Important Questions for CBSE Class 11 Maths Chapter 3 Trigonometric Functions 2024-25

1 Mark Questions

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

Ans- 

Converting the given value to a pure degree form

 $ {{5}^{\circ }}37'30''={{5}^{\circ }}37'\left( \dfrac{30}{60} \right)' $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{5}^{\circ }}\left( \dfrac{75}{2} \right)' $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{5}^{\circ }}{{\left( \dfrac{75}{2\left( 60 \right)} \right)}^{\circ }} $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{\left( \dfrac{45}{8} \right)}^{\circ }} $ 

Degree to Radian Conversion

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

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

Ans- 

Converting the given value from radian to degree form

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

Simplify degree form

$ {{\left( \dfrac{45}{4} \right)}^{\circ }}={{11}^{\circ }}15' $ 

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  $ \dfrac{5\pi }{12}cm $  using the formula  $ \text{Arc}\,\text{=}\,\pi \times \left( \theta  \right) $ .

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

Ans- 

We have  $ \tan \dfrac{19\pi }{3} $ 

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

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

                   $ =\tan \left( 3\times 2\pi +\dfrac{\pi }{3} \right) $ 

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

                   $ =\sqrt{3} $ 

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

Ans- 

We have  $ \sin \left( -{{1125}^{\circ }} \right) $ 

 $ \sin \left( -\dfrac{1125}{360}\times {{360}^{\circ }} \right) $ 

 $ =-\sin \left( \left( 3+\dfrac{45}{360} \right)\times {{360}^{\circ }} \right) $ 

 $ =-\sin \left( {{45}^{\circ }} \right) $ 

 $ =-\dfrac{1}{\sqrt{2}} $ 

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

Ans- 

We have  $ \tan {{15}^{\circ }} $ 

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

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

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

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

Ans- 

The condition  $ \dfrac{\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=\dfrac{-4}{5} $ 

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

Ans- 

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

                           $ =\dfrac{\dfrac{a}{a+1}+\dfrac{1}{2a+1}}{1-\dfrac{a}{a+1}\cdot \dfrac{1}{2a+1}} $ 

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

                          $ =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 \left( 8\theta +4\theta  \right)+\sin \left( 8\theta -4\theta  \right) $  

                                    $ =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 \left( 2x+4x \right)+\sin \left( 2x-4x \right) $ 

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

                                  $ =\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  $ \left[ -1,1 \right] $ .

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}-\left\{ (2n+1)\dfrac{\pi }{2};n\in \mathbb{Z} \right\} $  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=\dfrac{5\pi }{6},\dfrac{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)\dfrac{\pi }{2},n\in \mathbb{Z} $ .

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

Ans- 

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

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

                $ =1-2\times \dfrac{5}{9} $ 

                $ =-\dfrac{1}{9} $ 

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

Ans- 

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

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

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

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

 $ \Rightarrow {{\sin }^{2}}\dfrac{x}{2}=\dfrac{2}{3} $ 

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

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

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

Ans- 

Convert into pure degree form and then convert to radian

 $ -47{}^\circ 30'=-{{\left( 47+\dfrac{30}{60} \right)}^{{}^\circ }} $ 

                   $ =-{{\left( 47+\dfrac{1}{2} \right)}^{{}^\circ }} $ 

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

                   $ =-\dfrac{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 }}) $ 

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

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

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

Ans-  

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

We know,  $ \sin \left( a-b \right)=\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 \left[ 40+\theta -10-\theta  \right]=\sin 30 $ 

              $ =\dfrac{1}{2} $ 

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

Ans- 

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

21. Prove that  $ \cos \left( \dfrac{\pi }{4}+x \right)+\cos \left( \dfrac{\pi }{4}-x \right)=\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 \left( \dfrac{\pi }{4}+x \right)+\cos \left( \dfrac{\pi }{4}-x \right) $ 

               $ =2\cos \dfrac{\pi }{4}\cos x $ 

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

               $ =\sqrt{2}\cos x $ 

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

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

Ans- 

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

 $ -37{}^\circ 30'={{\left( 37+\dfrac{30}{60} \right)}^{{}^\circ }} $ 

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

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

                 $ =-\dfrac{5\pi }{24}\text{rad} $ 

23. Prove

 $ Sin\text{ }\left( n+1 \right)\text{ }x\text{ }Sin\text{ }\left( n+2 \right)\text{ }x\text{ }+\text{ }Cos\text{ }\left( n+1 \right)\text{ }x.\text{ }Cos\text{ }\left( n+2 \right)\text{ }x\text{ }=\text{ }Cos\text{ }x $   

Ans- 

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

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

                 $ =\cos \left( nx+x-n-2x \right) $ 

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

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

24. Find the value of  $ \operatorname{Sin}\dfrac{31\pi }{3} $  

Ans- 

We have  $ \sin \dfrac{31\pi }{3} $ 

 $ \operatorname{Sin}\dfrac{31\pi }{3}=\operatorname{Sin}\left( 10\pi +\dfrac{\pi }{3} \right) $ 

                   $ =\operatorname{Sin}\left( 2\pi \times 5+\dfrac{\pi }{3} \right)\,\,\,\,\,\,\left[ \text{Periodic Function} \right] $ 

