Class 11: Values of Trigonometric Functions at Multiples and Sub multiples of an Angle – Exercise 9.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Prove the following identities $(1-25)$

Question 1: $\sqrt{\frac{1 - \cos 2x}{1+\cos 2x} }$ $= \tan x$

Answer:

LHS $=$ $\sqrt{\frac{1 - \cos 2x}{1+\cos 2x} }$ $=$ $\sqrt{\frac{2 \sin^2 x}{2 \cos^2 x} }$ $=$ $\frac{\sin x}{\cos x}$ $= \tan x =$ RHS. Hence proved.

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Question 2: $\frac{\sin 2x}{1- \cos 2x}$ $= \cot x$

Answer:

LHS $=$ $\frac{\sin 2x}{1- \cos 2x}$ $=$ $\frac{2 \sin x \cos x}{2 \sin^2 x}$ $=$ $\frac{\cos x}{\sin x}$ $= \cot x =$ RHS. Hence proved.

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Question 3: $\frac{\sin 2x}{1+ \cos 2x}$ $= \tan x$

Answer:

LHS $=$ $\frac{\sin 2x}{1+ \cos 2x}$ $=$ $\frac{2 \sin x \cos x}{2 \cos^2 x}$ $=$ $\frac{\sin x}{\cos x}$ $= \tan x =$ RHS. Hence proved.

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Question 4: $\sqrt{2 + \sqrt{2 + 2 \cos 4x} } = 2 \cos x, 0 < x < \frac{\pi}{4}$

Answer:

LHS $= \sqrt{2 + \sqrt{2 + 2 \cos 4x} }$

$= \sqrt{2 + \sqrt{2(1 + \cos 4x)} }$

$= \sqrt{2 + \sqrt{2 \times 2 \cos^2 2x} }$

$= \sqrt{2 + 2 \cos 2x }$

$= \sqrt{2(1 + \cos 2x) }$

$= \sqrt{2 \times 2 \cos^2 x }$

$= 2 \cos x =$ RHS. Hence proved.

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Question 5: $\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x}$ $= \tan x$

Answer:

LHS $=$ $\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x}$

$=$ $\frac{2\sin^2 x + 2 \sin x \cos x}{2 \cos^2 x+ 2 \sin x \cos x}$

$=$ $\frac{2\sin x (\sin x + \cos x)}{2\cos x (\sin x + \cos x)}$

$= \tan x =$ RHS. Hence proved.

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Question 6: $\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x}$ $= \tan x$

Answer:

LHS $=$ $\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x}$

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

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

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

$=$ $\frac{\sin x }{\cos x }$

$= \tan x =$ RHS. Hence proved.

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Question 7: $\frac{\cos 2x}{1+ \sin 2x}$ $= \tan \Big($ $\frac{\pi}{4}$ $- x \Big)$

Answer:

LHS $=$ $\frac{\cos 2x}{1+ \sin 2x}$

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

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

$=$ $\frac{\cos x - \sin x }{\cos x + \sin x}$

$=$ $\frac{1 - \tan x}{1+ \tan x}$

$=$ $\frac{\tan \frac{\pi}{4} - \tan x}{1+ \tan \frac{\pi}{4} \tan x}$

$= \tan \Big($ $\frac{\pi}{4}$ $- x \Big)$ RHS. Hence proved.

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Question 8: $\frac{\cos x}{1- \sin x}$ $= \tan \Big($ $\frac{\pi}{4}$ $+$ $\frac{x}{2}$ $\Big)$

Answer:

LHS $=$ $\frac{\cos x}{1 - \sin x}$

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

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

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

$=$ $\frac{1 + \tan \frac{x}{2}}{1- \tan \frac{x}{2}}$

$=$ $\frac{\tan \frac{\pi}{4} + \tan \frac{x}{2}}{1- \tan \frac{\pi}{4} \tan \frac{x}{2}}$

$= \tan \Big($ $\frac{\pi}{4}$ $+$ $\frac{x}{2}$ $\Big)$ RHS. Hence proved.

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Question 9: $\cos^2$ $\frac{\pi}{8}$ $+ \cos^2$ $\frac{3\pi}{8}$ $+\cos^2$ $\frac{5\pi}{8}$ $+\cos^2$ $\frac{7\pi}{8}$ $= 2$

Answer:

LHS $= \cos^2$ $\frac{\pi}{8}$ $+ \cos^2$ $\frac{3\pi}{8}$ $+\cos^2$ $\frac{5\pi}{8}$ $+\cos^2$ $\frac{7\pi}{8}$

$= \cos^2$ $\frac{\pi}{8}$ $+ \cos^2$ $\frac{3\pi}{8}$ $+\cos^2 \big( \pi -$ $\frac{3\pi}{8}$ $\big) +\cos^2\big( \pi -$ $\frac{\pi}{8}$ $\big)$

$= \cos^2$ $\frac{\pi}{8}$ $+ \cos^2$ $\frac{3\pi}{8}$ $+\cos^2$ $\frac{3\pi}{8}$ $+\cos^2$ $\frac{\pi}{8}$

$= 2 \Big( \cos^2$ $\frac{\pi}{8}$ $+ \cos^2$ $\frac{3\pi}{8}$ $\Big)$

$= 2 \Big( \cos^2$ $\frac{\pi}{8}$ $+ \cos^2 \big($ $\frac{\pi}{2}$ $-$ $\frac{\pi}{8}$ $\big) \Big)$

$= 2 \Big( \cos^2$ $\frac{\pi}{8}$ $+ \sin^2$ $\frac{\pi}{8}$ $\Big)$

$= 2 =$ RHS. Hence proved.

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Question 10: $\sin^2$ $\frac{\pi}{8}$ $+ \sin^2$ $\frac{3\pi}{8}$ $+ \sin^2$ $\frac{5\pi}{8}$ $+\sin^2$ $\frac{7\pi}{8}$ $= 2$

Answer:

LHS $= \sin^2$ $\frac{\pi}{8}$ $+ \sin^2$ $\frac{3\pi}{8}$ $+ \sin^2$ $\frac{5\pi}{8}$ $+\sin^2$ $\frac{7\pi}{8}$

$= \sin^2$ $\frac{\pi}{8}$ $+ \sin^2$ $\big($ $\frac{\pi}{2} - \frac{\pi}{8}$ $\big) + \sin^2 \big( \pi -$ $\frac{3\pi}{8}$ $\big) +\sin^2$ $\big($ $\frac{\pi}{2} - \frac{\pi}{8}$ $\big)$

$= \sin^2$ $\frac{\pi}{8}$ $+ \cos^2$ $\frac{\pi}{8}$ $+ \sin^2$ $\frac{3\pi}{8}$ $+\sin^2$ $\frac{\pi}{8}$

$= 1+ \sin^2 \Big($ $\frac{\pi}{2}$ $-$ $\frac{\pi}{8}$ $\Big) + \sin^2$ $\frac{\pi}{8}$

$= 1+ \cos^2$ $\frac{\pi}{8}$ $+ \sin^2$ $\frac{\pi}{8}$

$= 2 =$ RHS. Hence proved.

