degrees. 

1° = 60′ reads as 1 degree is equal to 60 minutes. 

1′ = 60″ reads as 1 minute is equal to 60 seconds.

In the Circular system, the angles are measured in radians.

1 Radian = 180/π degree and 

1 Degree = π/180 radians

Functions of negative angles (A)

sin (-A) = - sinA,

cosec (-A) = - cosec A,

tan (-A) = - tan A,

cos (-A) = cos A,

cot (-A) = - cot A,

sec (-A) = sec A.

Note: The trigonometric functions cosine and secant are even functions. The remaining all trigonometric functions are odd functions.

Trigonometric formulas of compound angles:

sin (A + B) = sin A cos B + cos A sin B

sin (A – B) = sin A cos B – cos A sin B

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

2 sin A cos B = sin (A + B) + sin (A – B)

2 cos A sin B = sin (A + B) – sin (A – B)

2 cos A cos B = cos (A + B) + cos (A – B)

2 sin A sin B = cos (A – B) – cos (A + B)

Trigonometric equations are the equations which include trigonometric functions of unknown angles.

Solutions of a trigonometric equation are all the possible values of the unknown angle that satisfies the given equation. These solutions (possible values) can be an infinite number in some cases.

The general solution of a trigonometric equation is defined as the solution that involves an integer ‘n’ which gives all solutions of the given trigonometric equation.

The principal solution of a trigonometric equation is defined as the solutions which lie in the interval 0 ≤ x ≤ 2π.

We Cover All The Exercises In The Chapter Given Below

Key features of ncert solutions for class 11 maths chapter 3 - trigonometric functions.

Key features of NCERT Solutions for Class 11 Maths Chapter 3 are as follows:

Pythagorean Identities will be used to solve many trigonometric problems in which one trigonometric ratio is given and the other ratios must be calculated.

We will learn the relationship between the ratios of the sides of a right-angled triangle and their angles.

We will learn about trigonometric ratios such as sine, cosine, tangent, cotangent, secant, and cosecant. 

The solutions deal with angle measurement and angle problems. 

We use fundamental ratios to construct other key trigonometric functions: cotangent, secant, and cosecant. These functions serve as the foundation for all of the essential trigonometry topics.

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FAQs on NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions

1. What are the basic trigonometric functions?

In Mathematics, trigonometric functions are fundamental functions used to denote relation of the angles of a right-angled triangle to the length of its sides.

There are six trigonometric ratios in total, the basic three being sine, cosine, and tangent. The other three are described in relation to previously-discussed basic functions and are called cosecant, secant and cotangent.

In a given right-angled triangle, values of these ratios are calculated for a specific angle θ in terms of its sides in the following manner.

sin θ = opposite side / hypotenuse

cos θ = adjacent side / hypotenuse

tan θ = sin θ / cos θ = opposite site / adjacent side

cosec θ = 1 / sin θ

sec θ = 1 / cos θ

cot θ = cos θ / sin θ = 1 / tan θ

2. How are Radians and Degrees related?

A Radian (denoted by rad or c) is a standard unit of measuring angles whose value is based on the relationship between the length of an arc and radius of a circle.

rad = arc length / radius of circle

In a lot of cases, you might be required to convert given angle values in Radian into Degree. For that, the relation between these two units can be derived as follows.  

For a full circle with radius r and angle 360 o , the arc length is equal to circumference C.

Therefore, arc length = C = 2πr.

From the first relation, 360 o in radians will be

rad = 2πr / r = 2π

Therefore, 2π rad = 360 o ,

that is, 1 rad = 180 o / π.

This brings us to the final relationship that is, degree = radian x 180 / π.

3. What can you learn from Trigonometry Class 11 NCERT Solutions PDF?

Chapter 3 Maths Class 11 covers the vast and complex topic of trigonometric functions and their applications. This chapter comes with a total of four subsections dealing with concepts like measuring angles in degrees and radians and their interconversion, sine and cosine formulas in terms of variable angles x and y, finding solutions of trigonometric values, and so on.

NCERT Solutions for Class 11 Maths Chapter 3 comes with comprehensive answers to questions from all four exercises given in textbooks. Solutions to numerical sums have been explained in a step-by-step manner. You will learn about the applications of trigonometric equations in the vast field of science and be able to recognise these functions in Physics and Chemistry.

4. How do you solve trigonometric functions in Class 11 Maths?

Trigonometry can be a tricky topic. However, the students who have a good grasp of the basics of trigonometry seem to solve the problems in a better way. The first thing to remember while solving the trigonometric functions is to know the formulae and their applications. Vedantu offers a complete guide and the PDF of the solutions of the exercise given in the NCERT textbook. The solutions are provided free of cost in a step-by-step manner and are easy to comprehend.

5. Is Chapter 3 of Class 11 Maths hard?

A lot of students may find Trigonometry difficult. However, if the concepts are clear and they have a good hold on the basics of Trigonometry, then it seems easy. As trigonometry is all about the relation between the angles and sides of the triangle, students need to know the basic formulae and their applications to solve the problems. Vedantu offers solutions that are verified by the subject-matter-expert and are also solved comprehensively. 

6. What are the important topics in Chapter 3 of Class 11 Maths?

In Class 11 Mathematics, Chapter 3 is all about trigonometry. This chapter includes all about the relation between the angles and the sides of the triangles and their applications. Whenever there is a need to solve the trigonometric functions, a student must have a good grasp of the identities and the formula. The value table of all the trigonometric functions for different angles must be remembered for solving problems. Vedantu offers the solution to all the exercises provided in the NCERT Textbook in a step-by-step manner. 

7. What is Trigonometry for Class 11 Students?

Trigonometry is a section of Mathematics, which provides an understanding of the relation between the angles and the sides of the triangle. This chapter can be tricky for most students. But, the students can ace their exams by practising numerous questions and knowing the formulae and their applications. Vedantu offers the best learning guide and the PDF of the solutions that are verified by the subject-matter-experts. Students can either refer to it online or by downloading the PDF for free to refer to it offline. 

8. How can Trigonometry of Class 11 be made easy to study?

Trigonometry seems difficult for most of the students. Few of the ways to make it easy is to always choose the side that is complex, denote or represent all the trigonometric functions into sine and cosine, focus on the formulae and the signs, and grasp the basics of cancelling the terms, rationalizing, and expanding. Vedantu offers an explanation based solution that is easy to follow and score. 

9. What are the applications of Trigonometry in real life?

As trigonometry involves the relationship between the sides and angles, finding the height, and calculating the distance, it has a wide range of applications in marine biology, navigation, aviation department, etc. It is also used in solving major mathematical calculations such as Calculus and Algebra. Vedantu provides a solution guide suitable for all the students with solutions that are verified by experienced experts. The solutions are available on both Vedantu’s website and its app at free of cost.

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Chapter 3 Class 11 Trigonometric Functions

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NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below.

We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn

  • What is a positive or a negative angle
  • Measuring angles in Degree , Minutes and Seconds
  • Radian measure of an angle
  • Converting Degree to Radians , and vice-versa
  • Sign of sin, cos, tan in all 4 quadrants
  • Finding values of trigonometric functions when one value is given (Example: Finding value of sin, cot, cosec, tan, sec, when cos x = -3/5 is given)
  • Finding Value of trigonometric functions, given angle
  • Solving questions by formula like  (x + y) formula, 2x 3x formula, Cos x + cos y formula , 2 sin x sin y formula 
  • Finding principal and general solutions of a trigonometric equation
  • Sin and Cosine Formula with supplementary Questions

Important questions are marked, and Formula sheet is also provided. Click on an exercise or topic to begin.

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Class 11 Maths Sample Papers, NCERT Solutions | Mock Tests

Designed for the toppers, the course content of CBSE Class 11 Maths like Sample Papers, NCERT Solutions, Notes, and Important Questions can be availed as PDF on myCBSEguide.