                   $ =\operatorname{Sin}\dfrac{\pi }{3} $ 

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

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

Ans- 

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

26. Convert into radian measures $ 5{}^\circ \text{ }37'\text{ }30'' $  

Ans- 

Converting the given value to a pure degree form

 $ {{5}^{\circ }}37'30''={{5}^{\circ }}37'\left( \dfrac{30}{60} \right)' $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{5}^{\circ }}\left( \dfrac{75}{2} \right)' $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{5}^{\circ }}{{\left( \dfrac{75}{2\left( 60 \right)} \right)}^{\circ }} $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{\left( \dfrac{45}{8} \right)}^{\circ }} $ 

Degree to Radian Conversion

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

27.  Prove  $ Cos70{}^\circ .\text{ }Cos10{}^\circ +\text{ }Sin70{}^\circ .\text{ }Sin10{}^\circ =\dfrac{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 }} $ 

               $ =\dfrac{1}{2} $ 

28.  Evaluate  $ 2\operatorname{Sin}\dfrac{\pi }{12} $  

Ans- 

Use the trigonometric difference formula for the sine function and expand

 $ 2\sin \dfrac{\pi }{12}=2\sin \left[ \dfrac{\pi }{4}-\dfrac{\pi }{6} \right] $ 

                   $ =2\left[ \sin \dfrac{\pi }{4}\cos \dfrac{\pi }{6}-\cos \dfrac{\pi }{4}\sin \dfrac{\pi }{6} \right] $ 

                   $ =2\left[ \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2} \right] $ 

                    $ =\dfrac{\sqrt{3}-1}{\sqrt{2}} $ 

29.  Find the solution of  $ \operatorname{Sin}x=-\dfrac{\sqrt{3}}{2} $  

Ans- 

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

 $ \operatorname{Sin}x=-\dfrac{\sqrt{3}}{2} $ 

 $ \Rightarrow \operatorname{Sin}x=\operatorname{Sin}\left( \pi +\dfrac{\pi }{3} \right) $ 

 $ \Rightarrow \operatorname{Sin}x=\operatorname{Sin}\dfrac{4\pi }{3} $ 

When

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

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

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

30.  Prove that  $ \dfrac{\operatorname{Cos}9{}^\circ -\operatorname{Sin}9{}^\circ }{\operatorname{Cos}9{}^\circ +\operatorname{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 \left( {{45}^{\circ }}-{{9}^{\circ }} \right) $ 

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

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

                $ =\dfrac{\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 \dfrac{19\pi }{3} $  

Ans- 

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

 $ \tan \dfrac{19\pi }{3}=\tan \left( 6\pi -\dfrac{\pi }{3} \right) $ 

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

                  $ =\tan \dfrac{\pi }{3} $ 

                  $ =\sqrt{3} $ 

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

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

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

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

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

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

33.  Prove  $ \dfrac{\operatorname{Cos}(\pi +x).\operatorname{Cos}(-x)}{\operatorname{Sin}(\pi -x).\operatorname{Cos}\left( \dfrac{\pi }{2}+x \right)}=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{.}=\dfrac{\cos \left( \pi +x \right)\cos \left( -x \right)}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)} $ 

                 $ =\dfrac{-\cos x\cos x}{-\sin x\sin x} $ 

                 $ ={{\cot }^{2}}x $ 

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

34.  Prove that  $ \tan {{56}^{\circ }}=\dfrac{\operatorname{Cos}{{11}^{\circ }}+\operatorname{Sin}{{11}^{\circ }}}{\operatorname{Cos}{{11}^{\circ }}-\operatorname{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 }}) $ 

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

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

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

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

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

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

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

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

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

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

Ans- 

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

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

                               $ =\operatorname{Cos}\left[ 5\times 360+90 \right] $ 

                               $ =\operatorname{Cos}\dfrac{\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}=\dfrac{360}{60} $ 

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

 $ =12\pi  $ 

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

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

                $ =\sin 10x\sin 2x $ 

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

39. Prove that  $ \dfrac{\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{.}=\dfrac{\tan {{69}^{\circ }}+\tan {{66}^{\circ }}}{1-\tan {{69}^{\circ }}\tan {{66}^{\circ }}} $ 

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

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

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

                $ =-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{.}=\dfrac{\sin x}{1+\cos x} $ 

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

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

                 $ =\tan \dfrac{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 }60\min ={{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 =\dfrac{l}{r} $ 

 $ 240\times \dfrac{\pi }{180}=\dfrac{l}{1.5} $ 

      $ 2\times 3.14=l $ 

              $ 6.28=l $ 

                      $ l=6.28cm $ 

2. Show that  $ tan\text{ }3x.\text{ }tan\text{ }2x.\text{ }tan\text{ }x\text{ }=\text{ }tan\text{ }3x\text{ }\text{ }tan\text{ }2x\text{ }\text{ }tan\text{ }x $  

Ans- 

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

 $ \tan 3x=\tan (2x+x) $ 

 $ \dfrac{\tan 3x}{1}=\dfrac{\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 \dfrac{\pi }{8} $  

Ans- 

We know that

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

Therefore, we have

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

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

Put  $ \tan \dfrac{\pi }{8}=x $ 

 $ 1=\dfrac{2x}{1-{{x}^{2}}} $ 

 $ \Rightarrow 2x=1-{{x}^{2}} $ 

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

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

$ \tan \dfrac{\pi }{8}=\sqrt{2}-1 $ 

4. Prove that  $ \dfrac{\operatorname{Sin}(x+y)}{\operatorname{Sin}(x-y)}=\dfrac{\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{.}=\dfrac{\operatorname{Sin}(x+y)}{\operatorname{Sin}(x-y)} $ 

