NCERT Solutions for Miscellaneous Exercise Chapter 3 Class 11

Q: How many questions are there in Miscellaneous exercise chapter 3 ?

There are 10 questions total in Miscellaneous exercise chapter 3

Q: What is the application of Trigonometric Functions ?

It can be used to find the height of a point if the angle is known or vice versa.

Q: Are questions from previous year papers asked for examination from this chapter ?

Yes, in the final exam of class 11 the questions are present from the previous year.

Q: What is the difficulty level of problems asked from this unit ?

Moderate to difficult level pronlems are asked from this unit.

Q: Can one avoid Miscellaneous exercise during preparation of the chapter?

No, as it has questions that covers most of the topics of the chapter.

Q: What time it will take to complete all the problems of the miscellaneous exercise of this chapter for the first time ?

It takes around 3 to 4 hours to complete for the first time.

NCERT Solutions for Miscellaneous Exercise Chapter 3 Class 11 - trigonometric Functions

Edited By Vishal kumar | Updated on Nov 15, 2023 12:52 PM IST

NCERT Solutions for Class 11 Maths Chapter 3: Trigonometric Functions Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 3: Trigonometric Functions Miscellaneous Exercise- NCERT solutions for class 11 maths chapter 3 miscellaneous exercise has mainly a mixed type of questions which are asked in previous exercises. Class 11 maths chapter 3 miscellaneous exercise has proof related and value finding questions. Class 11 maths chapter 3 miscellaneous exercise is a good source for Boards as well as competitive examinations. Many concepts will be used simultaneously in a question. Hence it will cover the concepts holistically. Below provided is the NCERT book class 11 maths chapter 3 miscellaneous exercise solutions with other exercises. Students can refer to these for more practice in this chapter.

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NCERT Solutions for Class 11 Maths Chapter 3: Trigonometric Functions Miscellaneous Exercise- Download Free PDF
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Benefits of NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercises
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NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions Miscellaneous Exercise

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Access NCERT solutions for class 11 maths chapter 3 trigonometric functions-Miscellaneous Exercise

Question:1 Prove that $\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+\cos\frac{3\pi }{13}+\cos\frac{5\pi }{13}=0$

Answer:

We know that

cos A+ cos B = $2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$

we use this in our problem

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{(\frac{3\pi }{13}+\frac{5\pi}{13})}{2}\cos\frac{(\frac{3\pi}{13}-\frac{5\pi }{13})}{2}$

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{4\pi }{13}\cos\frac{-\pi}{13}$ ( we know that cos(-x) = cos x )

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{4\pi }{13}\cos\frac{\pi}{13}$
$\small 2\cos\frac{\pi }{13}(\cos\frac{9\pi }{13}+\cos\frac{4\pi }{13})$
again use the above identity

$\small 2\cos\frac{\pi }{13}(2\cos(\frac{\frac{9\pi }{13}+\frac{4\pi }{13}}{2})\cos(\frac{\frac{9\pi }{13}-\frac{4\pi }{13}}{2})$
$\small 2\cos\frac{\pi }{13}2\cos\frac{\pi }{2}\cos\frac{5\pi }{26}$
we know that

$\small \cos\frac{\pi }{2}$ = 0
So,
$\small 2\cos\frac{\pi }{13}2\cos\frac{\pi }{2}\cos\frac{5\pi }{26}$ = 0 = R.H.S.

Question:2 Prove that $\small (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x )\cos x = 0$

Answer:

We know that
$sin3x=3\sin x - 4\sin^{3}x$
and
$cos3x=4\cos^{3}x - 3\cos x$
We use this in our problem
$\small (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x )\cos x$
= $(3\sin x - 4\sin^{3}x+ sin x) sinx$ + $(4\cos^{3}x - 3\cos x- cos x)cos x$
= (4sinx - 4 $\small \sin^{3}x$ )sinx + (4 $\small \cos^{3}x$ - 4cos x)cosx
now take the 4sinx common from 1st term and -4cosx from 2nd term
= 4 $\small \sin^{2}x$ (1 - $\small \sin^{2}x$ ) - 4 $\small \cos^{2}x$ (1 - $\small \cos^{2}x$ )
= 4 $\small \sin^{2}x$ $\small \cos^{2}x$ - 4 $\small \cos^{2}x$ $\small \sin^{2}x$ $\small \because \ \ \ \cos^{2}x = 1 - \sin^2x\\ and\\ \sin^{2}x = 1 -\cos^{2}x$
= 0 = R.H.S.