  • CBSE Syllabus

CBSE Sample Papers

Cbse last year papers, user submitted papers, case study questions,   mock tests, relations and functions, trigonometric functions, complex numbers and quadratic equations, linear inequalities, permutations and combinations, binomial theorem, sequences and series, straight lines, conic sections, introduction to 3d geometry, limits and derivatives, probability.

  • CBSE Test Papers
  • CBSE Revision Notes

CBSE Important Questions

Other useful resourses,   online tests,   learning videos.

Regarded as a scoring subject, Mathematics is equally important, especially when it comes to the senior secondary level. One can master this subject with little effort. All they need to do is to apply the flow of reasons while proving a result or solving a problem. To attain this level of competence you need good practice. For having the good practice you need to have quality course material. You can get the best resources for CBSE class 11 Mathematics on myCBSEguide  as we gave a huge question bank with a variety of questions.  Download the latest Sample Papers for CBSE Class 11 Mathematics, chapter-related Revision Notes, NCERT solution, and test papers all in PDF files.

CBSE Class 11 Mathematics Resources

Trusted by millions of teachers and students alike, myCBSEguide is a one-stop solution for quality course content for CBSE subjects. Students must put some extra effort to excel in Mathematics at the senior secondary level, as it will be working as the foundation of their future careers and profession. We provide a wide range of products with the help of which you can prepare for senior secondary Mathematics on myCBSEguide beginning from the syllabus to the model test papers. There are online tests, mock tests, learning videos, revision notes, NCERT solutions, and some important questions for class 11 Mathematics.

Class 11 Mathematics Syllabus

The syllabus is regularly updated based on the latest development. Students of senior secondary level, i.e., Class 11 Mathematics are on the threshold of higher academics. The marks they score here will play a decisive role in choosing their career options. Therefore, the syllabus of class 11 Maths includes every such topic that helps the students to develop a positive attitude to think, analyze and articulate logically. Given below are the name and the weightage of chapters of Class 11 Mathematics.

The prescribed books for class 11 Mathematics are:

  • Mathematics Textbook for Class XI, NCERT Publications
  • Mathematics Lab Manual class XI, published by NCERT

To know the chapter and topic-related details check out our detailed blog on the  Class 11 Mathematics syllabus .

Class 11 Mathematics Notes and NCERT Solutions

All CBSE students should practice NCERT textbook solutions because most of the time the questions asked in the final board examination are based on them. As we know that the CBSE encourages following  NCERT  textbooks because they cover almost all possible topics of any central examination. Some of the topics of class 11 Mathematics are:

  • Binomial Theorem Miscellaneous
  • Complex Numbers and Quadratic Equations Miscellaneous
  • Conic Sections Miscellaneous
  • Limits and Derivatives Miscellaneous

To get the complete notes based on the exercise check out  Class 11 Mathematics NCERT Solutions . 

Although Mathematics is not a theory subject, an explanation of each chapter is needed. One cannot understand the chapter if they are not aware of the subject-specific terms and jargon along with the explanation of theorems. You can find a step-by-step explanation of each example sum of a particular topic of class 11 Mathematics. Do check our  Revision Notes for Class 11 Mathematics  and download them in PDF.

CBSE Class 11 Mathematics Case Study Questions

Case study questions are the latest type of questions included in the CBSE question paper. This change is to promote competency-based learning and motivate topics from real-life and other subject areas, greater emphasis has been laid on the application of various concepts. The class 11 Mathematics case study questions look forward to acquainting students with different aspects of Mathematics used in daily life. Being the newest question format, it is difficult to get an assured quality of CBQs in the market or over the internet. Check our  Class 11 Maths case study questions  to know the actual standard of the questions. These case-based questions have sub-questions which can be objective or subjective type depending on the CBSE sample question paper. To get these questions get a subscription to our  student dashboard.

CBSE Class 11 Mathematics Sample Papers

A student who aims to get good grades must solve sample papers religiously. Based on the official sample papers of the CBSE, myCBSEguide prepares around ten sample papers in each subject. Knowing the blueprint and MS can help the students to write their papers confidently. Even the school-level examinations follow the CBSE pattern, therefore it is advised to solve full-length class 11 Mathematic model test papers. Get the app to download the latest  Class 11 Maths Sample Papers 2023 . 

Sample papers are also helpful for periodic assessment too. This is because a sample paper has an assortment of questions. A standard sample paper has all types of questions such as very short answer (VSA), short answer (SA) and, long answer (LA) to effectively assess the knowledge, understanding, application, skills, analysis, evaluation and, synthesis. We can say that ideally, a sample paper matches the modalities of a proper pen-paper test.

CBSE Class 11 Maths Test Papers & Important Questions

Memorizing the rules and formulas is one thing and applying them is another. Good coordination between the two can make wonders. This can be achieved by enthusiastic practice. You can get the Test paper set for Class 11 Mathematics, which you can practice according to the level of your preparation. This helps in segmented learning. Test papers are good ways to learn and test simultaneously. The class 11 mathematics test papers have all types of questions. Besides, we make efforts to chalk out the most important questions from every chapter so that our students do not miss out on crucial information. Class 11 Mathematics important questions are available on myCBSEguide.

You can find such valuable study material for other CBSE subjects in our  CBSE module .

Class 11 Mathematics Case Study Questions

If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it …

CBSE Class 11 Mathematics Syllabus 2022-23

CBSE Class 11 Mathematics Syllabus 2022-23 includes Sets and Functions, Coordinate Geometry, Statistics and Probability, Calculus etc for the session 2022 – 2023. Here is the detailed syllabus. To download class 11 Mathematics CBSE latest sample question papers for the 2023 exams, please install the myCBSEguide App which is the …

CBSE Term-1 MCQ Sample Papers 2021-22

CBSE will ask only MCQs in Term-1 Examination this year. The board will hold this first term exam in Nov-Dec 2021. As CBSE has already cleared that 1st term exam will carry 90 minutes, now it’s time to start preparation. Students can download CBSE Term-1 MCQ Sample Papers from myCBSEguide …

CBSE to Conduct Two Term Exams in 2021-22

CBSE has issued a circular on 5th July 2021 regarding the Special Scheme of Assessment for Board Examination Classes X and XII for the Session 2021-22. Here is the complete text of the circular. COVID 19 pandemic caused almost all CBSE schools to function in a virtual mode for the …

CBSE Class-11 New Stream Selection Rules

The New Education Policy-2020 has suggested eliminating rigid stream selection rules. We know that there are three streams in class 11 and the choices to select subjects are limited. Students have to select subjects within the stream only. Thus, if you are taking Science Stream, you have no option to …

CBSE Reduced Syllabus by 30% for Session 2020-21

CBSE, New Delhi has reduced the syllabus for classes 9 to 12 by 30%. The CBSE Revised Syllabus 2020-21 is available in CBSE official website and myCBSEguide mobile app. Here is the complete circular regarding CBSE Revised Syllabus 2020-21: CBSE Reduced the Syllabus The prevailing health emergency in the country …

Multiple Choice Questions for CBSE Class 10 and 12

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case study questions trigonometry class 11

Syllabus for the Session 2023-24

CBSE Syllabus

Case Study Questions

Case Study on Sets      CS-2   CS-3   CS-4   CS-5

Case Study on Relations & Functions   CS-2   CS-3

Case Study on Trigonometric Functions

Case Study on Complex Numbers

Case Study on Linear Inequalities

Case Study on Permutations and Combinations

Case Study on Sequences & Series

Case Study on Straight Lines

Case Study on Conic Sections

Case Study on Statistics

Case Study on Probability

Pdf of  Case Studies

MCQs for Practice

Chapter 1 - Sets     

Chapter 2 - Relations & Functions

Chapter 3 - Trigonometric Functions

Chapter 4 - Complex Numbers & Quadratic Equations

Chapter 5 - Linear Inequalities

Chapter 6 - Permutations & Combinations

Chapter 7 - Binomial Theorem

Chapter 8 - Sequences & Series

Chapter 9 - Straight Line

Chapter 1 0 - Conic Sections

Chapter 1 1 - Introduction to Three-dimensional Geometry

Chapter 12 - Limits & D erivatives

Chapter 13 - Statistics

Chapter 14 - Probability

Answers of MCQs

Assertion & Reasoning Questions

Relations & Functions

Trigonometric Functions

Complex Numbers

Linear Inequalities

Permutations & Combinations

Binomial Theorem

Sequences & Series

Straight Lines

Conic Sections

Introduction to Three-Dimensional Geometry

Limits & derivatives

Probability

Topic Wise Assignments of Previous Year Questions

Relations and Functions

Straight Line

Limits & Derivatives

Question Papers - DoE, Delhi

Session 2015-2016

            SA2(QP)         SA2(MS)         COMP(QP)         COMP(MS)