                $ =\dfrac{\operatorname{Sin}x.\operatorname{Cos}y+\operatorname{Cos}x.\operatorname{Sin}y}{\operatorname{Sin}x.\operatorname{Cos}y-\operatorname{Cos}x.\operatorname{Sin}y} $ 

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

           $ =\dfrac{\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 =\dfrac{1}{{{r}_{1}}} $ 

Therefore, we have

 $ 60\times \dfrac{\pi }{18}=\dfrac{1}{{{r}_{1}}} $ 

 $ {{r}_{1}}=\dfrac{3l}{\pi } $      ….. $ (1) $ 

 $ \theta =\dfrac{1}{{{r}_{2}}} $ 

 $ 75\times \dfrac{\pi }{18}=\dfrac{1}{{{r}_{2}}} $ 

 $ {{r}_{2}}=\dfrac{12l}{5\pi } $     ….. $ (2) $ 

 $ (1)\div (2) $ 

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

 $ =\dfrac{31}{\pi }\times \dfrac{5\pi }{12l} $ 

 $ =5:4 $ 

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

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

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

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

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

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

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

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

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

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

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

                 $ =0 $ 

Now,

 $ \sin 4x=0 $ 

      $ 4x=n\pi  $ 

        $ x=\dfrac{n\pi }{4} $ 

Also,

 $ 2\cos 2x-1=0 $ 

        $ \cos 2x=\cos \dfrac{\pi }{3} $ 

             $ 2x=2n\pi \pm \dfrac{\pi }{3} $ 

              $ x=n\pi \pm \dfrac{\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 =\dfrac{l}{r} $ 

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

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

9. Prove that  $ \tan 4x=\dfrac{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 $ 

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

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

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

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

                 $ =\dfrac{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  $ {{\left( Cosx+Cosy \right)}^{2}}+{{\left( SinxSiny \right)}^{2}}=4Co{{s}^{2}}\left( \dfrac{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{.}={{\left( Cosx+Cosy \right)}^{2}}+{{\left( SinxSiny \right)}^{2}} $ 

                $ ={{\left( 2\operatorname{Cos}\dfrac{x+y}{2}.\operatorname{Cos}\dfrac{x-y}{2} \right)}^{2}}+{{\left( 2\operatorname{Cos}\left( \dfrac{x+y}{2} \right).\operatorname{Sin}\left( \dfrac{x-y}{2} \right) \right)}^{2}} $ 

                $ =4{{\operatorname{Cos}}^{2}}\dfrac{x+y}{2}.{{\operatorname{Cos}}^{2}}\left( \dfrac{x-y}{2} \right)+4{{\operatorname{Cos}}^{2}}\dfrac{x+y}{2}.{{\operatorname{Sin}}^{2}}\dfrac{x-y}{2} $ 

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

               $ =4{{\operatorname{Cos}}^{2}}\left( \dfrac{x+y}{2} \right) $ 

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

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

Ans- 

Given

 $ Cotx=-\dfrac{5}{12} $ 

Using some trigonometric identities, we obtain

 $ \tan x=-\dfrac{12}{5} $ 

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

 $ \operatorname{Sec}x=\pm \dfrac{13}{5} $ 

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

 $ \operatorname{Sec}x=-\dfrac{13}{5} $ 

 $ \operatorname{Cos}x=-\dfrac{5}{13} $ 

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

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

              $ =\dfrac{12}{13} $ 

 $ \operatorname{Csc}x=\dfrac{13}{12} $ 

12. Prove that  $ \dfrac{\operatorname{Sin}5x-2\operatorname{Sin}3x+\operatorname{Sin}x}{\operatorname{Cos}5x-\operatorname{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{.}=\dfrac{\operatorname{Sin}5x+\operatorname{Sin}x-2\operatorname{Sin}3x}{\operatorname{Cos}5x-\operatorname{Cos}x} $ 

                 $ =\dfrac{2\operatorname{Sin}3x.\operatorname{Cos}2x-2\operatorname{Sin}3x}{-2\operatorname{Sin}3x.\operatorname{Sin}2x} $ 

                 $ =\dfrac{2\operatorname{Sin}3x(\operatorname{Cos}2x-1)}{-2\operatorname{Sin}3x.\operatorname{Sin}2x} $ 

                 $ =\dfrac{-(1-\operatorname{Cos}2x)}{-\operatorname{Sin}2x} $ 

                 $ =\dfrac{2{{\operatorname{Sin}}^{2}}x}{2\operatorname{Sin}x.\operatorname{Cos}x} $ 

                 $ =\dfrac{\operatorname{Sin}x}{\operatorname{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 $ 

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

                 $ =2\operatorname{Sin}\left( \dfrac{x+7x}{2} \right).\operatorname{Cos}\left( \dfrac{x-7x}{2} \right)+2\operatorname{Sin}\left( \dfrac{3x+5x}{2} \right)\operatorname{Cos}\left( \dfrac{3x-5x}{2} \right) $ 

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

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

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

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

                 $ =4\operatorname{Cos}x.\operatorname{Cos}2x.\operatorname{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 $  

Ans- 

We know that the angle made by minute hand in  $ 15\min =15\times 6={{90}^{\circ }} $ 