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Question 11: $(\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 = 4 \cos^2 \Big($ $\frac{\alpha - \beta}{2}$ $\Big)$

Answer:

LHS $= (\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2$

$= \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta + \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta$

$= 2 + 2 \cos \alpha \cos \beta + 2 \sin \alpha \sin \beta$

$= 2 + 2 \cos ( \alpha - \beta)$

$= 2 [ 1 + \cos ( \alpha - \beta) ]$

$= 2 \times 2 \cos^2 \Big($ $\frac{\alpha - \beta}{2}$ $\Big)$

$= 4 \cos^2 \Big($ $\frac{\alpha - \beta}{2}$ $\Big) =$ RHS. Hence proved.

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Question 12: $\sin^2 \Big($ $\frac{\pi}{8}$ $+$ $\frac{x}{2}$ $\Big) - \sin^2 \Big($ $\frac{\pi}{8}$ $-$ $\frac{x}{2}$ $\Big) =$ $\frac{1}{\sqrt{2}}$ $\sin x$

Answer:

LHS $= \sin^2 \Big($ $\frac{\pi}{8}$ $+$ $\frac{x}{2}$ $\Big) - \sin^2 \Big($ $\frac{\pi}{8}$ $-$ $\frac{x}{2}$ $\Big)$

$=$ $\frac{1}{2}$ $\Big[ 1 - \cos 2 \Big($ $\frac{\pi}{8}$ $+$ $\frac{x}{2}$ $\Big) \Big] -$ $\frac{1}{2}$ $\Big[ 1 - \cos 2\Big($ $\frac{\pi}{8}$ $-$ $\frac{x}{2}$ $\Big) \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos 2\Big($ $\frac{\pi}{8}$ $-$ $\frac{x}{2}$ $\Big) - \cos 2 \Big($ $\frac{\pi}{8}$ $+$ $\frac{x}{2}$ $\Big) \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos \Big($ $\frac{\pi}{4}$ $- x \Big) - \cos \Big($ $\frac{\pi}{4}$ $+ x \Big) \Big]$

$=$ $\frac{1}{2}$ $\Big[ -2 \sin \Big($ $\frac{\frac{\pi}{4} - x \frac{\pi}{4}+x}{2}$ $\Big) \sin \Big($ $\frac{\frac{\pi}{4}-x-\frac{\pi}{4}-x}{2}$ $\Big) \Big]$

$=$ $\frac{1}{2}$ $\Big[ - 2 \sin$ $\frac{\pi}{4}$ $\sin ( -x) \Big]$

$=$ $\frac{1}{\sqrt{2}}$ $\sin x =$ RHS. Hence proved.

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Question 13: $1 + \cos^2 2x = 2 (\cos^4 x + \sin^4 x)$

Answer:

LHS $= 1 + \cos^2 2x$

$= 1 + ( \cos^2 x - \sin^2 x)^2$

$= 1 + \cos^4 x + \sin^4 x - 2 \cos^2 x \sin^2 x$

$= ( \cos^2 x + \sin^2 x)^2 + \cos^4 x + \sin^4 x - 2 \cos^2 x \sin^2 x$

$= \cos^4 x + \sin^4 x + 2 \cos^2 x \sin^2 x + \cos^4 x + \sin^4 x - 2 \cos^2 x \sin^2 x$

$= 2 ( \cos^4 x + \sin^4 x) =$ RHS. Hence proved.

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Question 14: $\cos^3 2x + 3 \cos 2x = 4 ( \cos^6 x - \sin^6 x)$

Answer:

RHS $= 4 ( \cos^6 x - \sin^6 x)$

$= 4 [ (\cos^2 x)^3 - (\sin^2 x)^3 ]$

$= 4 [ (\cos^2 x - \sin^2 x ) ( \cos^4 x + \sin^4 x + \sin^2 x \cos^2 x )]$

$= 4 \cos 2x ( \cos^4 x + \sin^4 x + \sin^2 x \cos^2 x )$

$= 4 \cos 2x [ (\cos^2 x - \sin^2 x)^2 + 2 \cos^2 x \sin^2 x+ \sin^2 x \cos^2 x ]$

$= 4 \cos 2x [ \cos^2 2x + 3 \sin^2 x \cos^2 x ]$

$= 4 \cos 2x \Big[ \cos^2 2x + 3 \Big($ $\frac{1 - \cos 2x}{2}$ $\Big) \Big($ $\frac{1 + \cos 2x}{2}$ $\Big) \Big ]$

$= 4 \cos 2x \Big[ \cos^2 2x +$ $\frac{3}{4}$ $(1 - \cos^2 2x) \Big]$

$= \cos 2x [ 4\cos^2 2x + 3 (1 - \cos^2 2x) ]$

$= \cos 2x [ \cos^2 2x + 3 ]$

$= \cos^3 2x + 3 \cos 2x =$ LHS. Hence proved.

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Question 15: $(\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0$

Answer:

LHS $= (\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x$

$= 2 \sin 2x \cos x \sin x - 2 \sin 2x \sin x \cos x$

$= 2 \sin 2x \cos x \sin x - 2 \sin 2x \cos x \sin x$

$= 0 =$ RHS. Hence proved.

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Question 16: $\cos^2 \Big($ $\frac{\pi}{4}$ $- x \Big) - \sin^2 \Big($ $\frac{\pi}{4}$ $- x \Big) = \sin 2x$

Answer:

LHS $= \cos^2 \Big($ $\frac{\pi}{4}$ $- x \Big) - \sin^2 \Big($ $\frac{\pi}{4}$ $- x \Big)$

Since $\cos 2x = \cos^2 x - \sin^2 x$

$= \cos 2 ($ $\frac{\pi}{4}$ $- x)$

$= \cos ($ $\frac{\pi}{2}$ $- 2x)$

$= \sin 2x =$ RHS. Hence proved.

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Question 17: $\cos 4x = 1 - 8 \cos^2 x + 8 \cos ^4 x$

Answer:

LHS $= \cos 4x$

$= \cos 2 \times 2x$

$= 2 \cos^2 2x - 1$

$= 2 [ ( 2 \cos^2 x - 1)^2 ] - 1$

$= 2 ( 4 \cos^4 x + 1 - 4 \cos^2 x ) - 1$

$= 8 \cos^4 x - 8 \cos^2 x + 1$

$=$ RHS. Hence proved.

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Question 18: $\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x$

Answer:

LHS $= \sin 4x$

$= \sin 2 .2x$

$= 2 \sin 2x \cos 2x$

$= 2 ( 2 \sin x \cos x)(\cos^2 x - \sin^2 x)$

$= 4 \sin x \cos^3 x - 4 \sin^3 x \cos x$

$=$ RHS. Hence proved.

