Class 11 Maths Exercise 3.3: Sum and Difference Formulas
Overview
This page provides comprehensive Class 11 Ch 3: Trigonometric Functions - Exercise 3.3 Practice. Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3-3. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.
Trigonometric Functions: Sum and Difference Formulas
About Class 11 Maths Trigonometric Functions Exercise 3.3
This NCERT Class 11 Maths exercise focuses on Trigonometric Functions: Sum and Difference Formulas 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 $\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} = -\frac{1}{2}$.
L.H.S. = $\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4}$
Substitute values: $\sin \frac{\pi}{6} = \frac{1}{2}, \cos \frac{\pi}{3} = \frac{1}{2}, \tan \frac{\pi}{4} = 1$.
$= (\frac{1}{2})^2 + (\frac{1}{2})^2 - (1)^2$
$= \frac{1}{4} + \frac{1}{4} - 1 = \frac{2}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$
= R.H.S.
L.H.S. = R.H.S.
Q2
00:00
Prove that $2\sin^2 \frac{\pi}{6} + \csc^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} = \frac{3}{2}$.
L.H.S. = $2(\frac{1}{2})^2 + \csc^2(\pi + \frac{\pi}{6}) (\frac{1}{2})^2$
We know $\csc(\pi + \theta) = -\csc \theta$. So $\csc(\frac{7\pi}{6}) = -\csc \frac{\pi}{6} = -2$.
$= 2(\frac{1}{4}) + (-2)^2 (\frac{1}{4})$
$= \frac{1}{2} + 4(\frac{1}{4}) = \frac{1}{2} + 1 = \frac{3}{2}$
= R.H.S.
L.H.S. = R.H.S.
Q3
00:00
Prove that $\cot^2 \frac{\pi}{6} + \csc \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6} = 6$.
L.H.S. = $(\sqrt{3})^2 + \csc(\pi - \frac{\pi}{6}) + 3(\frac{1}{\sqrt{3}})^2$
$= 3 + \csc \frac{\pi}{6} + 3(\frac{1}{3})$
$= 3 + 2 + 1 = 6$
= R.H.S.
L.H.S. = R.H.S.
Q4
00:00
Prove that $2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3} = 10$.
$\sin \frac{3\pi}{4} = \sin(\pi - \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
L.H.S. = $2(\frac{1}{\sqrt{2}})^2 + 2(\frac{1}{\sqrt{2}})^2 + 2(2)^2$
$= 2(\frac{1}{2}) + 2(\frac{1}{2}) + 2(4)$
$= 1 + 1 + 8 = 10$
= R.H.S.
L.H.S. = R.H.S.
Q5
00:00
Find the value of: (i) $\sin 75^\circ$ (ii) $\tan 15^\circ$.
(i) $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
(ii) $\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}$.
(i) $\frac{\sqrt{3}+1}{2\sqrt{2}}$, (ii) $2-\sqrt{3}$
Q6
00:00
Prove that $\cos(\frac{\pi}{4}-x)\cos(\frac{\pi}{4}-y) - \sin(\frac{\pi}{4}-x)\sin(\frac{\pi}{4}-y) = \sin(x+y)$.
Using $\cos A \cos B - \sin A \sin B = \cos(A+B)$.
L.H.S. = $\cos[(\frac{\pi}{4}-x) + (\frac{\pi}{4}-y)]$
$= \cos[\frac{\pi}{2} - (x+y)]$
$= \sin(x+y)$
= R.H.S.
L.H.S. = R.H.S.
Q7
00:00
Prove that $\frac{\tan(\frac{\pi}{4}+x)}{\tan(\frac{\pi}{4}-x)} = (\frac{1+\tan x}{1-\tan x})^2$.
$\tan(\frac{\pi}{4}+x) = \frac{1+\tan x}{1-\tan x}$
$\tan(\frac{\pi}{4}-x) = \frac{1-\tan x}{1+\tan x}$
L.H.S. = $\frac{(1+\tan x)/(1-\tan x)}{(1-\tan x)/(1+\tan x)} = \frac{1+\tan x}{1-\tan x} \times \frac{1+\tan x}{1-\tan x} = (\frac{1+\tan x}{1-\tan x})^2$
= R.H.S.
L.H.S. = R.H.S.
