CBSE Class 11 Maths Important Questions - Chapter 3 Trigonometric Functions

1 Mark Questions

1. Find the radian measure corresponding to  $5{}^{\circ }{37}^{\prime }{30}^{″}$  

Ans- 

Converting the given value to a pure degree form

 $5^{\circ }{37}^{\prime }{30}^{″}=5^{\circ }{37}^{\prime }{\left({\displaystyle \frac{30}{60}}\right)}^{\prime }$ 

 $\Rightarrow 5^{\circ }{37}^{\prime }{60}^{″}=5^{\circ }{\left({\displaystyle \frac{75}{2}}\right)}^{\prime }$ 

 $\Rightarrow 5^{\circ }{37}^{\prime }{60}^{″}=5^{\circ }{\left({\displaystyle \frac{75}{2\left(60\right)}}\right)}^{\circ }$ 

 $\Rightarrow 5^{\circ }{37}^{\prime }{60}^{″}={\left({\displaystyle \frac{45}{8}}\right)}^{\circ }$ 

Degree to Radian Conversion

 $\left({\displaystyle \frac{45}{8}}\right)\left({\displaystyle \frac{\pi }{180}}\right)={\displaystyle \frac{\pi }{32}}\text{rad}$ 

2. Find degree measure corresponding to  ${\left({\displaystyle \frac{\pi }{16}}\right)}^c$  

Ans- 

Converting the given value from radian to degree form

${\displaystyle \frac{\pi }{16}}\times {\displaystyle \frac{180}{\pi }}={\left({\displaystyle \frac{45}{4}}\right)}^{\circ }$ 

Simplify degree form

${\left({\displaystyle \frac{45}{4}}\right)}^{\circ }={11}^{\circ }{15}^{\prime }$ 

3. Find the length of an arc of a circle of radius  $5cm$  subtending a central angle measuring  $15{}^{\circ }$  

Ans- 

The arc of a circle with a radius of  $5\text{cm}$  with a central angle of  ${15}^{\circ }$  should be of the length  ${\displaystyle \frac{5\pi }{12}}cm$  using the formula  $\text{Arc}\text{=}\pi \times \left(\theta \right)$ .

4. Find the value of  ${\displaystyle \frac{19\pi }{3}}$  

Ans- 

We have  $\tan {\displaystyle \frac{19\pi }{3}}$ 

 $\tan {\displaystyle \frac{19\pi }{3}}=\tan \left(6{\displaystyle \frac{\pi }{3}}\right)$ 

                   $=\tan (6\pi +{\displaystyle \frac{\pi }{3}})$ 

                   $=\tan (3\times 2\pi +{\displaystyle \frac{\pi }{3}})$ 

                   $=\tan \left({\displaystyle \frac{\pi }{3}}\right)$ 

                   $=\sqrt{3}$ 

5. Find the value of  $\sin (-1125{}^{\circ })$  

Ans- 

We have  $\sin (-{1125}^{\circ })$ 

 $\sin (-{\displaystyle \frac{1125}{360}}\times {360}^{\circ })$ 

 $=-\sin ((3+{\displaystyle \frac{45}{360}})\times {360}^{\circ })$ 

 $=-\sin \left({45}^{\circ }\right)$ 

 $=-{\displaystyle \frac{1}{\sqrt{2}}}$ 

6. Find the value of  $\tan \left({15}^{\circ }\right)$  

Ans- 

We have  $\tan {15}^{\circ }$ 

 $\tan {15}^{\circ }=\tan ({60}^{\circ }-{45}^{\circ })$ 

                 $={\displaystyle \frac{\tan {60}^{\circ }-\tan {45}^{\circ }}{1+\tan {60}^{\circ }\times \tan {45}^{\circ }}}$ 

                 $={\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}}$ 

7. If  $\sin A={\displaystyle \frac{3}{5}}$  and  ${\displaystyle \frac{\pi }{2}}<A<$  find $\cos A$  

Ans- 

The condition  ${\displaystyle \frac{\pi }{2}}<A$  denotes that we need to take into account for the second quadrant, hence the cosine value will be negative. 

Therefore, 

 $\cos A={\displaystyle \frac{-4}{5}}$ 

8. If  $\tan A={\displaystyle \frac{a}{a+1}}$  and  $\tan B={\displaystyle \frac{1}{2a+1}}$  then find the value of  $A+B$  

Ans- 

 $\tan (A+B)={\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B}}$ 

                           $={\displaystyle \frac{{\displaystyle \frac{a}{a+1}}+{\displaystyle \frac{1}{2a+1}}}{1-{\displaystyle \frac{a}{a+1}}\cdot {\displaystyle \frac{1}{2a+1}}}}$ 

                          $={\displaystyle \frac{{\displaystyle \frac{2a^2+2a+1}{(a+1)(2a+1)}}}{{\displaystyle \frac{(a+1)(2a+1)-a}{(a+1)(2a+1)}}}}$ 

                          $=1$ 

Which can only be possible if  $A+B={45}^{\circ }$ .