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Question 19: $3(\sin x - \cos x)^4 + 6 (\sin x + \cos x)^2 + 4 (\sin^6 x + \cos^6 x) = 13$

Answer:

LHS $= 3(\sin x - \cos x)^4 + 6 (\sin x + \cos x)^2 + 4 (\sin^6 x + \cos^6 x)$

$= 3 ( \sin^4 x - 4 \sin^3x \cos x + 6 \sin^2 x \cos^2 x - 4 \sin x \cos^3 x + \cos^4 x) + 6 ( \sin^2 x + 2 \sin x \cos x + \cos^2 x) + 4 ( \sin^6 x + \cos^6 x)$

$= 3 [ \sin^4 x + \cos^4 x - 4 \sin x \cos x ( \sin^2 x + \cos^2 ) + 6 \sin^2 x \cos^2 x] + 6 (1 + 2 \sin x \cos x ) + 4[ (\cos^2 x +\sin^2 x)( \cos^4 x - \cos^2 x \sin^2 x+ \sin^4 x) ]$

$= 3 [ \sin^4 x + \cos^4 x - 4 \sin x \cos x + 6 \sin^2 x \cos^2 x] + 6 (1 + 2 \sin x \cos x ) + 4( \cos^4 x - \cos^2 x \sin^2 x+ \sin^4 x )$

$= 3 \sin^4 x + 3 \cos^4 x - 12 \sin x \cos x + 18 \sin^2 x \cos^2 x + 6 + 12 \sin x \cos x + 4 \cos^4 x - 4 \cos^2 x \sin^2 x+ 4\sin^4 x$

$= 7 \sin^4 x + 7 \cos^4 x + 14 \sin^2 x \cos^2 x + 6$

$= 7 (\sin^4 x + \cos^4 x + 2 \sin^2 x \cos^2 x) + 6$

$= 7 ( \sin^2 x + \cos^2 x )^2 + 6$

$= 7 + 6 = 13 =$ RHS. Hence proved.

$\\$

Question 20: $2 (\sin^6 x + \cos^6 x) - 3 (\sin^4 x + \cos^4 x) + 1 = 0$

Answer:

LHS $= 2 (\sin^6 x + \cos^6 x) - 3 (\sin^4 x + \cos^4 x) + 1$

$= 2 [ (\sin^3 x)^2 + (\cos^3 x)^2 ] - 3 ( \sin^4 x + \cos^4 x ) + 1$

$= 2 [ (\sin^2 x + \cos^2 x) ( \sin^4 x - \sin^2 x \cos^2 x + \cos^4x) ] - 3 ( \sin^4 x + \cos^4 x ) + 1$

$= 2 [ \sin^4 x - \sin^2 x \cos^2 x + \cos^4x ] - 3 ( \sin^4 x + \cos^4 x ) + 1$

$= 2 \sin^4 x - 2 \sin^2 x \cos^2 x + 2\cos^4x - 3 \sin^4 x -3 \cos^4 x + 1$

$= - \sin^4 x - \cos^4 x - 2 \sin^2 x \cos^2 x + 1$

$= - (\sin^2 x + \cos^2 x)^2 + 1$

$= -1 + 1 = 0$ = RHS. Hence proved.

$\\$

Question 21: $\cos^6 x - \sin^6 = \cos 2x \Big( 1 -$ $\frac{1}{4}$ $\sin^2 2x \Big)$

Answer:

LHS $= \cos^6 x - \sin^6 x$

$= (\cos^2 x)^3 - (\sin^2 x)^3$

$= (\cos^2 x - \sin^2 x) \Big[ \cos^4 x + \sin^2 x \cos^2 x + \sin^4 x \Big]$

$= \cos 2x \Big[ \cos^4 x + 2\sin^2 x \cos^2 x + \sin^4 x - \sin^2 x \cos^2 x \Big]$

$= \cos 2x \Big[ (\cos^2 x + \sin^2 x)^2 - \sin^2 x \cos^2 x \Big]$

$= \cos 2x \Big[ 1 - \sin^2 x \cos^2 x \Big]$

$= \cos 2x \Big[ 1 -$ $\frac{1}{4}$ $( 2\sin x \cos x)^2 \Big]$

$= \cos 2x \Big[ 1 -$ $\frac{1}{4}$ $\sin^ 2x \Big] =$ RHS. Hence proved.

$\\$

Question 22: $\tan \Big( \frac{\pi}{4} + x \Big) + \tan \Big( \frac{\pi}{4} - x \Big) = 2 \sec 2x$

Answer:

LHS $= \tan \Big( \frac{\pi}{4} + x \Big) + \tan \Big( \frac{\pi}{4} - x \Big)$

$=$ $\frac{\tan \frac{\pi}{4} + \tan x}{1 - \tan \frac{\pi}{4} \tan x}$ $+$ $\frac{\tan \frac{\pi}{4} 1 \tan x}{1 + \tan \frac{\pi}{4} \tan x}$

$=$ $\frac{1 + \tan x}{1 - \tan x} + \frac{1 - \tan x}{1 + \tan x}$

$=$ $\frac{(1 + \tan x)^2 - (1 - \tan x)^2}{1 - \tan^2 x}$

$=$ $\frac{1 + \tan^2 x + 2 \tan x + 1 + \tan^2 x - 2 \tan x}{1 - \tan^2 x}$

$=$ $\frac{2(1 + \tan^2 x )}{1 - \tan^2 x}$

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

$=$ $\frac{2}{\cos^2 x - \sin^2 x}$

$=$ $\frac{2}{\cos 2x}$

$= 2 \sec 2x =$ RHS. Hence proved.

$\\$

Question 23: $\cot^2 x - \tan^2 x = 4 \cot 2x \mathrm{cosec} 2x$

Answer:

LHS $= \cot^2 x - \tan^2 x$

$=$ $\frac{\cos^2 x}{\sin^2 x}$ $-$ $\frac{\sin^2 x}{\cos^2 x}$

$=$ $\frac{\cos^4 x - \sin^4 x}{\sin^2 x \cos^2 x}$

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

$=$ $\frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x}$

$=$ $\frac{4 \cos 2x}{(2\sin x \cos x)^2}$

$=$ $\frac{4 \cos 2x}{\sin^2 2x}$

$= 4 \cot 2x \ \mathrm{cosec} 2x =$ RHS. Hence proved.