Session 201 6 -20 17

            SA1(QP)      SA2(QP)         SA2(MS)         COMP(QP)        COMP(MS)

Session 201 7 -201 8

            SA2(QP)          SA2(MS)         COMP(QP)         COMP(MS)

Session 201 8 -201 9

            SA2(QP)          SA2(MS)         COMP(QP)         COMP(MS)

Session 201 9 -20 20

            SA2(QP)          SA2(MS)        

Session 2020-21

    Video Explanation:  Part A     Part B Section III     Part B Section IV   Part B Section V

Session 2021-22

Question Paper 

Video Explanation:  Section A     Section B     Section C

Session 2022-23

  Sample Question Paper for Mid-Term Exam

Video Explanation: Section A     Section B    Section C     Section D     Section E

Mid-Term Exam:  Question Paper

Video Explanation: Section A     Section B     Section C     Section D    Section E

Practice Paper for Final Exam

Video Explanation: Section A     Section B     Section C     Section D     Section E

Final Exam:  Question Paper     Marking Scheme

Video Explanation: Section A     Section B     Section C     Section D     Section E

Support Material Issued by DoE, Delhi

Session 202 3 -24

 Mid-Term Exam:  Question Paper

Video Explanation:   Section A    Section B     Section C     Section D     Section E

Practice Paper 1 for Final Exam: Question Paper

Practice Paper 2 for Final Exam: Question Paper

Short Capsules (Notes to Revise Concepts)

Chapter 1 - Sets

Chapter 4 - Mathematical Induction

Chapter 5 - Complex Numbers & Quadratic Equations

Chapter 6 - Linear Inequalities

Chapter 7 - Permutations & Combinations

Chapter 8 - Binomial Theorem

Chapter 9 - Sequences & Series

Chapter 10 - Straight Line

Chapter 11 - Conic Sections

Chapter 12 - Introduction to Three-dimensional Geometry

Chapter 13 - Limits & derivatives

Chapter 14 - Mathematical Reasoning

Chapter 15 - Statistics

Chapter 16 - Probability

Some Basic Concepts to Revise:

Number System

Mensuration Results

Quadratic Equations

Solution of Polynomial Inequality - Wavy Curve Method

Probability__Revision of the Topics studied in Earlier Classes  

Result Sheets

Trigonometric Identities

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Download Case Study Questions for Class 11 Maths

Download Case Study Questions for Class 11 Maths

[PDF] Download Case Study Questions for Class 11 Maths

Here we are providing case study questions for Class 11 Maths. In this article, we are sharing links for Class 11 Maths All Chapters. All case study questions of Class 11 Maths are solved so that students can check their solutions after attempting questions.

Click on the chapter to view.

Class 11 Maths Chapters

Chapter 1 Sets Chapter 2 Relations and Functions Chapter 3 Trigonometric Functions Chapter 4 Principle of Mathematical Induction Chapter 5 Complex Numbers and Quadratic Equations Chapter 6 Linear Inequalities Chapter 7 Permutation and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight Lines Chapter 11 Conic Sections Chapter 12 Introduction to Three-Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 15 Statistics Chapter 16 Probability

What is meant by Case Study Question?

In the context of CBSE (Central Board of Secondary Education), a case study question is a type of question that requires students to analyze a given scenario or situation and apply their knowledge and skills to solve a problem or answer a question related to the case study.

Case study questions typically involve a real-world situation that requires students to identify the problem or issue, analyze the relevant information, and apply their understanding of the relevant concepts to propose a solution or answer a question. These questions may involve multiple steps and require students to think critically, apply their problem-solving skills, and communicate their reasoning effectively.

Importance of Solving Case Study Questions for Class 11 Maths

Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills. Here are some reasons why case study questions are important in Class 11 maths education:

  • Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.
  • Higher-order thinking: Case study questions require students to think critically, analyze data, and make connections between different concepts. This helps develop higher-order thinking skills, which are essential for success in both academics and real-life situations.
  • Collaborative learning: Case study questions often require students to work in groups, which promotes collaborative learning and helps students develop communication and teamwork skills.
  • Problem-solving skills: Case study questions require students to apply their knowledge and skills to solve complex problems. This helps develop problem-solving skills, which are essential in many careers and in everyday life.
  • Exam preparation: Case study questions are included in exams and tests, so practicing them can help students prepare for these assessments.

Overall, case study questions are an important component of Class 11 mathematics education, as they help students develop critical thinking, problem-solving, and analytical skills, which are essential for success in both academics and real-life situations.

Feature of Case Study Questions on This Website

Here are some features of a Class 11 Maths Case Study Questions Booklet:

Many Case Study Questions: This website contains many case study questions, each with a unique scenario and problem statement.

Different types of problems: The booklet includes different types of problems, such as optimization problems, application problems, and interpretation problems, to test students’ understanding of various mathematical concepts and their ability to apply them to real-world situations.

Multiple-choice questions: Questions contains multiple-choice questions to assess students’ knowledge, understanding, and critical thinking skills.

Focus on problem-solving skills: The questions are designed to test students’ problem-solving skills, requiring them to identify the problem, select appropriate mathematical tools, and analyze and interpret the results.

Emphasis on practical applications: The case studies in the booklet focus on practical applications of mathematical concepts, allowing students to develop an understanding of how mathematics is used in real-life situations.

Comprehensive answer key: The booklet includes a comprehensive answer key that provides detailed explanations and step-by-step solutions for all the questions, helping students to understand the concepts and methods used to solve each problem.

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CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11 PDF

CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11 download here at free of cost. you can check here CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11 based on latest syllabus and examination pattern. CBSE Chapter 3 Trigonometric Functions HOTs Questions helps to improve conceptual understanding and develops thinking skills. CBSE Chapter 3 Trigonometric Functions HOTs Questions and answers for Class 11 are available for free download in pdf format.

CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11

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CBSE Board introduced questions based on - Higher Order Thinking Skills (HOTs). The primary objective here was to improve the evaluating and analytical skills of a student. A lot of students tend to only focus on memorizing information and relying on cramming learning. Therefore CBSE Board decided to dedicate a large part of the question paper to Chapter 3 Trigonometric Functions HOTs Questions as they typically concentrate on a student's ability to reason, analyze, process, justify,  and evaluate information. Since a large part of CBSE question papers is based on HOTS, it is necessary for students to get ample practice with these questions. Experts at SelfStudys have compiled HOTs question for Class 11 for Science and Maths. 

CBSE releases Chapter 3 Trigonometric Functions HOTs Questions for Class 11 every year. These papers are published prior to the examinations so that students can do the study. The students should practice Chapter 3 Trigonometric Functions HOTs Questions to increase perfection which will help him to get good marks in CBSE examination. Based on CBSE and NCERT guidelines same pattern as released every year.

In simple words, CBSE Class 11 HOTs Questions are the skills that involve learning that goes beyond mere comprehension. Higher Order Thinking Skills (HOTs) as the name suggests refers to thinking skills that correspond to the more complicated forms of learning within the cognitive domain. Mathematics HOTS Question HOTs Questions involve behaviour ranging from the simplest to the most difficult. 