9. Express  $\sin 12\theta +\sin 4\theta$  as the product of sines and cosine

Ans-

Using the trigonometric difference formula, we get

 $\sin 12\theta +\sin 4\theta =\sin (8\theta +4\theta )+\sin (8\theta -4\theta )$  

                                    $=2\sin 8\theta \cos 4\theta$ 

10. Express  $2\cos 4x\sin 2x$  as an algebraic sum of sines or cosine.

Ans- 

 $2\cos 4x\sin 2x=\sin (2x+4x)+\sin (2x-4x)$ 

                                 $=\sin 6x+\sin (-2x)$ 

                                  $=\sin 6x-\sin 2x$ 

11. Write the range of  $\cos \theta$  

Ans- 

The cosine function is a periodic function with a domain of  $\mathbb{R}$  and a range of  $[-1,1]$ .

12. What is domain of  $\sec \theta$  

Ans- 

The secant function is the reciprocal of the cosine function, it has a domain of  $\mathbb{R}-\{(2n+1){\displaystyle \frac{\pi }{2}};n\in \mathbb{Z}\}$  because those are the points where the cosine function equates to  $0$ .

13. Find the principal solution of  $\cot x=3$  

Ans- 

The principal solution of  $\cot x=3$  is for the following input values  $x={\displaystyle \frac{5\pi }{6}},{\displaystyle \frac{11\pi }{6}}$ .

14. Write the general solution of  $\cos \theta =0$  

Ans- 

The general solution for the equation  $\cos \theta =0$  is  $\theta =(2n+1){\displaystyle \frac{\pi }{2}},n\in \mathbb{Z}$ .

15. If  $\sin x={\displaystyle \frac{\sqrt{5}}{3}}$ and  $0<x<{\displaystyle \frac{\pi }{2}}$ find the value of $\cos 2x$  

Ans- 

We know that  $\cos 2x=1-{\sin }^{2}x$ 

 $\cos 2x=1-2{\left({\displaystyle \frac{\sqrt{5}}{3}}\right)}^2$ 

                $=1-2\times {\displaystyle \frac{5}{9}}$ 

                $=-{\displaystyle \frac{1}{9}}$ 

16. If  $\cos x=-{\displaystyle \frac{1}{3}}$  and  $x$  lies in quadrant  $\text{III}$ , find the value of  $\sin {\displaystyle \frac{x}{2}}$  

Ans- 

We know that  $\cos 2x=1-2{\sin }^{2}x$ 

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

 $\Rightarrow -{\displaystyle \frac{1}{3}}=1-2{\sin }^2{\displaystyle \frac{x}{2}}$ 

 $\Rightarrow 2{\sin }^2{\displaystyle \frac{x}{2}}=1+{\displaystyle \frac{1}{3}}$ 

 $\Rightarrow {\sin }^2{\displaystyle \frac{x}{2}}={\displaystyle \frac{2}{3}}$ 

 $\Rightarrow \sin {\displaystyle \frac{x}{2}}=\pm \sqrt{{\displaystyle \frac{2}{3}}}$ 

 $\Rightarrow \sin {\displaystyle \frac{x}{2}}=\sqrt{{\displaystyle \frac{2}{3}}}\left[\text{2nd Quadrant}\right]$ 

17. Convert into radian measures  $-47{}^{\circ }{30}^{\prime }$  

Ans- 

Convert into pure degree form and then convert to radian

 $-47{}^{\circ }{30}^{\prime }=-{(47+{\displaystyle \frac{30}{60}})}^{{}^{\circ }}$ 

                   $=-{(47+{\displaystyle \frac{1}{2}})}^{{}^{\circ }}$ 

                   $=-({\displaystyle \frac{95}{2}}\times {\displaystyle \frac{\pi }{180}})\text{rad}$ 

                   $=-{\displaystyle \frac{19\pi }{72}}\text{rad}$ 

18. Evaluate  $\tan 75{}^{\circ }$  

Ans- 

Use the trigonometric addition formula for the tangent function

 $\tan {75}^{\circ }=\tan ({45}^{\circ }+{30}^{\circ })$ 

                 $={\displaystyle \frac{\tan {45}^{\circ }+\tan {30}^{\circ }}{1-\tan {45}^{\circ }\tan {30}^{\circ }}}$ 

                 $={\displaystyle \frac{\sqrt{3}+1}{\sqrt{3}-1}}$ 

19. Prove that  $\sin (40+\theta )\cdot \cos (10+\theta )-\cos (40+\theta )\cdot \sin (10+\theta )={\displaystyle \frac{1}{2}}$  

Ans-  

Let us take the left-hand side of the equation and make some manipulations.