$\\$

Question 24: $\cos 4x - \cos 4 \alpha = 8( \cos x - \cos \alpha) ( \cos x + \cos \alpha) (\cos x - \sin \alpha)(\cos x + \sin \alpha)$

Answer:

LHS $= \cos 4x - \cos 4 \alpha$

$= 2 \cos^2 2x - 2 \cos^2 2\alpha$

$= 2 ( \cos 2x + \cos 2\alpha)(\cos 2x - \cos 2 \alpha)$

$= 2 ( 2 \cos^2 x - 1 + 1 - 2 \sin^2 \alpha)( 2 \cos^2 x - 1 - 2 \cos^2 \alpha + 1)$

$= 8 (\cos^2 x - \sin^2 \alpha) ( \cos^2 x - \cos^2 \alpha)$

$= 8 ( \cos x - \cos \alpha)(\cos x + \cos \alpha) ( \cos x + \cos\alpha)(\cos x - \cos \alpha)$

$=$ RHS. Hence proved.

$\\$

Question 25: $\sin 3x + \sin 2x - \sin x = 4 \sin x \cos$ $\frac{x}{2}$ $\cos$ $\frac{3x}{2}$

Answer:

LHS $= \sin 3x + \sin 2x - \sin x$

$= ( \sin 3x - \sin x) + \sin 2x$

$= 2 \cos 2x \sin x + 2 \sin x \cos x$

$= 2 \sin x ( \cos 2x + \cos x)$

$= 2 \sin x \Big( 2 \cos$ $\frac{3x}{2}$ $\cos$ $\frac{x}{2}$ $\Big)$

$= 4 \sin x \cos$ $\frac{3x}{2}$ $\cos$ $\frac{x}{2}$ $=$ RHS. Hence proved.

$\\$

Question 26: Prove that: $\tan 82$ $\frac{1}{2}$ $= (\sqrt{3}+\sqrt{2})(\sqrt{2}+1) = \sqrt{2}+ \sqrt{3}+ \sqrt{4}+ \sqrt{6}$

Answer:

We know $\tan$ $\frac{x}{2}$ $=$ $\frac{\sin x}{1+\cos x}$

$\Rightarrow \tan 7$ $\frac{1}{2}$ $^o =$ $\frac{\sin 15^o}{1+\cos 15^o}$

$=$ $\frac{\sin (45^o-30^o)}{1+\cos (45^o-30^o)}$

$=$ $\frac{\sin 45^o \cos 30^o - \sin 30^o \cos 45^o}{1 + 45^o \cos 30^o + \sin 45^o \sin 30^o}$

$=$ $\frac{\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} }$

$=$ $\frac{\frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2 \sqrt{2}} }{1 + \frac{\sqrt{3}}{2\sqrt{2}} \times \frac{1}{2 \sqrt{2}} }$

$=$ $\frac{\sqrt{3}-1}{2\sqrt{2} + \sqrt{3} + 1}$

$\therefore \tan 7$ $\frac{1}{2}$ $=$ $\frac{\sqrt{3}-1}{2\sqrt{2} + \sqrt{3} + 1}$

$\therefore \cot 7$ $\frac{1}{2}$ $=$ $\frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3}-1}$

$\Rightarrow \cot (90 - 82$ $\frac{1}{2}$ $) =$ $\frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3}-1}$

$\Rightarrow \tan 82$ $\frac{1}{2}$ $=$ $\frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3}-1}$

$=$ $\frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3}-1}$ $\times$ $\frac{\sqrt{3}+1}{\sqrt{3}+1}$

$=$ $\frac{2\sqrt{6} + 3 + \sqrt{3}+ 2\sqrt{2}+\sqrt{3}+1}{2}$

$=$ $\frac{2\sqrt{6} + 4 + 2\sqrt{3}+ 2\sqrt{2}}{2}$

$= \sqrt{6} + 2 + \sqrt{3}+ \sqrt{2}$

$= \sqrt{2} + \sqrt{3} + \sqrt{4}+ \sqrt{5} =$ RHS. Hence proved.

$\\$

Question 27: Prove that $\cot$ $\frac{\pi}{8}$ $= \sqrt{2}+1$

Answer:

LHS $= \cot$ $\frac{\pi}{8}$ $= \cot 22$ $\frac{1}{2}$

We know, $\cot 2x =$ $\frac{\cot^2 x - 1}{2 \cot x}$

$\Rightarrow \cot 45^o =$ $\frac{\cot^2 22\frac{1}{2} - 1}{2 \cot 22\frac{1}{2}}$

$\Rightarrow 2 \cot 22$ $\frac{1}{2}$ $= \cot^2 22$ $\frac{1}{2}$ $- 1$

Let $\cot 22$ $\frac{1}{2}$ $= a$

$\Rightarrow a^2 - 2a -1 = 0$

$\Rightarrow (a-1)^2 = 2$

$\Rightarrow a- 1 = \pm \sqrt{2}$

$\Rightarrow a = 1 + \pm \sqrt{2}$

Since $\cot 22\frac{1}{2}$ is in I quadrant, it is positive

$\therefore \cot 22\frac{1}{2} = 1 + \sqrt{2} =$ RHS. Hence proved.

$\\$

Question 28: i) If $\cos x = -$ $\frac{3}{5}$ and $x$ lies in III quadrant, find the value of $\cos$ $\frac{x}{2}$ , $\sin$ $\frac{x}{2}$ and $\sin 2x$

ii) If $\cos x = -$ $\frac{3}{5}$ and $x$ lies in II quadrant, find the value of $\sin 2x$ and $\sin$ $\frac{x}{2}$ .

Answer:

i) Given $\cos x = -$ $\frac{3}{5}$

We know $\cos 2x = \cos^2 x - \sin^2 x$

$\Rightarrow \cos x = \cos^2$ $\frac{x}{2}$ $- \sin^2$ $\frac{x}{2}$

$\Rightarrow -$ $\frac{3}{5}$ $= 2 \cos^2$ $\frac{x}{2}$ $- 1$

$\Rightarrow \cos^2$ $\frac{x}{2}$ $=$ $\frac{1}{2}$ $\Big[ -$ $\frac{3}{5}$ $+ 1 \Big]$

$\Rightarrow \cos^2$ $\frac{x}{2}$ $=$ $\frac{1}{5}$

$\Rightarrow \cos$ $\frac{x}{2}$ $= \pm$ $\frac{1}{\sqrt{5}}$

Given that $\frac{x}{2}$ lies in II Quadrant, $\cos$ $\frac{x}{2}$ is negative.

$\Rightarrow \cos$ $\frac{x}{2}$ $= -$ $\frac{1}{\sqrt{5}}$

Similarly,

$\cos x = \cos^2$ $\frac{x}{2}$ $- \sin^2$ $\frac{x}{2}$

$\Rightarrow -$ $\frac{3}{5}$ $= 1 - 2 \sin^2$ $\frac{x}{2}$

$\Rightarrow 2 \sin^2$ $\frac{x}{2}$ $= 1 +$ $\frac{3}{5}$

$\Rightarrow \sin^2$ $\frac{x}{2}$ $=$ $\frac{4}{5}$

$\Rightarrow \sin$ $\frac{x}{2}$ $= \pm$ $\frac{2}{\sqrt{5}}$

Given that $\frac{x}{2}$ lies in II Quadrant, $\sin$ $\frac{x}{2}$ is positive.