Q8
00:00
Prove that $\frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos(\frac{\pi}{2}+x)} = \cot^2 x$.
$\cos(\pi+x) = -\cos x$, $\cos(-x) = \cos x$.
$\sin(\pi-x) = \sin x$, $\cos(\frac{\pi}{2}+x) = -\sin x$.
L.H.S. = $\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} = \frac{-\cos^2 x}{-\sin^2 x} = \cot^2 x$
= R.H.S.
L.H.S. = R.H.S.
Q9
00:00
Prove that $\cos(\frac{3\pi}{2}+x)\cos(2\pi+x)[\cot(\frac{3\pi}{2}-x) + \cot(2\pi+x)] = 1$.
$\cos(\frac{3\pi}{2}+x) = \sin x$, $\cos(2\pi+x) = \cos x$.
$\cot(\frac{3\pi}{2}-x) = \tan x$, $\cot(2\pi+x) = \cot x$.
L.H.S. = $\sin x \cos x [\tan x + \cot x] = \sin x \cos x [\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}]$
$= \sin x \cos x [\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}] = 1$
= R.H.S.
L.H.S. = R.H.S.
Q10
00:00
Prove that $\sin(n+1)x \sin(n+2)x + \cos(n+1)x \cos(n+2)x = \cos x$.
Using $\cos A \cos B + \sin A \sin B = \cos(A-B)$.
Let $A = (n+1)x$ and $B = (n+2)x$.
L.H.S. = $\cos[(n+1)x - (n+2)x] = \cos[nx + x - nx - 2x] = \cos(-x)$
$= \cos x$
= R.H.S.
L.H.S. = R.H.S.
Q11
00:00
Prove that $\cos(\frac{3\pi}{4}+x) - \cos(\frac{3\pi}{4}-x) = -\sqrt{2}\sin x$.
Using $\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$.
L.H.S. = $-2\sin(\frac{3\pi}{4})\sin(x) = -2(\frac{1}{\sqrt{2}})\sin x = -\sqrt{2}\sin x$
= R.H.S.
L.H.S. = R.H.S.
Q12
00:00
Prove that $\sin^2 6x - \sin^2 4x = \sin 2x \sin 10x$.
Using $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$.
L.H.S. = $\sin(6x+4x)\sin(6x-4x) = \sin 10x \sin 2x$
= R.H.S.
L.H.S. = R.H.S.
Q13
00:00
Prove that $\cos^2 2x - \cos^2 6x = \sin 4x \sin 8x$.
Using $\cos^2 A - \cos^2 B = \sin(B+A)\sin(B-A)$.
L.H.S. = $\sin(6x+2x)\sin(6x-2x) = \sin 8x \sin 4x$
= R.H.S.
L.H.S. = R.H.S.
Q14
00:00
Prove that $\sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin 4x$.
L.H.S. = $(\sin 6x + \sin 2x) + 2\sin 4x$
$= 2\sin 4x \cos 2x + 2\sin 4x = 2\sin 4x (\cos 2x + 1)$
$= 2\sin 4x (2\cos^2 x) = 4\cos^2 x \sin 4x$
= R.H.S.
L.H.S. = R.H.S.
Q15
00:00
Prove that $\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x - \sin 3x)$.
L.H.S. = $\frac{\cos 4x}{\sin 4x} (2\sin 4x \cos x) = 2\cos 4x \cos x$.
R.H.S. = $\frac{\cos x}{\sin x} (2\cos 4x \sin x) = 2\cos 4x \cos x$.
L.H.S. = R.H.S.
L.H.S. = R.H.S.
Q16
00:00
Prove that $\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}$.
Numerator = $-2\sin 7x \sin 2x$.
Denominator = $2\cos 10x \sin 7x$.
L.H.S. = $\frac{-2\sin 7x \sin 2x}{2\cos 10x \sin 7x} = -\frac{\sin 2x}{\cos 10x}$
= R.H.S.
L.H.S. = R.H.S.
Q17
00:00
Prove that $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$.
Numerator = $2\sin 4x \cos x$.
Denominator = $2\cos 4x \cos x$.
L.H.S. = $\frac{2\sin 4x \cos x}{2\cos 4x \cos x} = \tan 4x$
= R.H.S.
L.H.S. = R.H.S.