We know,  $\sin (a-b)=\sin a\cos b-\cos a\sin b$ 

 $\text{L}\text{.H}\text{.S}=\sin (40+\theta )\cos (10+\theta )-\cos (40+\theta )\sin (10+\theta )$ 

              $=\sin [40+\theta -10-\theta ]=\sin 30$ 

              $={\displaystyle \frac{1}{2}}$ 

20. Find the principal solution of the eq.  $\sin x={\displaystyle \frac{\sqrt{3}}{2}}$  

Ans- 

The principal solution of  $\sin x={\displaystyle \frac{\sqrt{3}}{2}}$  is the input values of  $x={\displaystyle \frac{\pi }{3}},{\displaystyle \frac{2\pi }{3}}$ 

21. Prove that  $\cos ({\displaystyle \frac{\pi }{4}}+x)+\cos ({\displaystyle \frac{\pi }{4}}-x)=\sqrt{2}\cos x$ 

Ans- 

Let us start with the left-hand side and use the trigonometric differences formula for the cosine function

 $\text{L}\text{.H}\text{.S}=\cos ({\displaystyle \frac{\pi }{4}}+x)+\cos ({\displaystyle \frac{\pi }{4}}-x)$ 

               $=2\cos {\displaystyle \frac{\pi }{4}}\cos x$ 

               $=2\left({\displaystyle \frac{1}{\sqrt{2}}}\right)\cos x$ 

               $=\sqrt{2}\cos x$ 

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

22. Convert into radian measures  $-37{}^{\circ }{30}^{\prime }$  

Ans- 

Convert into pure degree form and then convert from degree to radian

 $-37{}^{\circ }{30}^{\prime }={(37+{\displaystyle \frac{30}{60}})}^{{}^{\circ }}$ 

                 $=-{\left({\displaystyle \frac{75}{2}}\right)}^{{}^{\circ }}$ 

                 $=-{\displaystyle \frac{75}{2}}\times {\displaystyle \frac{\pi }{180}}\text{rad}$ 

                 $=-{\displaystyle \frac{5\pi }{24}}\text{rad}$ 

23. Prove

 $Sin(n+1)xSin(n+2)x+Cos(n+1)x.Cos(n+2)x=Cosx$   

Ans- 

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

                 $=\cos \{(n+1)x-(n+2)x\}$ 

                 $=\cos (nx+x-n-2x)$ 

                 $=\cos (-x)$ 

                 $=\cos \left(x\right)$ 

24. Find the value of  $\mathrm{Sin}{\displaystyle \frac{31\pi }{3}}$  

Ans- 

We have  $\sin {\displaystyle \frac{31\pi }{3}}$ 

 $\mathrm{Sin}{\displaystyle \frac{31\pi }{3}}=\mathrm{Sin}(10\pi +{\displaystyle \frac{\pi }{3}})$ 

                   $=\mathrm{Sin}(2\pi \times 5+{\displaystyle \frac{\pi }{3}})\left[\text{Periodic Function}\right]$ 

                   $=\mathrm{Sin}{\displaystyle \frac{\pi }{3}}$ 

                   $={\displaystyle \frac{\sqrt{3}}{2}}$ 

25. Find the principal solution of the eq.  $\tan x=-{\displaystyle \frac{1}{\sqrt{3}}}$ . 

Ans- 

The principal solution of the equation  $\tan x=-{\displaystyle \frac{1}{\sqrt{3}}}$  will be the input values of  $x={\displaystyle \frac{5\pi }{6}},{\displaystyle \frac{11\pi }{6}}$ 

26. Convert into radian measures $5{}^{\circ }{37}^{\prime }{30}^{″}$  

Ans- 

Converting the given value to a pure degree form

 $5^{\circ }{37}^{\prime }{30}^{″}=5^{\circ }{37}^{\prime }{\left({\displaystyle \frac{30}{60}}\right)}^{\prime }$ 

 $\Rightarrow 5^{\circ }{37}^{\prime }{60}^{″}=5^{\circ }{\left({\displaystyle \frac{75}{2}}\right)}^{\prime }$ 

 $\Rightarrow 5^{\circ }{37}^{\prime }{60}^{″}=5^{\circ }{\left({\displaystyle \frac{75}{2\left(60\right)}}\right)}^{\circ }$ 

 $\Rightarrow 5^{\circ }{37}^{\prime }{60}^{″}={\left({\displaystyle \frac{45}{8}}\right)}^{\circ }$ 

Degree to Radian Conversion

 $\left({\displaystyle \frac{45}{8}}\right)\left({\displaystyle \frac{\pi }{180}}\right)={\displaystyle \frac{\pi }{32}}\text{rad}$ 

27.  Prove  $Cos70{}^{\circ }.Cos10{}^{\circ }+Sin70{}^{\circ }.Sin10{}^{\circ }={\displaystyle \frac{1}{2}}$  

Ans- 

Starting with the left-hand side and using the trigonometric differences formula for the cosine function.