Question:3 Prove that $\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2} = 4 \cos^{2}\left ( \frac{x+y}{2} \right )$

Answer:

We know that $(a+b)^{2} = a^{2} + 2ab + b^{2}$
and
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
We use these two in our problem

$(\sin x-\sin y)^{2} = \sin^{2}x - 2\sin x\sin y + \sin^{2}y$
and
$(\cos x+\cos y)^{2} = \cos^{2}x + 2\cos x\cos y + \cos^{2}y$

$\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2}x + 2\cos x\cos y + \cos^{2}y$ + $\sin^{2}x - 2\sin x\sin y + \sin^{2}y$
= 1 + 2cosxcosy + 1 - 2sinxsiny $\left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )$
= 2 + 2(cosxcosy - sinxsiny)
= 2 + 2cos(x + y)
= 2(1 + cos(x + y) )
Now we can write
$cos(x + y) =2cos^{2}\frac{(x + y)}{2} - 1$ $\left ( \because \cos2x = 2cos^{2}x - 1 \ \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1\right )$

= $2(1 + 2cos^{2}\frac{(x + y)}{2} - 1)$
$=4cos^{2}\frac{(x + y)}{2}$

= R.H.S.

Question:4 Prove that $\small (\cos x-\cos y)^{2} + (\sin x - \sin y)^{2} = 4\sin^{2}\left ( \frac{x-y}{2} \right )$

Answer:

We know that $(a+b)^{2} = a^{2} + 2ab + b^{2}$
and
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
We use these two in our problem

$(\sin x-\sin y)^{2} = \sin^{2}x - 2\sin x\sin y + \sin^{2}y$
and
$(\cos x-\cos y)^{2} = \cos^{2}x - 2\cos x\cos y + \cos^{2}y$

$\small (\cos x - \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2}x - 2\cos x\cos y + \cos^{2}y$ + $\sin^{2}x - 2\sin x\sin y + \sin^{2}y$
= 1 - 2cosxcosy + 1 - 2sinxsiny $\left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )$
= 2 - 2(cosxcosy + sinxsiny)
= 2 - 2cos(x - y) $\small (\because \cos(x-y) =\cos x \cos y + \sin x \sin y)$
= 2(1 - cos(x - y) )
Now we can write
$cos(x + y) = 1 -2sin^{2}\frac{(x + y)}{2}$ $\left ( \because \cos2x = 1 - 2\sin^{2}x \ \Rightarrow \cos x = 1 - 2\sin^{2}\frac{x}{2} \right )$

$2(1 - cos(x - y) ) = 2(1 - ( 1 -2sin^{2}\frac{(x + y)}{2}))$

$= 4sin^{2}\frac{(x - y)}{2}$ = R.H.S.

Question:5 Prove that $\small \sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos2x \sin4x$

Answer:
we know that
$sinA + sinB =2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this identity in our problem
If we notice we need sin4x in our final result so it is better if we made a combination of sin7x and sin x , sin3x and sin5x tp get sin4x

$(sin7x + sinx) + (sin5x + sin3x) = 2\sin\frac{7x+x}{2}\cos\frac{7x-x}{2}$ $+2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2}$
$=$ $2\sin4x\cos3x + 2\sin4x\cos x$
take 2sin4x common
= 2sin4x(cos3x + cosx)
Now,
We know that
$cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this
$cos3x + cosx =2\cos\frac{3x+x}{2}\cos\frac{3x-x}{2}$
= $2\cos2x\cos x$
= 2sin4x( $2\cos2x\cos x$ )
= 4cosxcos2xsin4x = R.H.S.

Question:6 Prove that $\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)} = \tan6x$

Answer:

We know that

$sinA + sinB = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
and
$cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$

We use these two identities in our problem

sin7x + sin5x = $2\sin\frac{7x+5x}{2}\cos\frac{7x-5x}{2}$ = $2\sin6x\cos x$

sin 9x + sin 3x = $2\sin\frac{9x+3x}{2}\cos\frac{9x-3x}{2}$ = $2\sin6x\cos 3x$

cos 7x + cos5x = $2\cos\frac{7x+5x}{2}\cos\frac{7x-5x}{2}$ = $2\cos6x\cos x$

cos 9x + cos3x = $2\cos\frac{9x+3x}{2}\cos\frac{9x-3x}{2}$ = $2\cos6x\cos 3x$

$\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)}$ = $\small \frac{(2\sin 6x\cos x) + (2\sin6x \cos3x)}{(2\cos6x cos x) + (2\cos6x cos3x)}$