We also know that the angle made by the hour hand in  $ 1hr={{30}^{\circ }} $ 

In  $ 60 $  minute  $ =\dfrac{30}{60} $ 

 $ =\dfrac{1}{2} $ 

 $ [\because  $ Angle Travelled by  $ hr $ hand in  $ 12hr={{360}^{\circ }}] $ 

In  $ 20 $ minutes  $ =\dfrac{1}{2}\times 20 $ 

                         $ ={{10}^{\circ }} $ 

Angle made  $ =90+10 $ 

                       $ ={{100}^{\circ }} $ 

15. Show that  $ \sqrt{2+\sqrt{2+2\operatorname{Cos}4\theta }}=2\operatorname{Cos}\theta  $  

Ans- 

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

 $ \text{L}\text{.H}\text{.S}\text{.}=\sqrt{2+\sqrt{2+2\operatorname{Cos}4\theta }} $ 

                 $ =\sqrt{2+\sqrt{2(1+\operatorname{Cos}4\theta )}} $ 

                 $ =\sqrt{2+\sqrt{2.2{{\operatorname{Cos}}^{2}}2\theta }} $ 

                 $ =\sqrt{2+2\operatorname{Cos}2\theta } $ 

                 $ =\sqrt{2(1+\operatorname{Cos}2\theta )} $ 

                 $ =\sqrt{2.2{{\operatorname{Cos}}^{2}}\theta } $ 

                 $ =2\operatorname{Cos}\theta  $ 

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

16. Prove that  $ Cot4x\left( Sin5x+Sin3x \right)=Cotx\left( Sin5xSin3x \right) $  

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{.}=Cot4x\left( Sin5x+Sin3x \right) $ 

                $ =\dfrac{\operatorname{Cos}4x}{\operatorname{Sin}4x}\left[ 2\operatorname{Sin}\dfrac{5x+3x}{2}.\operatorname{Cos}\dfrac{5x-3x}{2} \right] $ 

                $ =\dfrac{\operatorname{Cos}4x}{\operatorname{Sin}4x}2\operatorname{Sin}4x.\operatorname{Cos}x $ 

                $ =2\operatorname{Cos}4x.\operatorname{Cos}x $ 

Then, we move on to the right-hand side and using the trigonometric addition identity for the sine function, we obtain

   $ \text{R}\text{.H}\text{.S}\text{.}=Cotx\left( Sin5xSin3x \right) $ 

                   $ =\dfrac{\operatorname{Cos}x}{\operatorname{Sin}x}\left[ 2\operatorname{Cos}\dfrac{5x+3x}{2}.\operatorname{Sin}\dfrac{5x-3x}{2} \right] $ 

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

                   $ =2\operatorname{Cos}4x.\operatorname{Cos}x $ 

Therefore, 

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

6 Marks Questions

1. Find the general solution of  $ sin2x+sin4x+sin6x=0 $ 

Ans- 

We have that  $ \sin 2x+\sin 4x+\sin 6x=0 $ 

 $ \Rightarrow \left( \sin 2x+\sin 6x \right)+\sin 4x=0 $ 

 $ \Rightarrow \left( 2\sin \left( \dfrac{2x+6x}{2} \right)\cos \left( \dfrac{2x-6x}{2} \right) \right)+\sin 4x=0 $ 

 $ \Rightarrow 2\sin 4x\cos 2x+\sin 4x=0 $ 

 $ \Rightarrow \sin 4x\left( 2\cos 2x+1 \right)=0 $ 

Now

 $ \sin 4x=0 $ 

 $ \Rightarrow x=n\pi  $ 

 $ 2\cos 2x+1=0 $ 

 $ \Rightarrow x=n\pi \pm \dfrac{\pi }{3} $ 

2. Find the general solution of  $ \cos \theta \cos 2\theta \cos 3\theta =\dfrac{1}{4} $  

Ans- 

We have that  $ \cos \theta \cos 2\theta \cos 3\theta =\dfrac{1}{4} $ 

 $ \Rightarrow 4\cos \theta \cos 2\theta \cos 3\theta =1 $ 

Using the trigonometric addition identity for the cosine function, we obtain

 $ \Rightarrow 2\left( 2\cos \theta \cos 3\theta  \right)\cos 2\theta -1=0 $ 

 $ \Rightarrow 2\left( \cos 4\theta +\cos 2\theta  \right)\cos 2\theta -1=0 $ 

 $ \Rightarrow 2\left( 2{{\cos }^{2}}2\theta -1+\cos 2\theta  \right)\cos 2\theta -1=0 $ 

 $ \Rightarrow \left( 2{{\cos }^{2}}2\theta -1 \right)\left( 2\cos 2\theta +1 \right)=0 $ 

Now,

 $ 2{{\cos }^{2}}2\theta -1=0 $ 

 $ \Rightarrow \cos 4\theta =0 $ 

 $ \Rightarrow 4\theta =\left( 2n+1 \right)\dfrac{\pi }{2} $ 

 $ \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{8} $ 

Also,

 $ 2\cos 2\theta +1=0 $ 

 $ \Rightarrow \cos 2\theta =-\dfrac{1}{2} $ 

 $ \Rightarrow \theta =n\pi \pm \dfrac{\pi }{3} $ 

3. If  $ \operatorname{Sin}\alpha +\operatorname{Sin}\beta =a $  and  $ \operatorname{Cos}\alpha +\operatorname{Cos}\beta =b $  

Show that  $ \operatorname{Cos}(\alpha +\beta )=\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}+{{a}^{2}}} $  

Ans- 

Squaring both the equations and adding them together,

 $ {{b}^{2}}+{{a}^{2}}={{(\operatorname{Cos}\alpha +\operatorname{Cos}\beta )}^{2}}+{{(\operatorname{Sin}\alpha +\operatorname{Sin}\beta )}^{2}} $ 