 $\text{L}\text{.H}\text{.S}=\text{cos}\left({70}^{\circ }{10}^{\circ }\right)$ 

               $=\cos {60}^{\circ }$ 

               $={\displaystyle \frac{1}{2}}$ 

28.  Evaluate  $2\mathrm{Sin}{\displaystyle \frac{\pi }{12}}$  

Ans- 

Use the trigonometric difference formula for the sine function and expand

 $2\sin {\displaystyle \frac{\pi }{12}}=2\sin [{\displaystyle \frac{\pi }{4}}-{\displaystyle \frac{\pi }{6}}]$ 

                   $=2[\sin {\displaystyle \frac{\pi }{4}}\cos {\displaystyle \frac{\pi }{6}}-\cos {\displaystyle \frac{\pi }{4}}\sin {\displaystyle \frac{\pi }{6}}]$ 

                   $=2[{\displaystyle \frac{1}{\sqrt{2}}}\times {\displaystyle \frac{\sqrt{3}}{2}}-{\displaystyle \frac{1}{\sqrt{2}}}\times {\displaystyle \frac{1}{2}}]$ 

                    $={\displaystyle \frac{\sqrt{3}-1}{\sqrt{2}}}$ 

29.  Find the solution of  $\mathrm{Sin}x=-{\displaystyle \frac{\sqrt{3}}{2}}$  

Ans- 

We are required to find the general solution for the equation  $\sin x=-{\displaystyle \frac{\sqrt{3}}{2}}$ 

 $\mathrm{Sin}x=-{\displaystyle \frac{\sqrt{3}}{2}}$ 

 $\Rightarrow \mathrm{Sin}x=\mathrm{Sin}(\pi +{\displaystyle \frac{\pi }{3}})$ 

 $\Rightarrow \mathrm{Sin}x=\mathrm{Sin}{\displaystyle \frac{4\pi }{3}}$ 

When

 $\mathrm{Sin}\theta =\mathrm{Sin}\alpha$ 

 $\theta =n\pi +{(-1)}^n\cdot \alpha$ 

 $x=n\pi +{(-1)}^n\cdot {\displaystyle \frac{4\pi }{3}}$ 

30.  Prove that  ${\displaystyle \frac{\mathrm{Cos}9{}^{\circ }-\mathrm{Sin}9{}^{\circ }}{\mathrm{Cos}9{}^{\circ }+\mathrm{Sin}9{}^{\circ }}}=\tan 36{}^{\circ }$  

Ans- 

Let us start with the right-hand side and use the trigonometric differences formula for the tangent function.

 $\text{R}\text{.H}\text{.S}=\tan 36{}^{\circ }$ 

                $=\tan ({45}^{\circ }-9^{\circ })$ 

                $={\displaystyle \frac{\tan {45}^{\circ }-\tan 9^{\circ }}{1+\tan {45}^{\circ }\tan 9^{\circ }}}$ 

                $={\displaystyle \frac{1-\tan 9^{\circ }}{1+\tan 9^{\circ }}}$ 

                $={\displaystyle \frac{\cos 9^{\circ }-\sin 9^{\circ }}{\cos 9^{\circ }+\sin 9^{\circ }}}$ 

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

31.  Find the value of  $\tan {\displaystyle \frac{19\pi }{3}}$  

Ans- 

We have  $\tan \left({\displaystyle \frac{19\pi }{3}}\right)$ 

 $\tan {\displaystyle \frac{19\pi }{3}}=\tan (6\pi -{\displaystyle \frac{\pi }{3}})$ 

                  $=\tan [3\times 2\pi +{\displaystyle \frac{\pi }{3}}]\left[\text{Periodic Function}\right]$ 

                  $=\tan {\displaystyle \frac{\pi }{3}}$ 

                  $=\sqrt{3}$ 

32.  Prove  $\mathrm{Cos}4x=1-8{\mathrm{Sin}}^{2}x.{\mathrm{Cos}}^{2}x$ 

Ans- 

Starting with the left-hand side and using the trigonometric addition formula, $\cos 2x=1-2{\sin }^{2}x$ 

We get,

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

               $=1-2{\mathrm{Sin}}^{2}2x$ 

               $=1-2{(\mathrm{Sin}2x)}^2$ 

               $=1-2{(2\mathrm{Sin}x.\mathrm{Cos}x)}^2$ 

               $=1-2(4{\mathrm{Sin}}^{2}x.{\mathrm{Cos}}^{2}x)$ 

               $=1-8{\mathrm{Sin}}^{2}x.{\mathrm{Cos}}^{2}x$ 

33.  Prove  ${\displaystyle \frac{\mathrm{Cos}(\pi +x).\mathrm{Cos}(-x)}{\mathrm{Sin}(\pi -x).\mathrm{Cos}({\displaystyle \frac{\pi }{2}}+x)}}=Co{t}^{2}x$  

Ans- 

Starting with the left-hand side and using the trigonometric periodic identities, we obtain the following

 $\text{L}\text{.H}\text{.S}\text{.}={\displaystyle \frac{\cos (\pi +x)\cos (-x)}{\sin (\pi -x)\cos ({\displaystyle \frac{\pi }{2}}+x)}}$ 