= $\small \frac{2\sin6x(\cos x + \cos3x)}{2\cos6x (cos x + cos3x)} = \tan6x$ = R.H.S. $\small \left ( \because \frac{\sin x}{\cos x} = \tan x\right )$

Question:7 Prove that $\small \sin3x + \sin2x - \sin x = 4\sin x \cos\frac{x}{2}\cos\frac{3x}{2}$

Answer:

We know that
$cosA + cosB = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
$sinA - sinB = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$

we use these identities
$sin3x - sinx = 2\cos\frac{3x+x}{2}\sin\frac{3x-x}{2}$

$= 2\cos2x\sin x$

sin2x + $2\cos2x\sin x$ = 2sinx cosx + $2\cos2x\sin x$
take 2 sinx common
$2sinx ( cosx + cos2x) = 2sinx(2\cos\frac{2x+x}{2}\cos\frac{2x-x}{2})$

$= 2sinx(2\cos\frac{3x}{2}\cos\frac{x}{2})$
$= 4sinx\cos\frac{3x}{2}\cos\frac{x}{2}$

= R.H.S.

Question:8 Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \tan x = - \frac{4}{3}$ , x in quadrant II

Answer:

tan x = $-\frac{4}{3}$
We know that ,
$\sec^{2}x = 1 + \tan^{2}x$
$= 1 +\left ( -\frac{4}{3} \right )^{2}$
$= 1 + \frac{16}{9}$ = $\frac{25}{9}$
$sec x = \sqrt{\frac{25}{9}}$ = $\pm\frac{5}{3}$
x lies in II quadrant thats why sec x is -ve
So,

$sec x =-\frac{5}{3}$
Now, $cos x = \frac{1}{\sec x}$ = $-\frac{3}{5}$
We know that,
$cos x = 2\cos^{2}\frac{x}{2}- 1$ ( $\because \cos2x = 2\cos^{2}x - 1 \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1$ )
$-\frac{3}{5}+ 1 = 2$ $\cos^{2}\frac{x}{2}$

= $\frac{-3+5}{5}$ = $2\cos^{2}\frac{x}{2}$

$\frac{2}{5}$ = $2\cos^{2}\frac{x}{2}$
$\cos^{2}\frac{x}{2}$ = $\frac{1}{5}$
$\cos\frac{x}{2}$ = $\sqrt{\frac{1}{5}}$ = $\pm\frac{1}{\sqrt5}$
x lies in II quadrant so value of $\cos\frac{x}{2}$ is +ve

$\cos\frac{x}{2}$ = $\frac{1}{\sqrt5} = \frac{\sqrt5}{5}$
we know that
$cos x =1 - 2\sin^{2}\frac{x}{2}$

$2\sin^{2}\frac{x}{2}$ = 1 - $(-\frac{3}{5})$ = $\frac{8}{5}$

$\sin^{2}\frac{x}{2} = \frac{4}{5}\\ \\=\sin\frac{x}{2} = \sqrt{ \frac{4}{5}} = \pm \frac{2}{\sqrt{5}}$
x lies in II quadrant So value of sin x is +ve

$\sin\frac{x}{2} = \frac{2}{\sqrt{}5} = \frac{2\sqrt5}{5}$

$\tan \frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{2\sqrt5}{5}}{\left ( \frac{\sqrt5}{5} \right )} = 2$

Question:9 Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \cos x = -\frac{1}{3}$ , x in quadrant III

Answer:

$\pi < x < \frac{3\pi}{2}\\ \\ \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$

We know that
cos x = $2\cos^{2}\frac{x}{2} - 1$
$2\cos^{2}\frac{x}{2} =$ cos x + 1
= $\left ( -\frac{1}{3} \right )$ + 1 = $\left ( \frac{-1+3}{3} \right )$ = $\frac{2}{3}$

$\cos\frac{x}{2} = \sqrt{ \frac{1}{3}} = \pm \frac{1}{\sqrt3}$

$\cos\frac{x}{2} = - \frac{1}{\sqrt3} = - \frac{\sqrt3}{3}$
Now,
we know that
cos x = $1 - 2\sin^{2}\frac{x}{2}$
$2\sin^{2}\frac{x}{2} = 1 - \cos x$
= 1 - $\left ( -\frac{1}{3} \right )$ = $\frac{3+1}{3}$ = $\frac{4}{3}$

