Class 11 Maths Trigonometric Functions: Miscellaneous Exercise

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Overview

This page provides comprehensive Class 11 Ch 3: Trigonometric Functions - Miscellaneous Exercise Practice. Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.

Trigonometric Functions: Chapter Review

About Class 11 Maths Trigonometric Functions Miscellaneous Exercise

This NCERT Class 11 Maths exercise focuses on Trigonometric Functions: Chapter Review from the chapter Trigonometric Functions. Revise the relevant definitions, standard results, notation, and worked examples before attempting the questions.

Attempt each question independently and write the reasoning in complete steps. Check symbols, substitutions, diagrams, and final conditions carefully, then use the step-by-step solution to identify exactly what needs revision.

Q1
00:00
Prove that $2 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0$.
L.H.S. = $2 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$
Using $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:
$2 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} = \cos(\frac{10\pi}{13}) + \cos(\frac{8\pi}{13})$
L.H.S. = $\cos \frac{10\pi}{13} + \cos \frac{8\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$
Since $\cos(\pi - x) = -\cos x$:
$\cos \frac{10\pi}{13} = \cos(\pi - \frac{3\pi}{13}) = -\cos \frac{3\pi}{13}$
$\cos \frac{8\pi}{13} = \cos(\pi - \frac{5\pi}{13}) = -\cos \frac{5\pi}{13}$
L.H.S. = $-\cos \frac{3\pi}{13} - \cos \frac{5\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0$
= R.H.S.
L.H.S. = R.H.S.
Q2
00:00
Prove that $(\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0$.
L.H.S. = $\sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x$
Rearranging terms: $(\cos 3x \cos x + \sin 3x \sin x) - (\cos^2 x - \sin^2 x)$
Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and $\cos 2A = \cos^2 A - \sin^2 A$:
$= \cos(3x - x) - \cos 2x$
$= \cos 2x - \cos 2x = 0$
= R.H.S.
L.H.S. = R.H.S.
Q3
00:00
Prove that $(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4 \cos^2 \frac{x+y}{2}$.
Expand the squares:
L.H.S. = $(\cos^2 x + \cos^2 y + 2\cos x \cos y) + (\sin^2 x + \sin^2 y - 2\sin x \sin y)$
Group terms: $(\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2(\cos x \cos y - \sin x \sin y)$
$= 1 + 1 + 2\cos(x+y) = 2 + 2\cos(x+y)$
$= 2(1 + \cos(x+y))$
Using $1 + \cos A = 2\cos^2 \frac{A}{2}$:
$= 2(2\cos^2 \frac{x+y}{2}) = 4\cos^2 \frac{x+y}{2}$
= R.H.S.
L.H.S. = R.H.S.
Q4
00:00
Prove that $(\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 4 \sin^2 \frac{x-y}{2}$.
Expand the squares:
L.H.S. = $(\cos^2 x + \cos^2 y - 2\cos x \cos y) + (\sin^2 x + \sin^2 y - 2\sin x \sin y)$
$= (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) - 2(\cos x \cos y + \sin x \sin y)$
$= 1 + 1 - 2\cos(x-y) = 2 - 2\cos(x-y)$
$= 2(1 - \cos(x-y))$
Using $1 - \cos A = 2\sin^2 \frac{A}{2}$:
$= 2(2\sin^2 \frac{x-y}{2}) = 4\sin^2 \frac{x-y}{2}$
= R.H.S.
L.H.S. = R.H.S.
Q5
00:00
Prove that $\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x$.
Group terms: $(\sin 7x + \sin x) + (\sin 5x + \sin 3x)$
Using $\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$:
$= 2\sin 4x \cos 3x + 2\sin 4x \cos x$
Factor out $2\sin 4x$:
$= 2\sin 4x (\cos 3x + \cos x)$
Using $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$:
$= 2\sin 4x (2\cos 2x \cos x)$
$= 4 \cos x \cos 2x \sin 4x$
= R.H.S.
L.H.S. = R.H.S.
Q6
00:00