                 $ ={{\operatorname{Cos}}^{2}}\alpha +{{\operatorname{Cos}}^{2}}\beta +2\operatorname{Cos}\alpha .\operatorname{Cos}\beta +{{\operatorname{Sin}}^{2}}\alpha +{{\operatorname{Sin}}^{2}}\beta +2\operatorname{Sin}\alpha .\operatorname{Sin}\beta  $ 

                 $ =1+1+2(\operatorname{Cos}\alpha .\operatorname{Cos}\beta +\operatorname{Sin}\alpha .\operatorname{Sin}\beta ) $ 

            $ =2+2\operatorname{Cos}(\alpha -\beta ) $        $ (1) $ 

 $ {{b}^{2}}-{{a}^{2}}={{(\operatorname{Cos}\alpha +\operatorname{Cos}\beta )}^{2}}-{{(\operatorname{Sin}\alpha +\operatorname{Sin}\beta )}^{2}} $ 

                 $ =({{\operatorname{Cos}}^{2}}\alpha -{{\operatorname{Sin}}^{2}}\beta )+({{\operatorname{Cos}}^{2}}\beta -{{\operatorname{Sin}}^{2}}\alpha )+2\operatorname{Cos}(\alpha +\beta ) $ 

                 $ =\operatorname{Cos}(\alpha +\beta )\operatorname{Cos}(\alpha -\beta )+\operatorname{Cos}(\beta +\alpha )\operatorname{Cos}(\alpha -\beta )+2\operatorname{Cos}(\alpha +\beta ) $ 

                 $ =2\operatorname{Cos}(\alpha +\beta ).\operatorname{Cos}(\alpha -\beta )+2\operatorname{Cos}(\alpha +\beta ) $ 

                 $ =\operatorname{Cos}(\alpha +\beta )[2\operatorname{Cos}(\alpha -\beta )+2] $ 

              $ =\operatorname{Cos}(\alpha +\beta ).({{b}^{2}}+{{a}^{2}}) $       from  $ (1) $ 

Dividing equation  $ \left( 1 \right) $  with  $ {{b}^{2}}+{{a}^{2}} $ , we get

 $ \dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}+{{a}^{2}}}=\operatorname{Cos}(\alpha +\beta ) $ 

4. Prove  $ Cos\alpha +Cos\beta +Cos\gamma +Cos\left( \alpha +\beta +\gamma  \right)=4\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\beta +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\gamma +\alpha }{2} \right) $ 

Ans- 

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

 $ \text{L}\text{.H}\text{.S}\text{.}=Cos\alpha +Cos\beta +Cos\gamma +Cos\left( \alpha +\beta +\gamma  \right) $ 

                 $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha -\beta }{2} \right)+2\operatorname{Cos}\left( \dfrac{\alpha +\beta +\gamma +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha +\beta +\gamma -\gamma }{2} \right) $ 

                $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha -\beta }{2} \right)+2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha +\beta +2\gamma }{2} \right) $ 

                $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right)\left[ \operatorname{Cos}\left( \dfrac{\alpha -\beta }{2} \right)+\operatorname{Cos}\left( \dfrac{\alpha +\beta +2\gamma }{2} \right) \right] $ 

            $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\operatorname{Cos}\left( \dfrac{\dfrac{\alpha -\beta }{2}+\dfrac{\alpha +\beta +2\gamma }{2}}{2} \right).\operatorname{Cos}\left( \dfrac{\dfrac{\alpha +\beta +2\gamma }{2}-\dfrac{\alpha -\beta }{2}}{2} \right) \right] $ 

                $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\operatorname{Cos}\left( \dfrac{\alpha +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\beta +\gamma }{2} \right) \right] $ 

               $ =4\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\beta +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\gamma +\alpha }{2} \right) $ 

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

5. Prove that  $ \operatorname{Sin}3x+\operatorname{Sin}2x-\operatorname{Sin}2x=4\operatorname{Sin}x.\operatorname{Cos}\dfrac{x}{2}.\operatorname{Cos}\dfrac{3x}{2} $  

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{.}=\sin 3x+\sin x-\sin 2x $ 

                 $ =2\cos \left( \dfrac{3x+x}{2} \right).\operatorname{Sin}\left( \dfrac{3x+x}{2} \right)+\operatorname{Sin}2x $ 

                 $ =2\cos 2x.\sin x+\sin 2x $ 

                 $ =2\cos 2x.\sin x+2\sin x\cos x $ 

                 $ =2\sin x[\cos 2x+\cos x] $ 

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

                 $ =4\sin x\cos x\dfrac{3x}{2}\cos \dfrac{x}{2} $ 

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

6. Prove that  $ 2\cos \dfrac{\pi }{13}.\cos \dfrac{9\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13}=0 $  

Ans- 

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

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

                 $ =\cos \left( \dfrac{\pi }{13}+\dfrac{9\pi }{13} \right)+\cos \left( \dfrac{\pi }{13}-\dfrac{9\pi }{13} \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =\cos \dfrac{10\pi }{13}+\cos \dfrac{18\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =\cos \left( \pi -\dfrac{3\pi }{13} \right)+\cos \left( \pi -\dfrac{5\pi }{13} \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =-\cos \dfrac{3\pi }{13}-\cos \dfrac{5\pi }{13}+\dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =0 $ 

