NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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.

Hence proved.

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

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

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

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

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

L.H.S $=3+2+1$ 

$=1$ 

$=$ R.H.S.

Hence proved.

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

Ans:

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 $\text{sin x}$ repeats 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.

Hence proved.

5. Find the value of :

(i) Sin 75°

Ans: We have,

Sin 75°= Sin ( 45°+ 30° )

Sin 75°= Sin 45° Cos 30° + Cos 45° Sin 30° 

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

Therefore we have,

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

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

(ii) tan 15°

Ans: We have,

tan 15°= tan ( 45°- 30° )

= $\dfrac{( tan 45°- tan 30° )}{1+( tan 45° tan 30° )}$

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

Therefore we have,

$tan 15°=\dfrac{\text{1-}\dfrac{\text{1}}{\sqrt{\text{3}}}}{\text{1+1}\left( \dfrac{\text{1}}{\sqrt{\text{3}}} \right)}$

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

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

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

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.

Hence proved.

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

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

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

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

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

Hence proved.

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 the 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.

Hence proved.

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

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

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

= R.H.S

Hence proved.

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

Therefore we have,

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

$\text{=sin 10x sin 2x}$

= 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,                

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

=  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 $\text{=sin 2x+2sin 4x+sin 6x}$

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

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

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

Further computing,

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, $\text{cos 2x+1=2co}{{\text{s}}^{\text{2}}}\text{x}$ 

Therefore we have,

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

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

= R.H.S.

Hence proved.

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

Hence proved.

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

Ans: We know that,

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

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)

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

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

R.H.S.

Hence proved.

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

Ans: We know that

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

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

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

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

(Using the formula)

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

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

Further computing we have,

$\text{L}\text{.H}\text{.S=tan 4x}$ 

= R.H.S.

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

Ans: We  know that,

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

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

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

Hence proved.

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.

Hence proved.

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

Ans: We know that,

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

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.

Hence proved.

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

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

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

$\text{=cot 3x}$ 

= R.H.S.

Hence proved.

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

Ans:

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,

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

$\text{=1}$

= R.H.S.

Hence proved.

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]}$$$$$

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

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

= R.H.S.

Hence  proved.

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.

Hence proved.

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 $\text{=cos 6x}$

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

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

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

Further computing,

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

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

$\text{=4}\left[ \text{8co}{{\text{s}}^{\text{6}}}\text{x-1-12co}{{\text{s}}^{\text{4}}}\text{x+6co}{{\text{s}}^{\text{2}}}\text{x} \right]\text{-6co}{{\text{s}}^{\text{2}}}\text{x+3}$    $\text{=32co}{{\text{s}}^{\text{6}}}\text{x-48co}{{\text{s}}^{\text{4}}}\text{x+18co}{{\text{s}}^{\text{2}}}\text{x-1}$

Therefore we have,

L.H.S = R.H.S

Hence proved.

Conclusion

The Class 11 Maths Exercise 3.3 Solutions focuses on Trigonometric Functions, specifically their applications and identities. This exercise typically emphasizes solving trigonometric equations and verifying identities. From past exams, students can expect about 1-2 questions derived from ex 3.3 class 11, often involving proving identities or solving equations. Mastery of these concepts is crucial for understanding more advanced topics in mathematics, and thorough practice of Class 11 Maths Chapter 3 Exercise 3.3 will enhance problem-solving skills and exam readiness.

Class 11 Maths Chapter 3: Exercises Breakdown

CBSE Class 11 Maths Chapter 3 Other Study Materials

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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