CBSE Chapter 3 Trigonometric Functions HOTs Questions which are basically designed to improve the learning skill of students and test their comprehension, analytical thinking and problem-solving skill. In CBSE board exams, most of the questions which are of higher difficulty level are taken from NCERT books. So students must widely practice all the CBSE Chapter 3 Trigonometric Functions HOTs Questions so that they can easily attempt even the difficult questions asked in board examinations. While practising the CBSE Chapter 3 Trigonometric Functions HOTs Questions, students must keep a reliable source to seek precise solutions so that they may know the right way to approach the answers to Mathematics HOTS Question HOTs Questions. In this page, we are providing the CBSE Class 11 HOTs Questions for two major subjects such as Maths and Science. All the CBSE Chapter 3 Trigonometric Functions HOTs Questions provided here are prepared by experts to bring accurate study material for Class 11 students. All these CBSE Chapter 3 Trigonometric Functions HOTs Questions are designed in a step-wise manner to provide an easy and simple explanation so that students can easily learn the concepts and facts used.

Understanding of CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11

Class 11 requires a strategic approach and the CBSE Chapter 3 Trigonometric Functions HOTs Questions provided on our website for every chapter form a cornerstone for thorough preparation. Class 11 is one of the most important stages in students life. This Chapter 3 Trigonometric Functions HOTs Questions can help to get a high overall score and hence needs dedicated preparation.

In the run-up to board exams of Class 11, it is important to build the base itself. Getting a good mark in Class 11 examinations will not only boost your confidence but will also give you the motivation to study for the next class. In addition, our website also has important questions for Class 11 subject wise solutions which will help students to refer to the answers once they are done answering in order to self-evaluate their performance and correct any errors.

Our aim is to make this CBSE Mathematics HOTS Question available to every student. This will enable you to practice the subject better and get high marks in the exam. Practising these Chapter 3 Trigonometric Functions HOTs Questions is a good way to ensure that you have not left any topic and that you have complete coverage of the whole syllabus. Also, students can use these Mathematics HOTS Question to take a mock test and use it as a tool to gauge their preparation. The Chapter 3 Trigonometric Functions HOTs Questions and answers are developed by referring to previous year question papers given in the NCERT textbooks. The format of the answers is in accordance with the latest recommended format given by the CBSE. This allows students to study the answer and replicate it in the examination in similar language and get good marks.

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CBSE MCQ for Class 11 Maths Chapter 3 Trigonometric Functions Free PDF

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Guys, we are working very hard to provide you with TOPIC-WISE MCQs (as listed below). Till then, attached below is the Master PDF having all the topics. Hope you understand. Enjoy your preparation! All the Best!

CBSE MCQ for Class 11 Maths Chapter 3 Trigonometric Functions PDF

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The CBSE MCQ for Class 11 Maths Chapter 3 Trigonometric Functions  are provided above, in detailed and free to download PDF format. The solutions are latest , comprehensive , confidence inspiring , with easy to understand explanation . To download NCERT Class 11 Solutions PDF for Free, just click ‘ Download pdf ’.

Checkout our MCQs for Other Chapters

  • CBSE MCQ for Class 11 Maths Chapter 1 Sets
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  • CBSE MCQ for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

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NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are available at BYJU’S, which are prepared by our expert teachers. All these solutions are written as per the latest update on the CBSE Syllabus 2023-24 and its guidelines. BYJU’S provides step-by-step solutions by considering the different understanding levels of students and the marking scheme. Chapter 3, Trigonometric Functions, comes under NCERT Class 11 Maths and is an important chapter for students. Though the chapter has numerous mathematical terms and formulae, BYJU’S has made NCERT Solutions for Class 11 Maths easy for the students to understand and remember with the usage of tricks.

Trigonometry has been developed to solve geometric problems involving triangles. Students cannot skip Trigonometric chapters since this is used in many areas, such as finding the heights of tides in the ocean, designing electronic circuits, etc. In the earlier classes, students have already studied trigonometric identities and applications of trigonometric ratios. In Class 11 , trigonometric ratios are generalised to trigonometric function and their properties. However, the NCERT Solutions of BYJU’S help the students to attain more knowledge and score full marks in this chapter of the examination.

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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In this section, a few important terms are defined, such as principal solution and general solution of trigonometric functions, which are explained using examples too. Exercise 3.1 Solutions 7 Questions Exercise 3.2 Solutions 10 Questions Exercise 3.3 Solutions 25 Questions Exercise 3.4 Solutions 9 Questions Miscellaneous Exercise On Chapter 3 Solutions 10 Questions

Access NCERT Solutions for Class 11 Maths Chapter 3

Exercise 3.1 page: 54

1. Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 1

2. Find the degree measures corresponding to the following radian measures (Use π = 22/7)

Here π radian = 180°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 5

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

It is given that

No. of revolutions made by the wheel in

1 minute = 360

1 second = 360/60 = 6

We know that

The wheel turns an angle of 2π radian in one complete revolution.

In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12 π radian

Therefore, in one second, the wheel turns an angle of 12π radian.

4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 9

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

The dimensions of the circle are

Diameter = 40 cm

Radius = 40/2 = 20 cm

Consider AB be as the chord of the circle i.e. length = 20 cm

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 10

Radius of circle = OA = OB = 20 cm

Similarly AB = 20 cm

Hence, ΔOAB is an equilateral triangle.

θ = 60° = π/3 radian

In a circle of radius  r  unit, if an arc of length  l  unit subtends an angle  θ  radian at the centre

We get θ = 1/r

Therefore, the length of the minor arc of the chord is 20π/3 cm.

6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 12

7. Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length

(i) 10 cm (ii) 15 cm (iii) 21 cm

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r

We know that r = 75 cm

(i) l = 10 cm

θ = 10/75 radian

By further simplification

θ = 2/15 radian

(ii) l = 15 cm

θ = 15/75 radian

θ = 1/5 radian

(iii) l = 21 cm

θ = 21/75 radian

θ = 7/25 radian

Exercise 3.2 page: 63

Find the values of other five trigonometric functions in Exercises 1 to 5.

1. cos x = -1/2, x lies in third quadrant.

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 1

2. sin x = 3/5, x lies in second quadrant.

sin x = 3/5

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 2

sin 2 x + cos 2 x = 1

cos 2 x = 1 – sin 2 x

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 3

3. cot x = 3/4, x lies in third quadrant.

cot x = 3/4

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 4

1 + tan 2 x = sec 2 x

1 + (4/3) 2 = sec 2 x

Substituting the values

1 + 16/9 = sec 2 x

cos 2 x = 25/9

sec x = ± 5/3

Here x lies in the third quadrant so the value of sec x will be negative

sec x = – 5/3

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 5

4. sec x = 13/5, x lies in fourth quadrant.

sec x = 13/5

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 6

sin 2 x = 1 – cos 2 x

sin 2 x = 1 – (5/13) 2

sin 2 x = 1 – 25/169 = 144/169

sin 2 x = ± 12/13

Here x lies in the fourth quadrant so the value of sin x will be negative

sin x = – 12/13

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 7

5. tan x = -5/12, x lies in second quadrant.

tan x = – 5/12

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 9

1 + (-5/12) 2 = sec 2 x

1 + 25/144 = sec 2 x

sec 2 x = 169/144

sec x = ± 13/12

Here x lies in the second quadrant so the value of sec x will be negative

sec x = – 13/12

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 10

Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°

We know that values of sin x repeat after an interval of 2π or 360°

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 11

By further calculation

7. cosec (–1410°)

We know that values of cosec x repeat after an interval of 2π or 360°

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 12

= cosec 30 o = 2

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 14

We know that values of tan x repeat after an interval of π or 180°

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 15

Exercise 3.3 page: 73

Prove that:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 1

= 1/2 + 4/4

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 5

5. Find the value of:

(i) sin 75 o

(ii) tan 15 o

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 10

(ii) tan 15°

It can be written as

= tan (45° – 30°)