Prove that $\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x$.
Apply sum-to-product formulas:
Numerator: $2\sin 6x \cos x + 2\sin 6x \cos 3x = 2\sin 6x(\cos x + \cos 3x)$
Denominator: $2\cos 6x \cos x + 2\cos 6x \cos 3x = 2\cos 6x(\cos x + \cos 3x)$
L.H.S. = $\frac{2\sin 6x(\cos x + \cos 3x)}{2\cos 6x(\cos x + \cos 3x)}$
$= \frac{\sin 6x}{\cos 6x} = \tan 6x$
= R.H.S.
L.H.S. = R.H.S.
Q7
00:00
Prove that $\sin 3x + \sin 2x - \sin x = 4\sin x \cos \frac{x}{2} \cos \frac{3x}{2}$.
Group terms: $(\sin 3x - \sin x) + \sin 2x$
Using $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$:
$= 2\cos 2x \sin x + 2\sin x \cos x$
Factor out $2\sin x$:
$= 2\sin x (\cos 2x + \cos x)$
Using $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$:
$= 2\sin x (2\cos \frac{3x}{2} \cos \frac{x}{2})$
$= 4\sin x \cos \frac{x}{2} \cos \frac{3x}{2}$
= R.H.S.
L.H.S. = R.H.S.
Q8
00:00
Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ if $\tan x = -\frac{4}{3}$, $x$ in quadrant II.
Since $x$ is in quadrant II, $\frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$. Thus, $\sin \frac{x}{2}, \cos \frac{x}{2}, \tan \frac{x}{2}$ are all positive.
Given $\tan x = -\frac{4}{3}$. $\sec^2 x = 1 + \tan^2 x = 1 + \frac{16}{9} = \frac{25}{9}$.
Since $x$ is in quadrant II, $\sec x = -\frac{5}{3} \Rightarrow \cos x = -\frac{3}{5}$.
$\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - (-3/5)}{2}} = \sqrt{\frac{8/5}{2}} = \frac{2}{\sqrt{5}}$.
$\cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 + (-3/5)}{2}} = \sqrt{\frac{2/5}{2}} = \frac{1}{\sqrt{5}}$.
$\tan \frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2$.
$\sin \frac{x}{2} = \frac{2}{\sqrt{5}}, \cos \frac{x}{2} = \frac{1}{\sqrt{5}}, \tan \frac{x}{2} = 2$
Q9
00:00
Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ if $\cos x = -\frac{1}{3}$, $x$ in quadrant III.
Since $x$ is in quadrant III, $\pi < x < \frac{3\pi}{2} \Rightarrow \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$. Thus, $\sin \frac{x}{2} > 0$, while $\cos \frac{x}{2} < 0$ and $\tan \frac{x}{2} < 0$.
Given $\cos x = -\frac{1}{3}$.
$\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - (-1/3)}{2}} = \sqrt{\frac{4/3}{2}} = \sqrt{\frac{2}{3}}$.
$\cos \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}} = -\sqrt{\frac{1 + (-1/3)}{2}} = -\sqrt{\frac{2/3}{2}} = -\frac{1}{\sqrt{3}}$.
$\tan \frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)} = \frac{\sqrt{2/3}}{-1/\sqrt{3}} = -\sqrt{2}$.
$\sin \frac{x}{2} = \sqrt{\frac{2}{3}}, \cos \frac{x}{2} = -\frac{1}{\sqrt{3}}, \tan \frac{x}{2} = -\sqrt{2}$
Q10
00:00
Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ if $\sin x = \frac{1}{4}$, $x$ in quadrant II.
Since $x$ is in quadrant II, $\frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$. Thus, all are positive.
Given $\sin x = \frac{1}{4}$. Since $x$ is in quadrant II, $\cos x < 0$.
$\cos x = -\sqrt{1 - \sin^2 x} = -\sqrt{1 - \frac{1}{16}} = -\frac{\sqrt{15}}{4}$.
$\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 + \sqrt{15}/4}{2}} = \sqrt{\frac{4+\sqrt{15}}{8}}$.
$\cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 - \sqrt{15}/4}{2}} = \sqrt{\frac{4-\sqrt{15}}{8}}$.
$\tan \frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)} = \sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}} = 4+\sqrt{15}$.
$\sin \frac{x}{2} = \sqrt{\frac{4+\sqrt{15}}{8}}, \cos \frac{x}{2} = \sqrt{\frac{4-\sqrt{15}}{8}}, \tan \frac{x}{2} = 4+\sqrt{15}$
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