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

7.  Find the value of  $ \tan (\alpha +\beta ) $ given that  $ \cot \alpha =\dfrac{1}{2},\alpha \in \left( \pi ,\dfrac{3\pi }{2} \right) $  and  $ \operatorname{Sec}\beta =-\dfrac{5}{3},\beta \in \left( \dfrac{\pi }{2},\pi  \right) $  

Ans- 

We know that,

 $ \tan \left( \alpha +\beta  \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } $ 

Given,

 $ Cot\alpha =\dfrac{1}{2} $ 

 $ \tan \alpha =2 $ 

Now, let us find  $ \tan \beta  $ 

 $ 1+{{\tan }^{2}}\beta ={{\operatorname{Sec}}^{2}}\beta  $ 

 $ 1+{{\tan }^{2}}\beta ={{\left( \dfrac{-5}{3} \right)}^{2}}\left[ \because \operatorname{Sec}\beta =\dfrac{-5}{3} \right] $ 

 $ \tan \beta =\pm \dfrac{4}{3} $ 

 $ \tan \beta =-\dfrac{4}{3}\left[ \because \beta \in \left( \dfrac{\dfrac{\pi }{2}}{x} \right) \right] $ 

Therefore, we have that

 $ \tan \left( \alpha +\beta  \right)=\dfrac{2-\dfrac{4}{3}}{1-2\left( \dfrac{-4}{3} \right)} $ 

                     $ =\dfrac{2}{11} $ 

8. Prove that  $ \dfrac{\operatorname{Sec}8A-1}{\operatorname{Sec}4A-1}=\dfrac{\tan 8A}{\tan 2A} $  

Ans- 

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

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\sec 8A-1}{\sec 4A-1} $ 

                $ =\dfrac{\dfrac{1}{\operatorname{Cos}8A}-1}{\dfrac{1}{\operatorname{Cos}4A}-1} $ 

                $ =\dfrac{1-\operatorname{Cos}8A}{1-\operatorname{Cos}4A}\times \dfrac{\operatorname{Cos}4A}{\operatorname{Cos}8A} $ 

                $ =\dfrac{2{{\operatorname{Sin}}^{2}}4A}{2{{\operatorname{Sin}}^{2}}2A}.\dfrac{\operatorname{Cos}4A}{\operatorname{Cos}8A} $ 

                $ =\dfrac{(2\operatorname{Sin}4A.\operatorname{Cos}4A).\operatorname{Sin}4A}{2{{\operatorname{Sin}}^{2}}2A.\operatorname{Cos}8A} $ 

                $ =\dfrac{\operatorname{Sin}8A(2\operatorname{Sin}2A.\operatorname{Cos}2A)}{2{{\operatorname{Sin}}^{2}}2A.\operatorname{Cos}8A} $ 

                $ =\dfrac{\operatorname{Sin}8A\operatorname{Cos}2A}{\operatorname{Sin}2A.\operatorname{Cos}2A} $ 

                $ =\dfrac{\tan 8A}{\tan 2A} $ 

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

9. Prove that  $ {{\operatorname{Cos}}^{2}}x+{{\operatorname{Cos}}^{2}}\left( x+\dfrac{\pi }{3} \right)+{{\operatorname{Cos}}^{2}}\left( x-\dfrac{\pi }{3} \right)=\dfrac{3}{2} $  

Ans- 

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

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{1+\operatorname{Cos}2x}{2}+\dfrac{1+\operatorname{Cos}\left( 2x+\dfrac{2\pi }{3} \right)}{2}+\dfrac{1+\operatorname{Cos}\left( 2x-\dfrac{2\pi }{3} \right)}{2} $ 

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

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

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

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\operatorname{Cos}\dfrac{4\pi }{6} \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\operatorname{Cos}\dfrac{2\pi }{3} \right] $ 

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

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\left( \dfrac{-1}{2} \right) \right] $ 

                $ =\dfrac{3}{2} $ 

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

10.  Prove that  $ \operatorname{Cos}2x.\operatorname{Cos}\dfrac{x}{2}-\operatorname{Cos}3x.\operatorname{Cos}\dfrac{9x}{2}=\operatorname{Sin}5x\operatorname{Sin}\dfrac{5x}{2} $  

Ans- 

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

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{1}{2}\left[ 2\operatorname{Cos}2x.\operatorname{Cos}\dfrac{x}{2}-2\operatorname{Cos}3x.\operatorname{Cos}\dfrac{9x}{2} \right] $ 

                $ =\dfrac{1}{2}\left[ \operatorname{Cos}\left( 2x+\dfrac{x}{2} \right)+\operatorname{Cos}\left( 2x-\dfrac{x}{2} \right)-\operatorname{Cos}\left( \dfrac{9x}{2}+3x \right)-\operatorname{Cos}\left( \dfrac{9x}{2}-3x \right) \right] $ 

                $ =\dfrac{1}{2}\left[ \operatorname{Cos}\dfrac{5x}{2}+\operatorname{Cos}\dfrac{3x}{2}-\operatorname{Cos}\dfrac{15x}{2}-\operatorname{Cos}\dfrac{3x}{2} \right] $ 

               $ =\dfrac{1}{2}\left[ \operatorname{Cos}\dfrac{5x}{2}-\operatorname{Cos}\dfrac{15x}{2} \right] $ 

              $ =\dfrac{1}{2}\left[ -2\operatorname{Sin}\left( \dfrac{\dfrac{5x}{2}+\dfrac{15x}{2}}{2} \right).\operatorname{Sin}\left( \dfrac{\dfrac{5x}{2}-\dfrac{15x}{2}}{2} \right) \right] $ 