                 $={\displaystyle \frac{-\cos x\cos x}{-\sin x\sin x}}$ 

                 $={\cot }^{2}x$ 

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

34.  Prove that  $\tan {56}^{\circ }={\displaystyle \frac{\mathrm{Cos}{11}^{\circ }+\mathrm{Sin}{11}^{\circ }}{\mathrm{Cos}{11}^{\circ }-\mathrm{Sin}{11}^{\circ }}}$  

Ans- 

Starting with the left-hand side and using the trigonometric addition formula for the tangent function, we obtain

 $\text{L}\text{.H}\text{.S}\text{.}=\tan {56}^{\circ }$ 

                 $=\tan ({45}^{\circ }+{11}^{\circ })$ 

                 $={\displaystyle \frac{\tan {45}^{\circ }+\tan {11}^{\circ }}{1-\tan {45}^{\circ }\cdot \tan {11}^{\circ }}}$ 

                 $={\displaystyle \frac{1+\tan {11}^{\circ }}{1-\tan {11}^{\circ }}}$ 

                $={\displaystyle \frac{\cos {11}^{\circ }+\sin {11}^{\circ }}{\cos {11}^{\circ }-\sin {11}^{\circ }}}$ 

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

35.  Prove that  $\mathrm{Cos}{105}^{\circ }+\mathrm{Cos}{15}^{\circ }=\mathrm{Sin}{75}^{\circ }-\mathrm{Sin}{15}^{\circ }$ 

Ans- 

Starting with the left-hand side and using the trigonometric difference formula for the cosine function, we obtain

 $\text{L}\text{.H}\text{.S}\text{.}=\mathrm{Cos}{105}^{\circ }+\mathrm{Cos}{15}^{\circ }$ 

                 $=\mathrm{Cos}({90}^{\circ }+{15}^{\circ })+\mathrm{Cos}({90}^{\circ }-{75}^{\circ })$ 

                 $=-\mathrm{Sin}{15}^{\circ }+\mathrm{Sin}{75}^{\circ }$ 

                 $=\mathrm{Sin}{75}^{\circ }-\mathrm{Sin}{15}^{\circ }$ 

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

36.  Find the value of  $\mathrm{Cos}(-{1710}^{\circ })$ 

Ans- 

We have  $\cos (-{1710}^{\circ })$ . We also know  $\cos (-x)=\cos x$ 

 $\mathrm{Cos}(-{1710}^{\circ })=\mathrm{Cos}(1800-90)$ 

                               $=\mathrm{Cos}[5\times 360+90]$ 

                               $=\mathrm{Cos}{\displaystyle \frac{\pi }{2}}$ 

                               $=0$ 

37.   A wheel makes  $360$  revolutions in  $1$  minute. Through how many radians does it turn in  $1$  second.

Ans- 

Given,

 $\text{Number of revolutions made in 60s}=360$ 

 $\text{Number of revolutions made in 1s}={\displaystyle \frac{360}{60}}$ 

 $\text{Angle moved in 6 revolutions}=2\pi \times 6$ 

 $=12\pi$ 

38. Prove that  ${\mathrm{Sin}}^{2}6x-{\mathrm{Sin}}^{2}4x=\mathrm{Sin}2x.\mathrm{Sin}10x$ 

Ans- 

Starting with the left-hand side and using the trigonometric addition formula for the sine function, we obtain

 $\text{L}\text{.H}\text{.S}\text{.}={\mathrm{Sin}}^{2}6x-{\mathrm{Sin}}^{2}4x$ 

                $=\sin (6x+4x)\sin (6x-4x)$ 

                $=\sin 10x\sin 2x$ 

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

39. Prove that  ${\displaystyle \frac{\tan 69+\tan 66}{1\tan 69.\tan 66}}=-1$ 

Ans- 

Starting with the left-hand side and using the trigonometric difference identity for the tangent function, we obtain

 $\text{L}\text{.H}\text{.S}\text{.}={\displaystyle \frac{\tan {69}^{\circ }+\tan {66}^{\circ }}{1-\tan {69}^{\circ }\tan {66}^{\circ }}}$ 

                 $=\tan ({69}^{\circ }+{66}^{\circ })$ 

                 $=\tan \left({135}^{\circ }\right)$ 

                 $=\tan ({90}^{\circ }+{45}^{\circ })$ 

                $=-1$ 

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

40.  Prove that  $\dfrac{\operatorname{Sin}x}{1+\operatorname{Cos}x}=\tan 

\dfrac{x}{2} $  

Ans- 

Starting with the left-hand side and using the trigonometric addition identities for the sine and cosine function, we obtain

 $\text{L}\text{.H}\text{.S}\text{.}={\displaystyle \frac{\sin x}{1+\cos x}}$ 

                 $={\displaystyle \frac{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}{2{\cos }^2{\displaystyle \frac{x}{2}}}}$ 