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

Now $\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \big( - \frac{3}{5} \big)^2 } = \pm$ $\frac{4}{5}$

Given that $x$ lies in III Quadrant, $\sin x$ is negative.

$\therefore \sin x = -$ $\frac{4}{5}$

Also $\sin 2x = 2 \sin x \cos x = 2 \big( -$ $\frac{4}{5}$ $\big) \big( -$ $\frac{3}{5}$ $\big) =$ $\frac{24}{25}$

ii) Given $\cos x = -$ $\frac{3}{5}$ and $x$ lies in II Quadrant

$\therefore \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \big( -\frac{3}{5} \big)^2} = \pm$ $\frac{4}{5}$

Since $x$ lies in II Quadrant, $\sin x$ is positive

$\therefore \sin x =$ $\frac{4}{5}$

We know $\sin 2x = 2 \sin x \cos x$

$\Rightarrow \sin 2x = 2 \big($ $\frac{4}{5}$ $\big) \big( -$ $\frac{3}{5}$ $\big) = -$ $\frac{24}{25}$

We know $\cos x = 1 - 2 \sin^2$ $\frac{x}{2}$

$\Rightarrow \sin^2$ $\frac{x}{2}$ $=$ $\frac{1}{2}$ $( 1 - \cos x) =$ $\frac{1}{2}$ $\Big( 1- \big( -$ $\frac{3}{5} \big) \Big) =$ $\frac{4}{5}$

$\Rightarrow \sin$ $\frac{x}{2}= \pm$ $\frac{2}{\sqrt{5}}$

Since $\frac{x}{2}$ lies in I Quadrant, $\sin \frac{x}{2}$ is positive

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

$\\$

Question 29: If $\sin x =$ $\frac{\sqrt{5}}{3}$ and $x$ lies in II quadrant, find the value of $\cos$ $\frac{x}{2}$ , $\sin$ $\frac{x}{2}$ and $\tan$ $\frac{x}{2}$

Answer:

Given $\sin x =$ $\frac{\sqrt{5}}{3}$

$\Rightarrow \cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 -\frac{5}{9}} = \pm$ $\frac{2}{3}$

Since $x$ lies in II Quadrant , $\cos x$ is positive

$\therefore \cos x = -$ $\frac{2}{3}$

We know $\cos x = 1 - 2 \sin^2$ $\frac{x}{2}$

$\sin^2$ $\frac{x}{2}$ $=$ $\frac{1}{2}$ $\Big[ 1 - \big( -$ $\frac{2}{3}$ $\big) \Big] =$ $\frac{5}{6}$

$\sin$ $\frac{x}{2}$ $= \pm$ $\sqrt{\frac{5}{6}}$

Since $\frac{x}{2}$ lies in I Quadrant , $\sin$ $\frac{x}{2}$ is positive

$\therefore \sin$ $\frac{x}{2}$ $=$ $\sqrt{\frac{5}{6}}$

We know $\cos x = 2 \cos^2$ $\frac{x}{2}$ $- 1$

$\Rightarrow \cos^2$ $\frac{x}{2}$ $=$ $\frac{1}{2}$ $\Big[ \big( -$ $\frac{2}{3}$ $\big) + 1 \Big] =$ $\frac{1}{2}$ $\Big[$ $\frac{1}{3}$ $\Big] =$ $\frac{1}{6}$

$\therefore \cos$ $\frac{x}{2}$ $= \pm$ $\frac{1}{\sqrt{6}}$

Since $\frac{x}{2}$ lies in I Quadrant , $\cos$ $\frac{x}{2}$ is positive

$\therefore \cos$ $\frac{x}{2}$ $=$ $\frac{1}{\sqrt{6}}$

$\therefore \tan$ $\frac{x}{2}$ $=$ $\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$ $=$ $\frac{\sqrt{\frac{5}{6}} }{\frac{1}{\sqrt{6}}}$ $= \sqrt{5}$

$\\$

Question 30:

i) if $0 \leq x \leq \pi$ and $x$ lie in II quadrant such that $\sin x =$ $\frac{1}{4}$ , find the value of $\cos$ $\frac{x}{2}$ , $\sin$ $\frac{x}{2}$ and $\tan$ $\frac{x}{2}$

ii) If $\cos x =$ $\frac{4}{5}$ and $x$ is acute, find $\tan 2x$

iii) If $\sin x =$ $\frac{4}{5}$ and $0 < x <$ $\frac{\pi}{2}$ , find the value of $\sin 4x$

Answer:

i) Given $\sin x =$ $\frac{1}{4}$

$\therefore \cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \frac{1}{16}} = \pm$ $\frac{\sqrt{15}}{4}$

Since $x$ lie in II Quadrant, $\cos x$ is negative

$\therefore \cos x = -$ $\frac{\sqrt{15}}{4}$

Now we know, $\cos x = 2 \cos^2$ $\frac{x}{2}$ $- 1$

$\Rightarrow \cos^2$ $\frac{x}{2}$ $=$ $\frac{1}{2}$ $\Big[ -$ $\frac{\sqrt{15}}{4}$ $+ 1 \Big] =$ $\frac{4 - \sqrt{15}}{8}$

$\Rightarrow \cos$ $\frac{x}{2}$ $= \pm$ $\sqrt{ \frac{4 - \sqrt{15}}{8}}$

Since $x$ lies in II Quadrant, $\frac{x}{2}$ will lie in I Quadrant. Hence $\cos$ $\frac{x}{2}$ is positive.

$\Rightarrow \cos$ $\frac{x}{2}$ $=$ $\sqrt{ \frac{4 - \sqrt{15}}{8}}$

Now $\cos x = \cos^2$ $\frac{x}{2}$ $- \sin^2$ $\frac{x}{2}$

$\Rightarrow -$ $\frac{\sqrt{15}}{4}$ $= \Big($ $\sqrt{ \frac{4 - \sqrt{15}}{8}}$ $\Big)^2 - \sin^2$ $\frac{x}{2}$

$\sin^2$ $\frac{x}{2}$ $=$ $\frac{4 - \sqrt{15}}{8}$ $+$ $\frac{\sqrt{15}}{4}$

$\sin^2$ $\frac{x}{2}$ $=$ $\frac{4 + \sqrt{15}}{8}$

$\Rightarrow \sin$ $\frac{x}{2}$ $= \pm$ $\sqrt{ \frac{4 + \sqrt{15}}{8}}$

Since $x$ lies in II Quadrant, $\frac{x}{2}$ will lie in I Quadrant. Hence $\sin$ $\frac{x}{2}$ is positive.