              $ =-\operatorname{Sin}5x.\operatorname{Sin}\left( \dfrac{-5x}{2} \right) $ 

              $ =\operatorname{Sin}5x.\operatorname{Sin}\dfrac{5x}{2} $ 

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

11.  Prove that  $ \operatorname{Cos}20{}^\circ .\operatorname{Cos}40{}^\circ .\operatorname{Cos}60{}^\circ .\operatorname{Cos}80{}^\circ =\dfrac{1}{16} $  

Ans- 

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

 $ \text{L}\text{.H}\text{.S}\text{.}=\operatorname{Cos}{{20}^{{}^\circ }}.\operatorname{Cos}{{40}^{{}^\circ }}.\operatorname{Cos}{{60}^{{}^\circ }}.\operatorname{Cos}{{80}^{{}^\circ }} $ 

                 $ =\operatorname{Cos}{{60}^{{}^\circ }}.\operatorname{Cos}{{20}^{{}^\circ }}.\operatorname{Cos}{{40}^{{}^\circ }}.\operatorname{Cos}{{80}^{{}^\circ }} $ 

                 $ =\dfrac{1}{2}.\dfrac{1}{2}\operatorname{Cos}{{40}^{{}^\circ }}\left( 2\operatorname{Cos}{{20}^{{}^\circ }}.\operatorname{Cos}{{80}^{{}^\circ }} \right) $ 

                 $ =\dfrac{1}{4}\operatorname{Cos}{{40}^{{}^\circ }}\left[ \operatorname{Cos}(80+20)+\operatorname{Cos}(80-20) \right] $ 

                 $ =\dfrac{1}{4}\operatorname{Cos}{{40}^{{}^\circ }}\left[ \operatorname{Cos}{{100}^{{}^\circ }}+\operatorname{Cos}{{60}^{{}^\circ }} \right] $ 

                $ =\dfrac{1}{4}\operatorname{Cos}{{40}^{{}^\circ }}\left[ \operatorname{Cos}{{100}^{{}^\circ }}+\dfrac{1}{2} \right] $ 

                $ =\dfrac{1}{8}(2\operatorname{Cos}{{100}^{{}^\circ }}\operatorname{Cos}{{40}^{{}^\circ }})+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

           $ =\dfrac{1}{8}\left[ \operatorname{Cos}{{(100+40)}^{{}^\circ }}+\operatorname{Cos}{{(100-40)}^{{}^\circ }} \right]+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

               $ =\dfrac{1}{8}\left[ \operatorname{Cos}{{140}^{{}^\circ }}+\operatorname{Cos}{{60}^{{}^\circ }} \right]+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

               $ =\dfrac{1}{8}\left[ \operatorname{Cos}{{140}^{{}^\circ }}+\dfrac{1}{2} \right]+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

              $ =\dfrac{1}{8}\operatorname{Cos}{{(180-40)}^{{}^\circ }}+\dfrac{1}{16}+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

              $ =-\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }}+\dfrac{1}{16}+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

              $ =\dfrac{1}{16} $ 

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

12.  If  $ \tan x=\dfrac{3}{4},\pi <x<\dfrac{3\pi }{2}, $ Find the value of  $ \operatorname{Sin}\dfrac{x}{2},\operatorname{Cos}\dfrac{x}{2} $  and  $ \tan \dfrac{x}{2} $  

Ans- 

Given that

 $ \pi <x<\dfrac{3\pi }{2} $  implying that  $ x $  is in the third quadrant 

 $ \Rightarrow \dfrac{\pi }{2}<\dfrac{x}{2}<\dfrac{3\pi }{2} $ 

Therefore, we have that  $ \operatorname{Sin}\dfrac{x}{2} $  is positive and  $ \operatorname{Cos}\dfrac{x}{2} $  is negative.

Let us find for  $ \tan \dfrac{x}{2} $ 

We know

 $ 1+{{\tan }^{2}}x={{\operatorname{Sec}}^{2}}x\dfrac{5}{4} $ 

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

    $ {{\operatorname{Sec}}^{2}}x=\pm \dfrac{25}{16} $ 

      $ \operatorname{Cos}x=\pm \dfrac{4}{5} $ 

      $ \operatorname{Cos}x=-\dfrac{4}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \pi <x<\dfrac{3\pi }{2} \right] $ 

Let us find the required values 

 $ \operatorname{Sin}\dfrac{x}{2}=\sqrt{\dfrac{1-\operatorname{Cos}x}{2}} $ 

              $ =\sqrt{\dfrac{1+\dfrac{4}{5}}{2}} $ 

              $ =\sqrt{\dfrac{9}{10}} $ 

              $ =\dfrac{3}{\sqrt{10}} $ 

  $ \operatorname{Cos}\dfrac{x}{2}=-\sqrt{\dfrac{1-\operatorname{Cos}x}{2}} $ 

                  $ =-\sqrt{\dfrac{1-\dfrac{4}{5}}{2}} $ 

                  $ =-\sqrt{\dfrac{1}{10}} $ 

                 $ =\dfrac{-1}{\sqrt{10}} $ 

  $ \tan \dfrac{x}{2}=\dfrac{\dfrac{3}{\sqrt{10}}}{\dfrac{-1}{\sqrt{10}}} $ 

                $ =-3 $ 

CBSE Class 11 Maths Chapter-3 Important Questions - Free PDF Download

Class 11 Maths Chapter 3 Important Questions- What are Trigonometric Functions? 