                 $={\displaystyle \frac{\sin {\displaystyle \frac{x}{2}}}{\cos {\displaystyle \frac{x}{2}}}}$ 

                 $=\tan {\displaystyle \frac{x}{2}}$ 

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

4 Marks Questions

Prove the following identities

1. The minute hand of a watch is  $1.5cm$  long. How far does its tip move in  $40$  minute? 

Ans- 

Analysing the given information, we have 

 $r=1.5cm$ 

 $\text{Angle made in }60min={360}^{\circ }$ 

 $\text{Angle made in 1min}=6^{\circ }$ 

 $\text{Angle made in 40min}=6^{\circ }\times {40}^{\circ }={240}^{\circ }$ 

Calculating the arc distance

              $\theta ={\displaystyle \frac{l}{r}}$ 

 $240\times {\displaystyle \frac{\pi }{180}}={\displaystyle \frac{l}{1.5}}$ 

      $2\times 3.14=l$ 

              $6.28=l$ 

                      $l=6.28cm$ 

2. Show that  $tan3x.tan2x.tanx=tan3xtan2xtanx$  

Ans- 

Let us start with  $\tan 3x$  and we know  $3x=2x+x$ 

 $\tan 3x=\tan (2x+x)$ 

 ${\displaystyle \frac{\tan 3x}{1}}={\displaystyle \frac{\tan 2x+\tan x}{1-\tan 2x.\tan x}}$ 

 $\tan 3x(1-\tan 2x.\tan x)=\tan 2x+\tan x$ 

 $\tan 3x-\tan 3x.\tan 2x.\tan x=\tan 2x+\tan x$ 

 $\tan 3x.\tan 2x.\tan x=\tan 3x-\tan 2x-\tan x$ 

3. Find the value of  $\tan {\displaystyle \frac{\pi }{8}}$  

Ans- 

We know that

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

Therefore, we have

    $\tan \left(2{\displaystyle \frac{\pi }{8}}\right)={\displaystyle \frac{2\tan {\displaystyle \frac{\pi }{8}}}{1-{\tan }^2{\displaystyle \frac{\pi }{8}}}}$ 

 $\Rightarrow 1={\displaystyle \frac{2\tan {\displaystyle \frac{\pi }{8}}}{1-{\tan }^2{\displaystyle \frac{\pi }{8}}}}$ 

Put  $\tan {\displaystyle \frac{\pi }{8}}=x$ 

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

 $\Rightarrow 2x=1-x^2$ 

 $\Rightarrow x={\displaystyle \frac{-1\pm \sqrt{2}}{1}}$ 

Since,  ${\displaystyle \frac{\pi }{8}}$  lies in the first quadrant, the value must be positive, hence

$\tan {\displaystyle \frac{\pi }{8}}=\sqrt{2}-1$ 

4. Prove that  ${\displaystyle \frac{\mathrm{Sin}(x+y)}{\mathrm{Sin}(x-y)}}={\displaystyle \frac{\tan x+\tan y}{\tan x-\tan y}}$  

Ans- 

Starting with the left-hand side and using the trigonometric difference formula for the sine function, we get

 $\text{L}\text{.H}\text{.S}\text{.}={\displaystyle \frac{\mathrm{Sin}(x+y)}{\mathrm{Sin}(x-y)}}$ 

                $={\displaystyle \frac{\mathrm{Sin}x.\mathrm{Cos}y+\mathrm{Cos}x.\mathrm{Sin}y}{\mathrm{Sin}x.\mathrm{Cos}y-\mathrm{Cos}x.\mathrm{Sin}y}}$ 

Dividing numerator and denominator by  $\mathrm{Cos}x.\mathrm{Cos}y$ 

           $={\displaystyle \frac{\tan x+\tan y}{\tan x-\tan y}}$ 

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

5. If in two circles, arcs of the same length subtend angles  ${60}^{\circ }$  and  ${75}^{\circ }$  at the center find the ratio of their radii. 

Ans- 

We know that the length of the arc and its subtended angle is related using the following formula

 $\theta ={\displaystyle \frac{1}{r_1}}$ 

Therefore, we have

 $60\times {\displaystyle \frac{\pi }{18}}={\displaystyle \frac{1}{r_1}}$ 

 $r_1={\displaystyle \frac{3l}{\pi }}$      ….. $(1)$ 

 $\theta ={\displaystyle \frac{1}{r_2}}$ 

 $75\times {\displaystyle \frac{\pi }{18}}={\displaystyle \frac{1}{r_2}}$ 

 $r_2={\displaystyle \frac{12l}{5\pi }}$     ….. $(2)$ 

 $(1)÷(2)$ 

 ${\displaystyle \frac{r_1}{r_2}}={\displaystyle \frac{{\displaystyle \frac{3l}{\pi }}}{{\displaystyle \frac{12l}{5\pi }}}}$ 

 $={\displaystyle \frac{31}{\pi }}\times {\displaystyle \frac{5\pi }{12l}}$ 

 $=5:4$ 

6. Prove that  $\mathrm{Cos}6x=32{\mathrm{Cos}}^{2}x-48{\mathrm{Cos}}^{4}x+18{\mathrm{Cos}}^{2}x-1$  

Ans. 