$\Rightarrow \sin$ $\frac{x}{2}$ $=$ $\sqrt{ \frac{4 + \sqrt{15}}{8}}$

$\tan$ $\frac{x}{2}$ $=$ $\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$ $=$ $\frac{\sqrt{ \frac{4 + \sqrt{15}}{8}}}{\sqrt{ \frac{4 - \sqrt{15}}{8}}}$ $=$ $\sqrt{ \frac{4 + \sqrt{15}}{4 - \sqrt{15}} }$ $=$ $\sqrt{ \frac{4 + \sqrt{15}}{4 - \sqrt{15}} \times \frac{4 + \sqrt{15}}{4 + \sqrt{15}} }$ $= 4 + \sqrt{15}$

ii) Given $\cos x =$ $\frac{4}{5}$

$\tan x =$ $\frac{\sin x }{\cos x}$ $=$ $\frac{\sqrt{1 - \cos^2 x}}{\cos x}$ $=$ $\frac{\sqrt{1 - \big( \frac{4}{5} \big)^2}}{\frac{4}{5}}$ $=$ $\frac{3}{4}$

$\tan 2x =$ $\frac{2 \tan x}{1 - \tan^2 x}$ $=$ $\frac{2 \times \frac{3}{4}}{1 - \big( \frac{3}{4} \big)^2}$ $=$ $\frac{24}{7}$

iii) $\sin x =$ $\frac{4}{5}$ and $0 < x <$ $\frac{\pi}{2}$

$\cos x = \sqrt{1 - \sin^2 x} =$ $\sqrt{1 - \frac{16}{25}}$ $=$ $\frac{3}{5}$

$\therefore \sin 2x = 2 \sin x \cos x = 2 \times$ $\frac{4}{5}$ $\times$ $\frac{3}{5}$ $=$ $\frac{24}{25}$

$\cos 2x = \sqrt{1 - \sin^2 2x} =$ $\sqrt{1 - \big( \frac{24}{25} \big)^2}$ $= \pm$ $\frac{7}{25}$

Since $x$ likes in I Quadrant, $2x$ lies in II Quadrant $\therefore \cos 2x$ is negative

$\Rightarrow \cos 2x = -$ $\frac{7}{25}$

Now $\sin 4x = 2 \sin 2x \cos 2x = 2 \times$ $\frac{24}{7}$ $\times \big( -$ $\frac{7}{25}$ $\big) = -$ $\frac{336}{625}$

$\\$

Question 31: If $\tan x =$ $\frac{b}{a}$ , find the value of $\sqrt{\frac{a+b}{a-b}}$ $+$ $\sqrt{\frac{a-b}{a+b}}$

Answer:

$\sqrt{\frac{a+b}{a-b}}$ $+$ $\sqrt{\frac{a-b}{a+b}}$

$=$ $\frac{a+b + a - b}{\sqrt{a^2 - b^2}}$

$=$ $\frac{2a}{\sqrt{a^2 - b^2}}$

$=$ $\frac{2}{\sqrt{1 - \big(\frac{b}{a} \big)^2}}$

$=$ $\frac{2}{\sqrt{1 - \tan^2 x}}$

$=$ $\frac{2 \cos x}{\sqrt{\cos^2 x - \sin^2 x} }$

$=$ $\frac{2 \cos x}{\sqrt{\cos 2x}}$

$\\$

Question 32: If $\tan A =$ $\frac{1}{y}$ and If $\tan B =$ $\frac{1}{3}$ , show that $\cos 2A = \sin 4B$

Answer:

Given $\tan A =$ $\frac{1}{y}$ and If $\tan B =$ $\frac{1}{3}$

$\cos 2A =$ $\frac{1 - \tan^2 A}{1 + \tan^2 A}$ $=$ $\frac{1 - \big( \frac{1}{y} \big)^2}{1 + \big( \frac{1}{y} \big)^2}$ $=$ $\frac{49-1}{49+1}$ $=$ $\frac{48}{50}$ $=$ $\frac{24}{25}$

$\sin 4B = 2 \sin 2B \cos 2B$

$= 2 \Big($ $\frac{2\tan B}{1 + \tan^2 B}$ $\Big) \Big($ $\frac{1 - \tan^2 B}{1 + \tan^2 B}$ $\Big)$

$= 4 \Big($ $\frac{\frac{1}{3}}{1 + \big( \frac{1}{3}\big)^2 B}$ $\Big) \Big($ $\frac{1 - \big( \frac{1}{3}\big)^2}{1 + \big( \frac{1}{3}\big)^2}$ $\Big)$ $=$ $\frac{4 \times \frac{1}{3} \times \frac{8}{9}}{\frac{10}{9} \times \frac{10}{9} }$ $=$ $\frac{4 \times 3 \times 8}{10 \times 10} = \frac{24}{25}$

Hence $\cos 2A = \sin 4B$

$\\$

Question 33: Prove that $\cos 7^o \cos 14^o \cos 28^o \cos 56^o =$ $\frac{\sin 68^o}{16 \cos 83^o}$

Answer:

LHS $= \cos 7^o \cos 14^o \cos 28^o \cos 56^o$

$=$ $\frac{2 \sin 7^o \cos 7^o \cos 14^o \cos 28^o \cos 56^o}{2 \sin 7^o }$

$=$ $\frac{2 \sin 14^o \cos 14^o \cos 28^o \cos 56^o}{4 \sin 7^o }$

$=$ $\frac{2 \sin 28^o \cos 28^o \cos 56^o}{8 \sin 7^o }$

$=$ $\frac{2 \sin 56^o \cos 56^o}{16 \sin 7^o }$

$=$ $\frac{\sin 112^o}{16 \sin 7^o }$

$=$ $\frac{\sin (180^o - 68^o)}{16 \sin (90^o - 83^o) }$

$=$ $\frac{\sin 68^o}{16 \cos 83^o }$ $=$ RHS. Hence proved.

$\\$

Question 34: Prove that $\cos$ $\frac{2\pi}{15}$ $\cos$ $\frac{4\pi}{15}$ $\cos$ $\frac{8\pi}{15}$ $\cos$ $\frac{16\pi}{15}$ $=$ $\frac{1}{16}$

Answer:

LHS = $\cos$ $\frac{2\pi}{15}$ $\cos$ $\frac{4\pi}{15}$ $\cos$ $\frac{8\pi}{15}$ $\cos$ $\frac{16\pi}{15}$

$= \cos 24^o \cos 48^o \cos 96^o \cos 192^o$

$=$ $\frac{2 \sin 24^o \cos 24^o \cos 48^o \cos 96^o \cos 192^o}{2 \sin 24^o}$

$=$ $\frac{2 \sin 48^o \cos 48^o \cos 96^o \cos 192^o}{4 \sin 24^o}$

$=$ $\frac{2 \sin 96^o \cos 96^o \cos 192^o}{8 \sin 24^o}$

$=$ $\frac{2 \sin 192^o \cos 192^o}{16 \sin 24^o}$

$=$ $\frac{\sin 384^o}{16 \sin 24^o}$

$=$ $\frac{\sin (360^o+24^o)}{16 \sin 24^o}$

$=$ $\frac{\sin 24^o}{16 \sin 24^o}$

$=$ $\frac{1}{16}$ $=$ RHS. Hence proved.