Q18
00:00
Prove that $\frac{\sin x - \sin y}{\cos x + \cos y} = \tan \frac{x-y}{2}$.
Numerator = $2\cos\frac{x+y}{2}\sin\frac{x-y}{2}$.
Denominator = $2\cos\frac{x+y}{2}\cos\frac{x-y}{2}$.
L.H.S. = $\frac{\sin((x-y)/2)}{\cos((x-y)/2)} = \tan \frac{x-y}{2}$
= R.H.S.
L.H.S. = R.H.S.
Q19
00:00
Prove that $\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x$.
Numerator = $2\sin 2x \cos x$.
Denominator = $2\cos 2x \cos x$.
L.H.S. = $\frac{2\sin 2x \cos x}{2\cos 2x \cos x} = \tan 2x$
= R.H.S.
L.H.S. = R.H.S.
Q20
00:00
Prove that $\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2\sin x$.
Numerator = $-(\sin 3x - \sin x) = -(2\cos 2x \sin x)$.
Denominator = $-(\cos^2 x - \sin^2 x) = -\cos 2x$.
L.H.S. = $\frac{-2\cos 2x \sin x}{-\cos 2x} = 2\sin x$
= R.H.S.
L.H.S. = R.H.S.
Q21
00:00
Prove that $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$.
Group 4x and 2x terms.
Numerator = $(\cos 4x + \cos 2x) + \cos 3x = 2\cos 3x \cos x + \cos 3x = \cos 3x(2\cos x + 1)$.
Denominator = $(\sin 4x + \sin 2x) + \sin 3x = 2\sin 3x \cos x + \sin 3x = \sin 3x(2\cos x + 1)$.
L.H.S. = $\frac{\cos 3x}{\sin 3x} = \cot 3x$
= R.H.S.
L.H.S. = R.H.S.
Q22
00:00
Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$.
We know $\cot 3x = \cot(2x+x) = \frac{\cot 2x \cot x - 1}{\cot 2x + \cot x}$.
$\cot 3x(\cot 2x + \cot x) = \cot 2x \cot x - 1$
$\cot 3x \cot 2x + \cot 3x \cot x = \cot 2x \cot x - 1$
Rearranging: $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$
= R.H.S.
L.H.S. = R.H.S.
Q23
00:00
Prove that $\tan 4x = \frac{4\tan x(1-\tan^2 x)}{1-6\tan^2 x + \tan^4 x}$.
$\tan 4x = \tan 2(2x) = \frac{2\tan 2x}{1-\tan^2 2x}$.
Substitute $\tan 2x = \frac{2\tan x}{1-\tan^2 x}$.
Numerator = $2(\frac{2\tan x}{1-\tan^2 x}) = \frac{4\tan x}{1-\tan^2 x}$.
Denominator = $1 - (\frac{2\tan x}{1-\tan^2 x})^2 = 1 - \frac{4\tan^2 x}{(1-\tan^2 x)^2} = \frac{(1-\tan^2 x)^2 - 4\tan^2 x}{(1-\tan^2 x)^2}$.
Simplifying: $\frac{4\tan x(1-\tan^2 x)}{1 + \tan^4 x - 2\tan^2 x - 4\tan^2 x} = \frac{4\tan x(1-\tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}$
= R.H.S.
L.H.S. = R.H.S.
Q24
00:00
Prove that $\cos 4x = 1 - 8\sin^2 x \cos^2 x$.
$\cos 4x = \cos 2(2x) = 1 - 2\sin^2 2x$
$= 1 - 2(2\sin x \cos x)^2$
$= 1 - 2(4\sin^2 x \cos^2 x)$
$= 1 - 8\sin^2 x \cos^2 x$
= R.H.S.
L.H.S. = R.H.S.
Q25
00:00
Prove that $\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$.
$\cos 6x = \cos 3(2x) = 4\cos^3 2x - 3\cos 2x$
Substitute $\cos 2x = 2\cos^2 x - 1$.
$= 4(2\cos^2 x - 1)^3 - 3(2\cos^2 x - 1)$
$= 4(8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1) - 6\cos^2 x + 3$
$= 32\cos^6 x - 48\cos^4 x + 24\cos^2 x - 4 - 6\cos^2 x + 3$
$= 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$
= R.H.S.
L.H.S. = R.H.S.