Using formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 11

Prove the following:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 13

= sin x cos x (tan x + cot x)

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 23

10. sin ( n  + 1) x  sin ( n  + 2) x  + cos ( n  + 1) x  cos ( n  + 2) x  = cos  x

LHS = sin ( n  + 1) x  sin ( n  + 2) x  + cos ( n  + 1) x  cos ( n  + 2) x

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 24

Using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 27

12. sin 2  6 x  – sin 2  4 x  = sin 2 x  sin 10 x

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 28

13. cos 2  2 x  – cos 2  6 x  = sin 4 x  sin 8 x

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 30

= [2 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]

= [2 cos 4 x  cos 2 x ] [–2 sin 4 x  (–sin 2 x )]

= (2 sin 4 x  cos 4 x ) (2 sin 2 x  cos 2 x )

= sin 8x sin 4x

14. sin 2x + 2sin 4x + sin 6x = 4cos 2  x sin 4x

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 31

= 2 sin 4x cos (– 2x) + 2 sin 4x

= 2 sin 4x cos 2x + 2 sin 4x

Taking common terms

= 2 sin 4x (cos 2x + 1)

= 2 sin 4x (2 cos 2  x – 1 + 1)

= 2 sin 4x (2 cos 2  x)

= 4cos 2  x sin 4x

15. cot 4 x  (sin 5 x  + sin 3 x ) = cot  x  (sin 5 x  – sin 3 x )

LHS = cot 4x (sin 5x + sin 3x)

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 32

= 2 cos 4x cos x

Hence, LHS = RHS.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 34

22. cot  x  cot 2 x  – cot 2 x  cot 3 x  – cot 3 x  cot  x  = 1

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 50

LHS = tan 4x = tan 2(2x)

By using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 52

24. cos 4 x  = 1 – 8sin 2  x  cos 2  x

LHS = cos 4x

= cos 2(2 x )

Using the formula cos 2 A  = 1 – 2 sin 2   A

= 1 – 2 sin 2  2 x

Again by using the formula sin2 A  = 2sin  A  cos A

= 1 – 2(2 sin  x  cos  x ) 2

= 1 – 8 sin 2 x  cos 2 x

25. cos 6 x  = 32 cos 6   x  – 48 cos 4   x  + 18 cos 2   x  – 1

L.H.S. = cos 6 x

= cos 3(2 x )

Using the formula cos 3 A  = 4 cos 3   A  – 3 cos   A

= 4 cos 3  2 x  – 3 cos   2 x

Again by using formula cos 2 x  = 2 cos 2   x  – 1

= 4 [(2 cos 2   x  – 1) 3  – 3 (2 cos 2   x  – 1)

= 4 [(2 cos 2   x ) 3  – (1) 3  – 3 (2 cos 2   x ) 2  + 3 (2 cos 2   x )] – 6cos 2   x  + 3

= 4 [8cos 6 x  – 1 – 12 cos 4 x  + 6 cos 2 x ] – 6 cos 2 x  + 3

By multiplication

= 32 cos 6 x  – 4 – 48 cos 4 x  + 24 cos 2   x  – 6 cos 2 x  + 3

On further calculation

= 32 cos 6 x  – 48 cos 4 x  + 18 cos 2 x  – 1

Exercise 3.4 PAGE: 78

Find the principal and general solutions of the following equations:

1. tan x = √3

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 1

2. sec x = 2

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 2

3. cot x = – √3

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 4

4. cosec x = – 2

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 6

Find the general solution for each of the following equations:

5. cos 4x = cos 2x

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 8

6. cos 3x + cos x – cos 2x = 0

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 9

7. sin 2x + cos x = 0

sin 2x + cos x = 0

2 sin x cos x + cos x = 0

cos x (2 sin x + 1) = 0

cos x = 0 or 2 sin x + 1 = 0

Let cos x = 0

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 10

8. sec 2 2x = 1 – tan 2x

sec 2 2x = 1 – tan 2x

1 + tan 2 2x = 1 – tan 2x

tan 2 2x + tan 2x = 0

tan 2x (tan 2x + 1) = 0

tan 2x = 0 or tan 2x + 1 = 0

If tan 2x = 0

tan 2x = tan 0

2x = nπ + 0, where n ∈ Z

x = nπ/2, where n ∈ Z

tan 2x + 1 = 0

tan 2x = – 1

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 11

2x = nπ + 3π/4, where n ∈ Z

x = nπ/2 + 3π/8, where n ∈ Z

Hence, the general solution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.

9. sin x + sin 3x + sin 5x = 0

sin x + sin 3x + sin 5x = 0

(sin x + sin 5x) + sin 3x = 0

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 12

2 sin 3x cos (-2x) + sin 3x = 0

2 sin 3x cos 2x + sin 3x = 0

By taking out the common terms

sin 3x (2 cos 2x + 1) = 0

sin 3x = 0 or 2 cos 2x + 1 = 0

If sin 3x = 0

3x = nπ, where n ∈ Z

x = nπ/3, where n ∈ Z

If 2 cos 2x + 1 = 0

cos 2x = – 1/2

= – cos π/3

= cos (π – π/3)

cos 2x = cos 2π/3

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 13

Miscellaneous Exercise page: 81

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 1

2. (sin 3 x  + sin  x ) sin  x  + (cos 3 x  – cos  x ) cos  x  = 0

LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

= sin 3x sin x + sin 2 x + cos 3x cos x – cos 2 x

Taking out the common terms

= cos 3x cos x + sin 3x sin x – (cos 2 x – sin 2 x)

cos (A – B) = cos A cos B + sin A sin B

= cos (3x – x) – cos 2x

= cos 2x – cos 2x

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 5

LHS = (cos x + cos y) 2 + (sin x – sin y) 2

By expanding using formula we get

= cos 2 x + cos 2 y + 2 cos x cos y + sin 2 x + sin 2 y – 2 sin x sin y

Grouping the terms

= (cos 2 x + sin 2 x) + (cos 2 y + sin 2 y) + 2 (cos x cos y – sin x sin y)

Using the formula cos (A + B) = (cos A cos B – sin A sin B)

= 1 + 1 + 2 cos (x + y)

= 2 + 2 cos (x + y)

Taking 2 as common

= 2 [1 + cos (x + y)]

From the formula cos 2A = 2 cos 2 A – 1

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 6

LHS = (cos x – cos y) 2 + (sin x – sin y) 2

By expanding using formula

= cos 2 x + cos 2 y – 2 cos x cos y + sin 2 x + sin 2 y – 2 sin x sin y

= (cos 2 x + sin 2 x) + (cos 2 y + sin 2 y) – 2 (cos x cos y + sin x sin y)

Using the formula cos (A – B) = cos A cos B + sin A sin B

= 1 + 1 – 2 [cos (x – y)]

= 2 [1 – cos (x – y)]

From formula cos 2A = 1 – 2 sin 2 A

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 8

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 9

8. Find sin x/2, cos x/2 and tan x/2 in each of the following:

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 17

cos x = -3/5

From the formula

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 19

9. cos x = -1/3, x in quadrant III

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 22

10. sin x = 1/4, x in quadrant II

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 25

This chapter has 6 exercises and a miscellaneous exercise to help students understand the concepts related to Trigonometric Functions clearly. BYJU’S provides all the concepts and solutions for Chapter 3 of Class 11 Maths with clear explanations and formulas. The PDF of Maths  NCERT Solutions for Class 11 Chapter 3 includes the topics and sub-topics listed below. 3.1 Introduction The basic trigonometric ratios and identities are given here, along with the applications of trigonometric ratios in solving the word problems related to heights and distances. 3.2 Angles 3.2.1 Degree measure 3.2.2 Radian measure 3.2.3 Relation between radian and real numbers 3.2.4 Relation between degree and radian In this section, different terms related to trigonometry are discussed, such as terminal side, initial sides, measuring an angle in degrees and radian, etc. 3.3 Trigonometric Functions 3.3.1 Sign of trigonometric functions 3.3.2 Domain and range of trigonometric functions After studying this section, students are able to understand the generalised trigonometric functions with signs. Also, they can gain knowledge on domain and range of trigonometric functions with examples. 3.4 Trigonometric Functions of Sum and Difference of Two Angles This section contains formulas related to the sum and difference of two angles in trigonometric functions.

Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Studying the Trigonometric Functions of Class 11 enables the students to understand the following:

  • Introduction to Trigonometric Functions
  • Positive and negative angles
  • Measuring angles in radians and in degrees and conversion of one into other
  • Definition of trigonometric functions with the help of unit circle
  • Truth of the sin 2x + cos 2x = 1, for all x
  • Signs of trigonometric functions
  • Domain and range of trigonometric functions
  • Graphs of Trigonometric Functions
  • Expressing sin (x±y) and cos (x±y) in terms of sin x, sin y, cos x & cosy and their simple application, identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x
  • The general solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a

Disclaimer – 

Dropped Topics – 

3.5 Trigonometric Equations (up to Exercise 3.4) Last five points in the Summary 3.6 Proofs and Simple Applications of Sine and Cosine Formulae

Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 3

What are the topics discussed in chapter 3 of ncert solutions for class 11 maths, as per the latest update of the cbse syllabus, how many exercises are there in chapter 3 of ncert solutions for class 11 maths, what will i learn in chapter 3 trigonometric functions of ncert solutions for class 11 maths, leave a comment cancel reply.

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MCQ Questions For Class 11 Trigonometric Functions With Answers

Students can refer to the following Trigonometric Functions MCQ Questions for Class 11 Maths with Answers provided below based on the latest curriculum and examination pattern issued by CBSE and NCERT . Our teachers have provided here a collection of multiple choice questions for Trigonometric Functions Class 11 covering all topics in your textbook so that students can assess themselves on all important topics and thoroughly prepare for their exams

Trigonometric Functions MCQ Questions for Class 11 Maths with Answers

We have provided below Trigonometric Functions MCQ Questions for Class 11 Maths with answers which will help the students to go through the entire syllabus and practice multiple choice questions provided here with solutions. As Trigonometric Functions MCQs in Class 11 Mathematics pdf download can be really scoring for students, you should go through all problems provided below so that you are able to get more marks in your exams.

Question. Difference between the greatest and the least angle is (a) cos-1 4/5 (b) tan-1 3/4 (c) cos-1 3/5 (d) None of these

Question: The maximum value of

case study questions trigonometry class 11

(a) 4 +√2 (b) 3 +√2 (c) 9 (d) 4 

Question: cos2θ + 2cos θ is always (a) greater πthan -3/2 (b) less than or equal to 3/2 (c) greater than or equal to -3/2 and less than or equal to 3 (d) None of the above 

Question: The greatest and least value of sinX cos x  are respectively (a) 1, 1 (b) 1/2,1/2 (c) 1/4,-1/4 (d) 2, -2 

Question: The minimum value of 3 cosX+4 sin x +  8 is (a) 5 (b) 9 (c) 7 (d) 3 

Question: 

case study questions trigonometry class 11

Question. When the elevation of sun changes from 45° to 30°, the shadow of a tower increases by 60 m, then the height of the tower is (a) 30 √3 m (b) 30( √2 + 1) m (c) 30( √3 – 1) m (d) 30( √3 + 1) m

Question. A flag staff 20 m long standing on a wall 10 m high subtends an angle whose tangent is 0.5 at a point on the ground. If q is the angle subtended by the wall at this point, then (a) tan θ = 1 (b) tan θ = 3 (c) tan θ = 1/2 (d) None of these

Question. ABis a vertical pole with Bat the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7m. From D the angle of elevation of the point A is 45°. Then, the height of the pole is (a) 7√3/2(1/√3+1)m (b) 7√3/2(1/√3-1)m (c) 7√3/2(√3+1)m (d) 7√3/2(√3-1)m

Question. A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB(= a) subtends an angle of 60° at the foot of the tower and the angle of elevation of the top of the tower from A or B is 30°. The height of the tower is (a) 2a/√3 (b) 2a√3 (c) a/√3 (d) √3

Question. A ladder rests against a wall making an anglea with the horizontal. The foot of the ladder is pulled away from the wall through a distance x, so that it slides a distance y down the wall making an angle b with the horizontal. The correct relation is (a) x = y tan(α+β/2) (b) y = x tan (α+β/2) (c) x = y tan(α+β) (d) y = x tan (α+β)

Question. Consider the following statements  I. If in a ΔABC , sin A/sin C = sin(A-B)sin(B-C) then a 2 ,b 2  and c 2  are in AP. II. If exradius r1, r2 and r3 of a DABC are in HP, then the sides a , b and c are in AP. Which of these is/are correct ? (a) Only (I) (b) Only (II) (c) Both (I) and (II) (d) None of these

Question. In a triangle (1-r 1 /r 2 )(1-r 1 /r 3 )=2 then the triangle is (a) right angled (b) equilateral (c) isosceles (d) None of these

Question. In a ΔABC, cosec A (sin Bcos C + cos Bsin C) is equal to (a) c/a (b) a/c (c) 1 (d) c/ab

Question. sin A : sin C = sin (A – B) : sin (B – C), then a 2  , b 2  , c 2   , and are in (a) AP (b) GP (c) HP (d) None of these

Question. At a distance 2 hm from the foot of a tower of height h m the top of the tower and a pole at the top of the tower subtend equal angles. Height of the pole should be (a) (5h/3)m (b) (4h/3)m (c) (7h/5)m (d) (3h/2)m

Question. In a ∠ABC, a : b : c = 4 :5 :6. The ratio of the radius of the circumcircle to that of the incircle is (a) 16/9 (b) 16/7 (c) 11/7 (d) 7/16

Question. If x, y and z are perpendicular drawn from the vertices of triangle having sides a , b and c, then the value of bx/c +cy/a +az/b will be (a) (a 2  + b 2  + c 2 )/2R (b) (a 2  + b 2  + c 2 )/R (c) (a 2  + b 2  + c 2 )/4R (d) 2(a 2  + b 2  + c 2 )/R

Question. A house of height 100msubtends a right angle at the window of an opposite house. If the height of the window be 64 m, then the distance between the two houses is (a) 48 m (b) 36 m (c) 54 m (d) 72 m

Question. An observer standing on a 300 m high tower observes two boats in the same direction their angles of depression are 60° and 30°, respectively. The distance between boats is (a) 173.2 m (b) 346.4 m (c) 25 m (d) 72 m

Question. An observer on the top of tree, finds the angle of depression of a car moving towards the tree to be 30°.After 3 min this angle becomes 60°. After how much more time, the car will reach the tree ? (a) 4 min (b) 4.5 min (c) 1.5 min (d) 2 min

Question. The horizontal distance between two towers is 60 m and the angle of depression of the top of the first tower as seen from the top of the second is 30°. If the height of the second tower be 150 m, then the height of the first tower is (a) (150 – 60√3) m (b) 90 m (c) (150 – 20√3) m (d) None of these

Question. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60° and after 10 s the elevation is observed to be 30°. The uniform speed of the aeroplane (in km/h) is (a) 240 (b) 240√3 (c) 60√3 (d) None of the above

Question. Points D, E are taken on the side BC of a ΔABC such that BD = DE = EC. If ∠BAD =x , ∠DAE =y , ∠EAC = z, then the value of  sin(x+y) sin (y+z) /sinx sinz is equal to (a) 1 (b) 2 (c) 4 (d) None of these

Question. In a ΔABC, if a = 2 x, b = 2y and ÐC = 120°, then the area of the triangle is (a) xy sq unit (b) xy √3 sq unit (c) 3xy sq unit (d) 2 xy sq unit