$2\sin^{2}\frac{x}{2} = \frac{4}{3} \\ \\ \sin^{2}\frac{x}{2} = \frac{2}{3}\\ \sin\frac{x}{2} = \sqrt{ \frac{2}{3}} = \pm \sqrt{ \frac{2}{3}} = \frac{\sqrt6}{3}$
Because $\sin\frac{x}{2}$ is +ve in given quadrant

$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt6}{3}}{\frac{-\sqrt3}{3}} = - \sqrt2$

Question:10 Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \sin x = \frac{1}{4}$ ,x in quadrant II

Answer:

$\frac{\pi}{2} < x < \pi\\ \\ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ all functions are positive in this range
We know that
$\cos^{2}x = 1 - \sin^{2}x$
= 1 - $\left ( \frac{1}{4} \right )^{2}$ = $1 - \frac{1}{16}$ = $\frac{15}{16}$

cos x = $\sqrt\frac{15}{16} = \pm \frac{\sqrt15}{4} = - \frac{\sqrt15}{4}$ (cos x is -ve in II quadrant)

We know that
cosx = $2\cos^{2}\frac{x}{2} - 1$
$2\cos^{2}\frac{x}{2} = \cos x + 1 = -\frac{\sqrt15}{4} + 1 = \frac{-\sqrt15+4}{4}$

$\cos^{2}\frac{x}{2} = \frac{-\sqrt15+4}{8}$
$\cos\frac{x}{2} = \pm \sqrt\frac{-\sqrt15+4}{8} = \frac{\sqrt{-\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8-2\sqrt15}}{4}$ (because all functions are posititve in given range)

similarly,
cos x = $1-2\sin^{2}\frac{x}{2}$
$2\sin^{2}\frac{x}{2} = 1 - \cos x\\ \\ 2\sin^{2}\frac{x}{2} = 1 -\left (\frac{-\sqrt15}{4} \right ) = \frac{4+\sqrt15}{4}$
$\sin\frac{x}{2} = \pm \sqrt\frac{\sqrt15+4}{8} = \frac{\sqrt{\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8+2\sqrt15}}{4}$ (because all functions are posititve in given range)
$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt{8+2\sqrt15}}{4}}{\frac{\sqrt{8-2\sqrt15}}{4}} = \frac{{8+2\sqrt15}}{\sqrt{64 - 15\times4}} = \frac{{8+2\sqrt15}}{\sqrt{4}} = 4 + \sqrt15$

Benefits of NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercises

The NCERT syllabus class 11th maths chapter 3 miscellaneous exercise has some good questions which should be practised before examination to score good marks.
Class 11 maths chapter 3 miscellaneous exercises provides a very broad understanding of trigonometry which will be used in upcoming classes as well.
These class 11 maths chapter 3 miscellaneous exercises and the ideas used here are are useful to solve problems in Physics aswell.

Key Features of NCERT Class 11 Maths Ch 3 Miscellaneous Exercise Solutions

Comprehensive Approach: Miscellaneous exercise class 11 chapter 3 offers a thorough explanation of miscellaneous exercise topics, ensuring a well-rounded understanding.
Stepwise Clarity: Miscellaneous exercise chapter 3 class 11 Breaks down each problem with step-by-step solutions, aiding in a logical and clear progression of concepts.
Accessible Language: Class 11 chapter 3 maths miscellaneous solutions Presented in clear and concise language, making intricate mathematical concepts more accessible for students.
Variety of Problems: Class 11 maths miscellaneous exercise chapter 3 Provides a diverse range of problems, catering to different difficulty levels and scenarios.
Free PDF Access: Class 11 maths ch 3 miscellaneous exercise solutions provide free PDF solution.

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NCERT solutions of class 11 subject wise

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Frequently Asked Questions (FAQs)

1. How many questions are there in Miscellaneous exercise chapter 3 ?

There are 10 questions total in Miscellaneous exercise chapter 3

2. What is the application of Trigonometric Functions ?

It can be used to find the height of a point if the angle is known or vice versa.

3. Are questions from previous year papers asked for examination from this chapter ?

Yes, in the final exam of class 11 the questions are present from the previous year.

4. What is the difficulty level of problems asked from this unit ?

Moderate to difficult level pronlems are asked from this unit.

5. Can one avoid Miscellaneous exercise during preparation of the chapter?

No, as it has questions that covers most of the topics of the chapter.

6. What time it will take to complete all the problems of the miscellaneous exercise of this chapter for the first time ?

It takes around 3 to 4 hours to complete for the first time.

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