$\\$

Question 35: Prove that $\cos$ $\frac{\pi}{5}$ $\cos$ $\frac{2\pi}{5}$ $\cos$ $\frac{4\pi}{5}$ $\cos$ $\frac{8\pi}{5}$ $= -$ $\frac{1}{16}$

Answer:

LHS = $\cos$ $\frac{\pi}{5}$ $\cos$ $\frac{2\pi}{5}$ $\cos$ $\frac{4\pi}{5}$ $\cos$ $\frac{8\pi}{5}$

$= \cos 12^o \cos 24^o \cos 48^o \cos 96^o$

$=$ $\frac{2 \sin 12^o \cos 12^o \cos 24^o \cos 48^o \cos 96^o}{2 \sin 12^o}$

$=$ $\frac{2 \sin 24^o \cos 24^o \cos 48^o \cos 96^o}{4 \sin 12^o}$

$=$ $\frac{2 \sin 48^o \cos 48^o \cos 96^o}{8 \sin 12^o}$

$=$ $\frac{2 \sin 96^o \cos 96^o}{16 \sin 12^o}$

$=$ $\frac{\sin 192^o}{16 \sin 12^o}$

$=$ $\frac{\sin (180^o+12^o)}{16 \sin 12^o}$

$=$ $\frac{-\sin 12^o}{16 \sin 12^o}$

$= -$ $\frac{1}{16}$ $=$ RHS. Hence proved.

$\\$

Question 36: Prove that $\cos$ $\frac{\pi}{65}$ $\cos$ $\frac{2\pi}{65}$ $\cos$ $\frac{4\pi}{65}$ $\cos$ $\frac{8\pi}{65}$ $\cos$ $\frac{16\pi}{65}$ $\cos$ $\frac{32\pi}{65}$ $=$ $\frac{1}{64}$

Answer:

LHS = $\cos$ $\frac{\pi}{65}$ $\cos$ $\frac{2\pi}{65}$ $\cos$ $\frac{4\pi}{65}$ $\cos$ $\frac{8\pi}{65}$ $\cos$ $\frac{16\pi}{65}$ $\cos$ $\frac{32\pi}{65}$

$=$ $\frac{2 \sin \frac{\pi}{65} \cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65}}{2 \sin \frac{\pi}{65}}$

$=$ $\frac{2 \sin \frac{2\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65} }{4 \sin \frac{\pi}{65}}$

$=$ $\frac{2 \sin \frac{4\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65}}{8 \sin \frac{\pi}{65}}$

$=$ $\frac{2 \sin \frac{8\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65} }{16 \sin \frac{\pi}{65}}$

$=$ $\frac{2 \sin \frac{16\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65} }{32 \sin \frac{\pi}{65}}$

$=$ $\frac{2 \sin \frac{32\pi}{65} \cos \frac{32\pi}{65} }{64 \sin \frac{\pi}{65}}$

$=$ $\frac{ \sin \frac{64\pi}{65} }{64 \sin \frac{\pi}{65}}$

$=$ $\frac{ \sin \big( \pi - \frac{\pi}{65} \big) }{64 \sin \frac{\pi}{65}}$

$=$ $\frac{ \sin \frac{\pi}{65} }{64 \sin \frac{\pi}{65}}$

$=$ $\frac{1}{64}$ $=$ RHS. Hence proved.

$\\$

Question 37: If $2 \tan \alpha = 3 \tan \beta$ , then prove that $\tan (\alpha - \beta) =$ $\frac{\sin 2 \beta}{5- \cos 2\beta}$

Answer:

Given $2 \tan \alpha = 3 \tan \beta \Rightarrow \tan \alpha =$ $\frac{3}{2}$ $\tan \beta$

LHS $= \tan (\alpha - \beta)$

$=$ $\frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

$=$ $\frac{\frac{3}{2} \tan \beta - \tan \beta}{1 + \frac{3}{2} \tan \beta \tan \beta}$

$=$ $\frac{\tan \beta}{2 + 3 \tan^2 \beta}$

$=$ $\frac{\frac{\sin \beta}{\cos \beta}}{2 + 3 \big( \frac{\sin \beta}{\cos \beta} \big)^2}$

$=$ $\frac{\sin \beta \cos \beta}{2 \cos^2 \beta + 3 \sin^2 \beta}$

$=$ $\frac{\sin \beta \cos \beta}{2 + \sin^2 \beta}$

$=$ $\frac{2\sin \beta \cos \beta}{4 + 2 \sin^2 \beta}$

$=$ $\frac{\sin 2\beta}{4 + 2 (1- \cos^2 \beta)}$

$=$ $\frac{\sin 2\beta}{6 - 2 \cos^2 \beta}$ $=$ RHS. Hence proved.

$\\$

Question 38: If $\sin \alpha + \sin \beta = a$ and $\cos \alpha + \cos \beta = b$ , prove that:

i) $\sin (\alpha + \beta) =$ $\frac{2ab}{a^2 + b^2}$

ii) $\cos (\alpha - \beta) =$ $\frac{a^2 + b^2 - 2}{2}$

Answer:

i) Given $\sin \alpha + \sin \beta = a$

$\Rightarrow 2 \sin$ $\frac{\alpha + \beta}{2}$ $\cos \frac{\alpha - \beta}{2}$ $= a$ … … … … … i)

Given $\cos \alpha + \cos \beta = b$

$\Rightarrow 2 \cos$ $\frac{\alpha + \beta}{2}$ $\cos$ $\frac{\alpha - \beta}{2}$ $= b$ … … … … … ii)

Dividing i) by ii) we get

$\tan$ $\frac{\alpha + \beta}{2}$ $=$ $\frac{a}{b}$

We know $\sin (\alpha + \beta) =$ $\frac{2 \tan \big( \frac{\alpha + \beta}{2} \big)}{1+ \tan^2 \big( \frac{\alpha + \beta}{2} \big)}$ $=$ $\frac{2 \frac{a}{b}}{1+\frac{a^2}{b^2} }$ $=$ $\frac{2ab}{a^2+b^2}$

ii) Squaring and adding

$( \sin \alpha + \sin \beta)^2 + ( \cos \alpha + \cos \beta)^2 = a^2 +b^2$

$\Rightarrow \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta + \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta = a^2 + b^2$

$\Rightarrow 2 + 2 \sin \alpha \sin \beta + 2 \cos \alpha \cos \beta = a^2 + b^2$