Question. In ΔABC, if tan A/2 tan C/2 = 1/2 then a , band c are in (a) AP (b) GP (c) HP (d) None of these

Question. A flag staff is upon the top of a building. If at a distance of 40 m from the base of building the angles of elevation of the topes of the flag staff and building are 60° and 30° respectively, then the height of the flag staff is (a) 46.19 m (b) 50 m (c) 25 m (d) None of these

Question. At a point on the ground the angle of elevation of a tower is such that its cotangent is 3/5 . On walking 32 m towards the tower the cotangent of the angle of elevation is 2/5. The height of the tower is (a) 160 m (b) 120 m (c) 64 m (d) None of these

Question. From the bottom of a pole of height h the angle of elevation of the top of a tower is a and the pole subtends an angle b at the top of the tower. The height of the tower is (a) h tan(α-β) /tan(α-β) -tanα (b) h cot(α-β) /cot(α-β) -cotα (c) cot(α-β) / cot(α-β) -cotα (d) None of these

Question. Each side of a square subtends an angle of 60° at the top of a towerh metres high standing in the centre of the square. If ais the length of each side of the square, then (a) 2a 2 = h 2 (b) 2h 2 = h 2 (c) 3a 2 = 2h 2 (d) 2h 2 = 3a 2

Question. The top of a hill observed from the top and bottom of a building h is at angles of elevation p and q, respectively. The height of hill is (a) h cot q/ cotq – cotp (b) h cot p/ cotp – cotq (c) h tan p /tanp – tanq (d) None of these

Question. ABC is a triangular park with AB = AC = 100 m. A clock tower is situated at the mid-point of BC. The angles of elevation of the top of the tower at A and B are cot-1 3.2 and cosec-1 2.6, respectively. The height of the tower is (a) 50 m (b) 25 m (c) 40 m (d) None of these

Question. The angle of elevation of the top of a tower from a point A due South of the tower is a and from a point Bdue East of the tower isb. If AB= d, then the height of the tower is (a) d/√tan 2 α – tan 2 β (b) d/√tan 2 α + tan 2 β (c) d/√cot 2 α + cot 2 β (d) d/√cot 2 α – cot 2 β

Question. ABCD is a square plot. The angle of elevation of the top of a pole standing at Dfrom A orC is 30° and that from B is q, then tan q is equal to (a) 6 (b) 1/√6 (c) √3/2 (d) √(2 /3)

Question. A vertical tower stands on a declivity which is inclined at 15° to the horizon. From the foot of the tower a man ascends the declivity for 80 ft and then, finds that the tower subtends an angle of 30°. The height of tower is (a) 20(√6 – √2) ft (b) 40(√6 – √2) ft (c) 40(√6 + √2) ft (d) None of these

Question. If A and B are two points on one bank of a straight river and C, D are two other points on the other bank of river. If direction from A to B is same as that from C to D and AB = a, ∠CAD = a, ∠DAB = b, ∠CBA = g, then CD is equal to (a) a sinβ sinγ /sinαsin(α + β + γ) (b) a sinα sinγ /sinβsin(α + β + γ) (c) a sinα sinβ /sinβsin(α + β + γ) (d) None of these

Question. The angle of elevation of the top of a hill from a point is a. After walking b m towards the top up a slope inclined at an ∠β to the horizon, the angle of elevation of the top becomes γ. Then, the height of the hill is (a) b sinα sin (γ – β)/sin (γ – α) (b) b sinα sin (γ – α)/sin (γ – β) (c) b sin(γ – β) /sin (γ – α) (d) sin(γ – β)/b sinα sin(γ – α)

Question. The angle of elevation of a cloud at a point 2500 m high above a lake is 15° and the angle of depression of its image in the lake is 45°, the height of the cloud above the surface of the lake is (a) 2500√3 m (b) 2500 m (c) 500√3 m (d) 500 m

Question. In an isosceles D ABC, AB = AC. If vertical ΔA is 20°, then a 3  + b 3  is equal to (a) 3a 2 b (b) 3b 2 c (c) 3c 2 a (d) abc

Question. If A, B, C, D are the angles of a quadrilateral, then ∑tanA/∑cotA is equal to (a) π tan A (b) π cot A (c) ∑ tan 2  A (d) ∑ cot 2  A

Question. In a ΔABC, (a + b + c)(b + c – a) = kbc, if (a) k < 0 (b) k > 6 (c) 0 < k < 4 (d) k > 4

Question. In an acute angled ΔABC, r + r1 = r1+ r2 + r3 and ∠B >π/3, then (a) b + 2c < 2a < 2b + 2c (b) b + 4c < 4a < 2b + 4c (c) b + 4c < 4a < 4b + 4c (d) b + 3c < 3a < 3b + 3c

Question. For a ΔABC, R = 5/2 and r = 1. Let I be the incentre of the triangle and D, E and F be the feet of the perpendiculars from I to BC, CA and AB,respectively. The value of ID X IE X IF/IA X IB X IC is equal to (a) 5/2 (b) 5/4 (c) 1/10 (d) 1/5

Question. The area of the circle and the area of a regular polygon of n sides and of perimeter equal to that of the circle are in the ratio of (a) tan(π/n) : π/n (b) cos (π/n) : π/n (c) sin (π/n) : π/n (d) cot (π/n) : π/n

Question. The radii r r r 1 2 3 , , of the described circles of the DABC are in HP. If the area of the triangle is 24 2 cm and its perimeter is 24 cm, then the length of its largest side is (a) 10 (b) 9 (c) 8 (d) 7

Question. From the top of a cliff of height a , the angle of depression of the foot of a certain tower is found to be double the angle of elevation of the top of the tower of height h. If q be the angle of elevation, then its value is (a) rcosec    sinβ    (b) rcosec α sin (a) cos -1 √(2h/a) (b) sin -1  √(2h/a) (c) sin -1  √(a/2-h) (d) tan-1√(3 -2h/a)

Question. From a point a metres above a lake the angle of elevation of a cloud is α and the angle of depression of its reflection is β. The height of the cloud is (a) asin(α+β) /sin(α-β) m (b) asin(α+β)/sin(β-α) m (c) asin(β-α)/sin(α+β) m (d) None of these

Question. If PQbe a vertical tower subtending anglesa, b and g at the points A, B and C respectively on the line in the horizontal plane through the foot Dof tower and on the same side of it, then BC cota – CA cotb + ABcot g is equal to (a) 0 (b) 1 (c) 2 (d) None of these

Question. A flag staff stands in the centre of a rectangular field whose diagonal is 1200 m and subtends angles 15° and 45° at the mid-points of the sides of the field. The height of the flag staff is (a) 200 m (b) 300 √2 + √3 m (c) 300 √2 – √3 m (d) 400 m

Question. At each end of a horizontal line of length 2a, the angular elevation of the peak of a vertical tower is q and that at its middle point it is f. The height of the peak is (a) a sin θ sin Φ (b) a sin θ sin Φ /√sin(θ+Φ)sin(θ-Φ) (c) a cosθ cosΦ /√cos(θ+Φ)cos(θ-Φ) (d) None of the above

Question. In a cubical hall ABCDPQRS with each side 10 m,G is the centre of the wall BCRQand T is the mid-point of the side AB. The angle of elevation of G at the point T is (a) sin-1 (1/√3)  (b) cos-1 (1/√3)  (c) cot-1 (1/√3)  (d) None of these

Question. Two vertical poles 20 m and 80 m stands apart on a horizontal plane. The height of the point of intersection of the lines joining the top of each pole to the foot of the other is (a) 15 m (b) 16 m (c) 18 m (d) 50 m

Question. Let ABC be an isosceles triangle with base BC. If r is the radius of the circle inscribed in the ΔABC and r1 be the radius of the circle escribed opposite to the angle A, then the product r1r can be equal to (a) R 2 sin 2 A (b) R 2 sin 2 B (c) 1/2a 2 (d) a 2 /4 where R is the radius of the circumcircle of the ΔABC.