Starting with the left-hand side and using the trigonometric identities for the cosine function, we obtain

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

                 $=\mathrm{Cos}2(3x)=2{\mathrm{Cos}}^{2}3x-1$ 

                 $=\mathrm{Cos}2(3x)$ 

                 $=2{(4co{s}^{3}x-3\cos x)}^2-1$ 

                 $=2[16{\mathrm{Cos}}^{6}x+9{\mathrm{Cos}}^{2}x-24{\mathrm{Cos}}^{4}x]-1$ 

                  $=32{\mathrm{Cos}}^{6}x+18{\mathrm{Cos}}^{2}x-48{\mathrm{Cos}}^{4}x-1$ 

               $=32{\mathrm{Cos}}^{6}x-48{\mathrm{Cos}}^{4}x+18{\mathrm{Cos}}^{2}x1$ 

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

7. Solve  $\mathrm{Sin}2x-\mathrm{Sin}4x+\mathrm{Sin}6x=0$  

Ans- 

Starting with the left-hand side and using the trigonometric addition identity for the sine function, we obtain

 $\text{L}\text{.H}\text{.S}\text{.}=\mathrm{Sin}6x+\mathrm{Sin}2x-\mathrm{Sin}4x$ 

                 $=2\sin \left({\displaystyle \frac{6x+2x}{2}}\right)\cos \left({\displaystyle \frac{6x-2x}{2}}\right)-\sin 4x$ 

                 $=\sin 4x(2\cos 2x-1)$ 

                 $=0$ 

Now,

 $\sin 4x=0$ 

      $4x=n\pi$ 

        $x={\displaystyle \frac{n\pi }{4}}$ 

Also,

 $2\cos 2x-1=0$ 

        $\cos 2x=\cos {\displaystyle \frac{\pi }{3}}$ 

             $2x=2n\pi \pm {\displaystyle \frac{\pi }{3}}$ 

              $x=n\pi \pm {\displaystyle \frac{\pi }{6}}$ 

8. In a circle of diameter  $40cm$ , the length of a chord is  $20cm$ . Find the length of the minor area of the chord. 

Ans- 

Given,

 $\theta ={\displaystyle \frac{l}{r}}$ 

 $\Rightarrow 60\times {\displaystyle \frac{\pi }{180}}={\displaystyle \frac{l}{20}}$ 

 $\Rightarrow l={\displaystyle \frac{20\pi }{3}}\text{cm/s}$ 

9. Prove that  $\tan 4x={\displaystyle \frac{4\tan x(1-{\tan }^{2}x)}{1-6{\tan }^{2}x+{\tan }^{4}x}}$  

Ans- 

Starting with the left-hand side and using the trigonometric addition identities for the tangent function, we obtain

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

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

                 $={\displaystyle \frac{2.{\displaystyle \frac{2\tan 2x}{1-{\tan }^{2}2x}}}{1-{\left({\displaystyle \frac{2\tan 2x}{1-{\tan }^{2}2x}}\right)}^2}}$ 

                $={\displaystyle \frac{{\displaystyle \frac{4\tan x}{1-{\tan }^{2}x}}}{{\displaystyle \frac{{(1-{\tan }^{2}x)}^2-4{\tan }^{2}x}{{(1-{\tan }^{2}x)}^2}}}}$ 

                 $={\displaystyle \frac{4\tan x}{(1-{\tan }^{2}x)}}\times {\displaystyle \frac{(1-{\tan }^{2}x)}{1+{\tan }^{4}x-2{\tan }^{2}x-4{\tan }^{2}x}}$ 

                 $={\displaystyle \frac{4\tan x(1-{\tan }^{2}x)}{1-6{\tan }^{2}x+{\tan }^{4}x}}$ 

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

10. Prove that  ${(Cosx+Cosy)}^2+{\left(SinxSiny\right)}^2=4Co{s}^2\left({\displaystyle \frac{x+y}{2}}\right)$  

Ans- 

Starting with the left-hand side and using the trigonometric addition identities for the cosine and sine function, we obtain

 $\text{L}\text{.H}\text{.S}\text{.}={(Cosx+Cosy)}^2+{\left(SinxSiny\right)}^2$ 

                $={(2\mathrm{Cos}{\displaystyle \frac{x+y}{2}}.\mathrm{Cos}{\displaystyle \frac{x-y}{2}})}^2+{(2\mathrm{Cos}\left({\displaystyle \frac{x+y}{2}}\right).\mathrm{Sin}\left({\displaystyle \frac{x-y}{2}}\right))}^2$ 

