Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

NCERT Solutions for Class 11 Maths Chapter 3: Trigonometric Functions - Exercise 3.3

ffImage
Last updated date: 29th Mar 2024
Total views: 566.1k
Views today: 11.66k
MVSAT 2024

CBSE NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 - Trignometry

Chapter 3 of Maths explains the concept of trigonometric ratios to trigonometric functions. The students will study the properties of trigonometric functions like sin theta, cos theta, tan theta, and more in a comprehensive way. In class 11 maths chapter 3 exercise 3.3, students have to solve problems on radian measure, arc length, conversion, finding the values of trigonometric functions, and more. Additionally, students will learn general solutions like sine and cosine, their identity, and another formula in ex 3.3 class 11 maths NCERT. 


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 3 - Trigonometric Functions

Exercise:

Exercise - 3.3

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

Competitive Exams after 12th Science

Access 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.


NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 Trigonometry-

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3.3 have been provided here for students to prepare well for their academic and competitive examinations. The steps given in the examples have been followed while providing the NCERT Solutions for Class 11 Exercise 3.3. for all the questions present in this exercise. These solutions have been prepared by the subject experts at Vedantu and are in accordance with the latest syllabus issued by the CBSE Board.


Topic Covered In Class 11 Maths Chapter 3 Exercise 3.3

Class 11 Maths Chapter 3 Exercise 3.3. Is based on the topic of the Trigonometry Function of  sum and difference of two angles.


Free PDF Download

The experts offer these solutions after a detailed analysis of the marking pattern plus the model answer sheet issued by CBSE. It is recommendable for class 11 students to go through with these solutions and enhance their speed for attempting Trigonometry questions. Class 11 Maths exercise 3.3 solutions can help students throughout the journey of their board and competitive exams.  

 

NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 - Trigonometry

In class 11 Trigonometry exercise 3.3, students will learn about the representation and properties of trigonometric functions. Students will get to know about the empty set, equal trigonometric functions, finite and infinite trigonometric functions, and more. The chapter educates students about principal solutions, general solutions, (x + y) formula, sine and cosine formula, and more. In chapter 3 exercise 3.3 Trigonometric Functions, students will acknowledge the concepts like a sign of trigonometric functions, domain and range of trigonometric functions, sum and difference of two angles formed by trigonometric functions, and more. There are four quadrants; all trigonometric functions are positive in the first quadrant. In the second quadrant, sine and cosec functions are positive. In the third, tan and cot functions are positive, and in the last quadrant, cos and sec functions are positive. 

The chapter introduces students to sub trigonometric functions, intersection, and union of trigonometric functions, the complement of a set, and more. In the section domain and range of trigonometric functions, students will study about domain and range of functions ranging from 0 to 1. The sine function increases from -1 to 0 and from 0 to 1. The cosine function increases from 0 to 1 and -1 to 0. The tan function ranges from 0 to infinity. Students should practice maths class 11 chapter 3 exercise 3.3 questions based on the concepts like trigonometric functions, properties of complement trigonometric functions, and more. 

By using the identity and properties of trigonometric functions, students will study how to derive expressions for the sum and difference of two numbers. Apart from it, students will come across problems based on trigonometric identities, basic applications, related expressions, and more. With the help of class 11 maths NCERT solutions chapter 3 exercise 3.3, students can practice a greater number of questions.

NCERT Solutions for Class 11 Maths Chapters


Chapter 3 - Trigonometric Functions Exercises in PDF Format

Exercise 3.1

7 Questions & Solutions

Exercise 3.2

10 Questions & Solutions

Exercise 3.3

25 Questions & Solutions

Exercise 3.4

9 Questions & Solutions

Miscellaneous Exercise

10 Questions & Solutions

 

Along with this, students can also download additional study materials provided by Vedantu, for Chapter 3 of CBSE Class 11 Maths Solutions –

Class 11 Maths Chapter 3 Exercise 3.3- Weightage Marks

Out of 80 marks, the Sets and Functions chapter hold a weight of 23 marks. Trigonometric Function is a part of this unit. 

 

Benefits of Maths Chapter 3 Class 11 NCERT Solutions

Class 11th leads to several career-making decisions as students have to go through some crucial competitive exams to get the desired college. Class 11 Maths NCERT solutions chapter 3 exercise 3.3 offers a better knowledge of the chapter that will help students to study deeper into basics. These solutions can be of great value for those trying to outshine in their regular school examinations. Some of the benefits of referring class 11 maths exercise 3.3 solutions include:

  • The students can gain the confidence to attempt an advanced level of questions asked from the exercise 3.3 maths class 11. 