Question. Circumradius R is equal to (a) 2.5 (b) 3.5 (c) 1.5 (d) 4.2

Question. For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is (a) there is a regular polygon with r/R = 1/2 (b) there is a regular polygon with r/R = 1/√2 (c) there is a regular polygon with r/R = 2/3 (d) there is a regular polygon with r/R =√3/2

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  1. NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.2

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  2. TRIGONOMETRY FUNCTION HANDWRITTEN NOTES FOR 11TH CLASS

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  3. Trigonometry Class 11 Notes

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  4. 2 case studies of chapter 3 Trigonometry of class 11thmaths book ncert

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  5. Important Formulas of Trigonometry

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  6. Worksheets for Trigonometry Formulas For Class 11 Pdf

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  4. IMPORTANT QUESTIONS TRIGONOMETRY CLASS X

  5. Trigonometry Case Based Question 11th Maths l In a class test of Class XI , a teacher asked to…

  6. Important multiple choice questions trigonometry class 10 CBSE

COMMENTS

  1. Case Study Questions for Class 11 Maths Chapter 3 Trigonometric

    Here are some reasons why case study questions are important in Class 11 maths education: Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.

  2. Class 11 Mathematics Case Study Questions

    Class 11 Mathematics case study question 1. Read the Case study given below and attempt any 4 sub parts: In drilling world's deepest hole, the Kola Superdeep Borehole, the deepest manmade hole on Earth and deepest artificial point on Earth, as a result of a scientific drilling project, it was found that the temperature T in degree Celsius, x ...

  3. CBSE Case Study Questions for Class 11 Maths Sets Free PDF

    First, learn to sit for at least 2 hours at a stretch. Level 2. Solve every question of NCERT by hand, without looking at the solution. Level 3. Solve NCERT Exemplar (if available) Level 4. Sit through chapter wise FULLY INVIGILATED TESTS. Level 5. Practice MCQ Questions (Very Important)

  4. Case Studies on the Topic Trigonometric Functions

    Case Studies on the Topic Trigonometric Functions - Class 11 MathematicsThis video lecture discusses 2 case study questions on the topic of Trigonometric Fun...

  5. Important Questions For Class 11 Maths Chapter 3 with Solutions

    Practice Questions For Class 11 Maths Chapter 3 Trigonometric Functions. Find the value of the below expression. Hint: Simplify the expression to. Find the general solution of the equation 5cos 2 θ + 7sin 2 θ - 6 = 0. If θ lies in the first quadrant and cos θ = 8/17, then find the value of cos (30° + θ) + cos (45° - θ) + cos (120 ...

  6. CBSE 11th : Case study Based Questions (5) :Trigonometric functions

    Case study Based Question from chapter "Trigonometric functions"Next in playlist:https://youtu.be/vuFLIpZGmtk

  7. CBSE 11th: Case study based question (19th) : "Trigonometric ...

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  8. Important Questions for CBSE Class 11 Maths Chapter 3 Trigonometric

    The trigonometric function is one of the important chapters of class 11 maths. The concept of trigonometry was mainly developed to solve geometric problems that incorporate triangles. By practising the important questions of maths class 11 trigonometry functions, students can easily score high marks in the examinations.

  9. Trigonometric functions

    Class 11 math (India) 15 units · 180 skills. Unit 1 Sets. Unit 2 Relations and functions. Unit 3 Trigonometric functions. Unit 4 Complex numbers. Unit 5 Linear inequalities. Unit 6 Permutations and combinations. Unit 7 Binomial theorem. Unit 8 Sequence and series.

  10. CBSE Question Bank For Class 11 Maths Chapter 3 Trigonometric Functions

    First, learn to sit for at least 2 hours at a stretch. Level 2. Solve every question of NCERT by hand, without looking at the solution. Level 3. Solve NCERT Exemplar (if available) Level 4. Sit through chapter wise FULLY INVIGILATED TESTS. Level 5. Practice MCQ Questions (Very Important)

  11. NCERT Solutions for Class 11 Maths Chapter 3

    In Class 11 Mathematics, Chapter 3 is all about trigonometry. This chapter includes all about the relation between the angles and the sides of the triangles and their applications. Whenever there is a need to solve the trigonometric functions, a student must have a good grasp of the identities and the formula.

  12. Chapter 3 Class 11 Trigonometric Functions

    NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below. We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn. Finding values of trigonometric functions ...

  13. Class 11 Maths Sample Papers, NCERT Solutions

    Class 11 Mathematics Case Study Questions. If you're seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations.

  14. Trigonometry for Class 11

    Trigonometry for class 11 with basic formulas as well as important functions of sum and product of angles. Solve trigonometric functions related questions here. ... and 'metry' means 'measurement'. So basically, trigonometry is a study of triangles, which has angles and lengths on its side. Trigonometry basics consist of sine, cosine ...

  15. NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions

    NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions. NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions Q4. If cos (α + ) =4/5 and sin (α- )=5/13 , where α lie between […]

  16. Conquer Mathematics

    Class 11. Syllabus for the Session 2023-24. CBSE Syllabus. Case Study Questions. Case Study on Sets CS-2 CS-3 CS-4 CS-5. Case Study on Relations & Functions CS-2 CS-3. Case Study on Trigonometric Functions. Case Study on Complex Numbers. Case Study on Linear Inequalities.

  17. Download Case Study Questions for Class 11 Maths

    Class 11 Maths Chapters. Chapter 1 Sets. Chapter 2 Relations and Functions. Chapter 3 Trigonometric Functions. Chapter 4 Principle of Mathematical Induction. Chapter 5 Complex Numbers and Quadratic Equations. Chapter 6 Linear Inequalities. Chapter 7 Permutation and Combinations. Chapter 8 Binomial Theorem.

  18. CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11 PDF

    CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11. CBSE Board introduced questions based on - Higher Order Thinking Skills (HOTs). The primary objective here was to improve the evaluating and analytical skills of a student. A lot of students tend to only focus on memorizing information and relying on cramming learning.

  19. CBSE MCQ for Class 11 Maths Chapter 3 Trigonometric Functions Free PDF

    MCQ Set 1 -. Download PDF. The CBSE MCQ for Class 11 Maths Chapter 3 Trigonometric Functions are provided above, in detailed and free to download PDF format. The solutions are latest, comprehensive, confidence inspiring, with easy to understand explanation. To download NCERT Class 11 Solutions PDF for Free, just click ' Download pdf '.

  20. NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

    Ex 3.2 Class 11 Maths Question 1: Find the values of other five trigonometric functions if cos x = - 1 2 x lies in third quadrant. Ans: Ex 3.2 Class 11 Maths Question 2: Find the values of other five trigonometric functions if sin x = 35, x lies in second quadrant. Ans: sin x = 3 5. cosec x = 1 sin x = 1(3 5) = 5 3.

  21. Class 11 Maths Chapter 3 Trigonometric Functions MCQs

    MCQs for Chapter 3 Trigonometric Functions Class 11 with Answers. 1. If sin θ and cos θ are the roots of ax2 - bx + c = 0, then the relation between a, b and c will be. Given that sin θ and cos θ are the roots of the equation ax 2 - bx + c = 0, so sin θ + cos θ = b/a. 2.

  22. NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions

    The PDF of Maths NCERT Solutions for Class 11 Chapter 3 includes the topics and sub-topics listed below. 3.1 Introduction. The basic trigonometric ratios and identities are given here, along with the applications of trigonometric ratios in solving the word problems related to heights and distances. 3.2 Angles.

  23. MCQ Questions For Class 11 Trigonometric Functions With Answers

    A false statement among the following is. (a) there is a regular polygon with r/R = 1/2. (b) there is a regular polygon with r/R = 1/√2. (c) there is a regular polygon with r/R = 2/3. (d) there is a regular polygon with r/R =√3/2. Answer. We hope you liked Trigonometric Functions MCQ Questions for Class 11 Maths provided above.