                $=4{\mathrm{Cos}}^2{\displaystyle \frac{x+y}{2}}.{\mathrm{Cos}}^2\left({\displaystyle \frac{x-y}{2}}\right)+4{\mathrm{Cos}}^2{\displaystyle \frac{x+y}{2}}.{\mathrm{Sin}}^2{\displaystyle \frac{x-y}{2}}$ 

               $=4{\mathrm{Cos}}^2\left({\displaystyle \frac{x+y}{2}}\right)[{\mathrm{Cos}}^2{\displaystyle \frac{x-y}{2}}+{\mathrm{Sin}}^2{\displaystyle \frac{x-y}{2}}]$ 

               $=4{\mathrm{Cos}}^2\left({\displaystyle \frac{x+y}{2}}\right)$ 

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

11. If  $Cotx=-{\displaystyle \frac{5}{12}},x$  lies in second quadrant find the values of other five trigonometric functions 

Ans- 

Given

 $Cotx=-{\displaystyle \frac{5}{12}}$ 

Using some trigonometric identities, we obtain

 $\tan x=-{\displaystyle \frac{12}{5}}$ 

 ${\mathrm{Sec}}^{2}x=1+{\tan }^{2}x$ 

 $\mathrm{Sec}x=\pm {\displaystyle \frac{13}{5}}$ 

Since  $x$  lies in the second quadrant, the cosine value will be negative

 $\mathrm{Sec}x=-{\displaystyle \frac{13}{5}}$ 

 $\mathrm{Cos}x=-{\displaystyle \frac{5}{13}}$ 

 $\mathrm{Sin}x=\tan x.\mathrm{Cos}x$ 

             $={\displaystyle \frac{-12}{5}}\times \left({\displaystyle \frac{-5}{13}}\right)$ 

              $={\displaystyle \frac{12}{13}}$ 

 $\mathrm{Csc}x={\displaystyle \frac{13}{12}}$ 

12. Prove that  ${\displaystyle \frac{\mathrm{Sin}5x-2\mathrm{Sin}3x+\mathrm{Sin}x}{\mathrm{Cos}5x-\mathrm{Cos}x}}=\tan x$  

Ans- 

Starting with the left-hand side and using the trigonometric difference identities for the sine function, we obtain

 $\text{L}\text{.H}\text{.S}\text{.}={\displaystyle \frac{\mathrm{Sin}5x+\mathrm{Sin}x-2\mathrm{Sin}3x}{\mathrm{Cos}5x-\mathrm{Cos}x}}$ 

                 $={\displaystyle \frac{2\mathrm{Sin}3x.\mathrm{Cos}2x-2\mathrm{Sin}3x}{-2\mathrm{Sin}3x.\mathrm{Sin}2x}}$ 

                 $={\displaystyle \frac{2\mathrm{Sin}3x(\mathrm{Cos}2x-1)}{-2\mathrm{Sin}3x.\mathrm{Sin}2x}}$ 

                 $={\displaystyle \frac{-(1-\mathrm{Cos}2x)}{-\mathrm{Sin}2x}}$ 

                 $={\displaystyle \frac{2{\mathrm{Sin}}^{2}x}{2\mathrm{Sin}x.\mathrm{Cos}x}}$ 

                 $={\displaystyle \frac{\mathrm{Sin}x}{\mathrm{Cos}x}}$ 

                 $=\tan x$ 

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

13. Prove that  $Sinx+Sin3x+Sin5x+Sin7x=4Cosx.Cos2x.Sin4x$  

Ans- 

Starting with the left-hand side and using the trigonometric addition identities for the sine function, we obtain

 $\text{L}\text{.H}\text{.S}\text{.}=Sinx+Sin3x+Sin5x+Sin7x$ 

                 $=\mathrm{Sin}x+\mathrm{Sin}7x+\mathrm{Sin}3x+\mathrm{Sin}5x$ 

                 $=2\mathrm{Sin}\left({\displaystyle \frac{x+7x}{2}}\right).\mathrm{Cos}\left({\displaystyle \frac{x-7x}{2}}\right)+2\mathrm{Sin}\left({\displaystyle \frac{3x+5x}{2}}\right)\mathrm{Cos}\left({\displaystyle \frac{3x-5x}{2}}\right)$ 

                 $=2\mathrm{Sin}4x.\mathrm{Cos}3x+2\mathrm{Sin}4x.\mathrm{Cos}x$ 

                 $=2\mathrm{Sin}4x[\mathrm{Cos}3x+\mathrm{Cos}x]$ 

                 $=2\mathrm{Sin}4x[2\mathrm{Cos}\left({\displaystyle \frac{3x+x}{2}}\right).\mathrm{Cos}\left({\displaystyle \frac{3x-x}{2}}\right)]$ 

                 $=2\mathrm{Sin}4x[2\mathrm{Cos}2x.\mathrm{Cos}x]$ 

                 $=4\mathrm{Cos}x.\mathrm{Cos}2x.\mathrm{Sin}4x$ 

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

14. Find the angle between the minute hand and hour hand of a clock when the time is  $7.20$