  • Maths Class 11 NCERT solutions are a one-stop destination for all the possible questions with proved answers that will help students to secure a high percentage. 

  • The students can understand the properties of Trigonometry in a better way by going through these solutions. They get all the crucial contents required in one place. 

  • Exercise 3.3 class 11 maths solutions offered by Vedantu are in a well-organized format. It helps students to analyze the paper attempting method for Trigonometry questions asked in the exam.

  • The use of user-friendly language makes it trouble-free for students to study the problems based on Trigonometry. 

FAQs on NCERT Solutions for Class 11 Maths Chapter 3: Trigonometric Functions - Exercise 3.3

1. How to prove 2(𝜋/6) + (7𝜋 6)(𝜋/3)  =3/2?

Let us first solve the Left-hand side of the problem mentioned. 

Given equation at the left-hand side is 2(π/6) +(7π b)(π/3).

Using the properties of trigonometric functions: 

sin sin (𝜋/6) = ½ 𝜋/3 = ½ and = (π+ 𝜋/6) 

Hence, (𝜋/6) = (½)2=¼ (𝜋/3)  = (½)2= ¼ 

The given equation becomes 2(¼) + (π+ π/6)(¼) 

Hence, the equation becomes 2(¼) + (-π/6)(¼)

L.H.S. becomes 1/2 + (-2)2 (1/4)

Here, 1/2 + 4/4 = 1/2 + 1= 3/2 = R.H.S.

Hence proved L.H.S = R.H.S

2. How to find the value of tan 15o?

tan 15o can be written as tan (45o – 30o)

By using the Trigonometry formula for tan (x – y):

tan (x – y) = (tan x – tan y)/ (1 + tan x tan y)

tan (45o – 30o) = (tan 45o – tan 30o)/ (1 + tan 45o tan 30o)

Now, tan 45o = 1 and tan 30o = 1/√3; we get:

tan (45o – 30o) = (1 - 1/√3) / (1 + 1/√3)

tan 15o = ((√3 – 1)/√3)/ ((√3 + 1)/√3)

tan 15o = (√3 – 1)/ (√3 + 1) = (√3– 1)/ (√3 + 1) (√3 – 1)

By further calculations:

tan 15o = 4 - 2√3/ 3 – 1 = 2 - √3

3. How do I solve Class 11 Maths Chapter 3 Exercise 3.3?

Ans: You can refer to the NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 available on Vedantu to solve the sums. This NCERT Solutions to Trigonometric Functions is available for downloading as well so now, you can solve this exercise even without the internet! Hurry up and download the NCERT Solutions from the page NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.3 on the Vedantu website at free of cost. You can also download these solutions from the Vedantu app for free.

4. What is a radian measure?

A radian measure, in simple words, is the angle in a circle. For example, in a circle of one unit radius, the angle made at the centre by a one-unit arc will be equal to one radian. A radian measure is used to express the measure of an angle in a circle. For more clarity on the topic, you can consider using the NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.3. 

5. Where can I get the full solutions to Class 11 Maths Chapter 3 Exercise 3.3?

Vedantu is the best place for you. Here you can get the full solutions to Class 11 Maths Chapter Exercise 3.3. You can check out the NCERT Solutions for complete step by step answers to all the questions present in the exercise. Practice the complete solution to get full marks on your test. Since they are explained in simple language and in detail, the students will be able to grasp them easily. 

6. Is Class 11 Maths Chapter 3 very difficult?

No, Class 11 Maths Chapter 3 “Trigonometric Functions” is not at all difficult. Success can be easily achieved from practice. The best guide for practice is Vedantu’s NCERT Solutions Class 11 Maths Chapter 3. You will get a complete solution to all the questions from all the different exercises that are there in the chapter. These answers are precise and step by step so you will learn a lot while practising. 

7. How to complete the Class 11 Maths 3.3 Exercise?

You can successfully complete Class 11 Maths Chapter 3 Exercise 3.3. By referring to Vedantu’s NCERT Solutions. This solution is the best that you will get as it has all questions solved by experts, who have provided step by step precise answers. Moreover, all the important questions from the miscellaneous exercise have also been explained in detail. So, download the notes today and practise well.