In simple language, the trigonometric functions are the functions of an angle of triangles. It defines the relationship between sides and angles of a triangle is given on basis of these functions. The trigonometric functions consist of the sine, cosine, sectant, cosecant, tangent, and cotangent. It is also known as circular functions. There are several trigonometric formulas and identities that help to define the relationship between the angles and also the functions. At the end of chapter 3, students will find trigonometric functions class 11 extra questions. 

Tips to Score Marks in Trigonometric Functions?

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. When students prepare these important questions from Vedantu, they can also learn several tricks and shortcuts to solve the questions fast. Besides, students need to focus on the formulas of trigonometric functions that are crucial to solve the sums. Students should not skip this chapter at any cost as there are many significant areas like designing electronic circuits, finding the heights of tides, etc. To get a deeper insight into class 11 maths ch 3 important questions, students should practice the resources available from Vedantu. 

Discuss the Trigonometric Tables and Formulas?

In class 11 trigonometric functions, important questions about trigonometric tables and formulas constitute a vital part of the chapter. Let’s discuss both these concepts in detail below. 

The Formula for Function of Trigonometric Ratios

Trigonometric Table

Important Questions for Class 11 Maths Chapter 3 Based on Exercise

Q. An engine produces 360 revolutions in one minute. Through how many radians will it turn in one second?

Solution:

Provided,

The total number of revolutions made by an engine in one minute = 360

1 minute = 60 seconds

Therefore, number of revolutions in 1 second = 360/60 = 6

Angle formed in 1 revolution = 360°

Angles formed in 6 revolutions = 6 × 360°

Radian measure of the angle in a total of six revolutions = 6 × 360 × π/180

= 6 × 2 × π

= 12π

So, the engine turns 12π radians in one second.

Importance of Downloading Class 11 Maths Chapter 3 Important Questions PDF?

By downloading the important questions for class 11 maths chapter 3 students will get exposure to the concept of trigonometric functions in depth. Here are some benefits that students will get when they have the PDF version.

• They can prepare important notes for the examination. 

• They will get access to trigonometry functions class 11 extra questions. 

• They can use it as a ready resource for reference. 

• It will help them understand the question pattern of the examination. 

Practice Questions

• Prove that:
  (sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x

• Find the value of tan 765° cot 675° + tan 225° cot 405°

• Solve the equation: tan² θ + cot² θ = 2

• Write the value of 2sin 75° sin 15°

• Show that:
  tan 4A = (cos8Acos5A - cos12Acos9A) / (sin8Acos5A + cos12Asin9A)

• Find the general solution of the following equation:
  tan2θ +(1 – √3) tan θ – √3 = 0

• Prove that:
  3sinπ/6secπ/3 - 4sin5π/6cotπ/4 = 1

• Find the value:
  cos4π/8 + cos43π/8 + cos45π/8 + cos47π/8

• Show that:
  tan 15° + cot 15° = 4

• Find the most general value of θ satisfying the equation tan θ = -1 and cos θ = 1/√2

Advantages of Opting Vedantu for Important Questions of Class 11 Maths: Chapter 3 Trigonometric

Vedantu offers several benefits to students using their platform for the "Important Questions for CBSE Class 11 Maths Chapter 3 - Trigonometric Functions (2024-25)":

• Comprehensive Coverage: Vedantu's important questions are curated to cover a wide spectrum of topics within the chapter, ensuring a comprehensive understanding of trigonometric functions.

• Strategic Exam Preparation: These questions are strategically selected to align with the CBSE curriculum and examination patterns, preparing students effectively for their exams.

• Conceptual Clarity: Vedantu's platform emphasizes conceptual clarity by providing in-depth explanations and solutions for each question, helping students grasp the underlying principles.

• Variety of Problem Types: The diverse range of questions offered by Vedantu challenges students to apply trigonometric concepts in various problem-solving scenarios, enhancing their problem-solving skills.

• Self-Assessment and Practice: Students can use these questions for self-assessment and regular practice, enabling them to gauge their progress and identify areas that need improvement.

• Flexibility and Convenience: Vedantu's platform allows students to access these questions anytime, anywhere, providing flexibility in their study routine.

• Personalized Learning: Vedantu's adaptive learning approach tailors the learning experience to each student's pace and needs, ensuring effective comprehension and retention.

• Interactive Sessions: Vedantu offers interactive live sessions where students can clarify doubts and seek guidance from experienced educators, ensuring a holistic learning experience.

• Peer Learning: Students can engage in discussions and peer learning through Vedantu's platform, sharing insights and strategies with their peers.

• Holistic Support: Beyond just important questions, Vedantu provides additional study material, revision notes, and comprehensive study plans to support students' overall exam preparation.

Prepare Well with Vedantu Important Questions for CBSE Class 11 Maths Chapter Trigonometric Function 2024-25

The compilation of important questions for CBSE Class 11 Maths Chapter 3 - "Trigonometric Functions" (2024-25) offers a strategic tool to enhance students' grasp of fundamental trigonometric concepts. These questions delve into various aspects of trigonometry, aiding in the development of problem-solving skills and conceptual clarity. By addressing a diverse range of problem types, these questions prepare students comprehensively for examinations and instil a deeper understanding of the subject matter. This resource not only facilitates exam preparation but also promotes a robust foundation for more advanced mathematical exploration, ensuring students' competence in applying trigonometric principles in various scenarios.

Important Related Links for CBSE Class 11