$\Rightarrow 2 + 2 ( \sin \alpha \sin \beta + \cos \alpha \cos \beta) = a^2 + b^2$

$\Rightarrow 2 + 2 \cos ( \alpha - \beta) = a^2 + b^2$

$\Rightarrow \cos ( \alpha - \beta) =$ $\frac{a^2 + b^2-2}{2}$

$\\$

Question 39: If $2\tan$ $\frac{\alpha}{2}$ $= \tan$ $\frac{\beta}{2}$ , prove that $\cos \alpha =$ $\frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}$

Answer:

Given $2\tan$ $\frac{\alpha}{2}$ $= \tan$ $\frac{\beta}{2}$ ,

RHS $=$ $\frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}$

We know $\cos \beta =$ $\frac{1 - \tan^2 \frac{\beta}{2} }{1 + \tan^2 \frac{\beta}{2} }$

$=$ $\frac{3 + 5 \big( \frac{1 - \tan^2 \frac{\beta}{2} }{1 + \tan^2 \frac{\beta}{2} } \big) }{5 + 3 \big(\frac{1 - \tan^2 \frac{\beta}{2} }{1 + \tan^2 \frac{\beta}{2} } \big)}$

$=$ $\frac{3 + 3 \tan^2 \frac{\beta}{2}+ 5 - 5 \tan^2 \frac{\beta}{2}}{5 + 5\tan^2 \frac{\beta}{2} + 3 - 3 \tan^2 \frac{\beta}{2}}$

$=$ $\frac{8 - 2 \tan^2 \frac{\beta}{2}}{8 + 2 \tan^2 \frac{\beta}{2}}$

Since $\tan$ $\frac{\beta}{2}$ $= 2$ $\tan$ $\frac{\alpha}{2}$

$=$ $\frac{8 - 2 \big( 2 \tan \frac{\alpha}{2} \big)^2 }{8 + 2 \big( 2 \tan \frac{\alpha}{2} \big)^2 }$

$=$ $\frac{8 - 8 \tan^2 \frac{\alpha}{2}}{8 + 8 \tan^2 \frac{\alpha}{2}}$

$=$ $\frac{1 - \tan^2 \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}}$

$= \cos \alpha =$ LHS. Hence proved.

$\\$

Question 40: If $\cos x =$ $\frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}$ prove that $\tan$ $\frac{x}{2}$ $= \pm \tan$ $\frac{\alpha}{2}$ $\tan$ $\frac{\beta}{2}$

Answer:

Given $\cos x =$ $\frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}$

$\Rightarrow$ $\frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}$ $=$ $\frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}$

Applying Componendo and Dividendo

$\frac{1 - \tan^2 \frac{x}{2}+ 1 + \tan^2 \frac{x}{2}}{1 - \tan^2 \frac{x}{2} -1 - \tan^2 \frac{x}{2}}$ $=$ $\frac{\cos \alpha + \cos \beta + 1 + \cos \alpha \cos \beta}{\cos \alpha + \cos \beta - 1 - \cos \alpha \cos \beta}$

$\Rightarrow$ $\frac{2}{-2 \tan^2 \frac{x}{2}}$ $=$ $\frac{(1 + \cos \alpha)(1 + \cos \beta)}{- (1 - \cos \alpha)(1 - \cos \beta)}$

$\Rightarrow \tan^2$ $\frac{x}{2}$ $=$ $\frac{(1 - \cos \alpha)(1 - \cos \beta)}{ (1 + \cos \alpha)(1 + \cos \beta)}$

$\Rightarrow \tan^2$ $\frac{x}{2}$ $=$ $\frac{2 \sin^2 \frac{\alpha}{2} \sin^2 \frac{\beta}{2} }{ 2 \cos^2 \frac{\alpha}{2} \cos^2 \frac{\beta}{2} }$

$\Rightarrow \tan^2$ $\frac{x}{2}$ $= \tan^2$ $\frac{\alpha}{2}$ $\tan^2$ $\frac{\beta}{2}$

$\Rightarrow \tan$ $\frac{x}{2}$ $= \pm \tan$ $\frac{\alpha}{2}$ $\tan$ $\frac{\beta}{2}$

$\\$

Question 41: If $\sec (x + \alpha) + \sec ( x - \alpha) = 2 \sec x$ , prove that $\cos x = \pm \sqrt{2} \cos$ $\frac{\alpha}{2}$

Answer:

Given $\sec (x + \alpha) + \sec ( x - \alpha) = 2 \sec x$ , prove that $\cos x = \pm \sqrt{2} \cos$ $\frac{\alpha}{2}$

$\Rightarrow$ $\frac{1}{\cos (x + \alpha)}$ $+$ $\frac{1}{\cos (x - \alpha)}$ $=$ $\frac{2}{\cos x}$

$\Rightarrow$ $\frac{\cos (x + \alpha) + \cos (x - \alpha)}{\cos (x + \alpha)\cos (x - \alpha)}$ $=$ $\frac{2}{\cos x}$

$\Rightarrow$ $\frac{2 \cos x \cos \alpha}{\cos^2 x \cos^2 \alpha - \sin^2 x \sin^2 \alpha }$ $=$ $\frac{2}{\cos x}$

$\Rightarrow$ $\frac{\cos^2 x \cos \alpha}{\cos^2 x \cos^2 \alpha - \sin^2 x \sin^2 \alpha }$ $= 1$

$\Rightarrow \cos^2 x \cos \alpha = \cos^2 x \cos^2 \alpha - \sin^2 x \sin^2 \alpha$

$\Rightarrow \cos^2 x \cos \alpha = \cos^2 x \cos^2 \alpha - (1 - \cos^2 x) \sin^2 \alpha$

$\Rightarrow \cos^2 x \cos \alpha = \cos^2 x \cos^2 \alpha - \sin^2 \alpha + \cos^2 x \sin^2 \alpha$

$\Rightarrow \cos^2 x \cos \alpha = \cos^2 x ( \cos^2 \alpha + \sin^2 \alpha) - \sin^2 \alpha$

$\Rightarrow \cos^2 x \cos \alpha = \cos^2 x - \sin^2 \alpha$

$\Rightarrow \cos^2 x \cos \alpha - \cos^2 x = - \sin^2 \alpha$

$\Rightarrow \cos^2 x ( \cos \alpha -1) = - \sin^2 \alpha$

$\Rightarrow \cos^2 x ( 1 - \cos \alpha ) = \sin^2 \alpha$

$\Rightarrow \cos^2 x =$ $\frac{\sin^2 \alpha}{ 1 - \cos \alpha }$

$\Rightarrow \cos^2 x=$ $\frac{4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2} }{ 2 \sin^2 \frac{\alpha}{2} }$

$\Rightarrow \cos^2 x = 2 \cos^2$ $\frac{\alpha}{2}$

$\Rightarrow \cos x = \pm \sqrt{2} \cos$ $\frac{\alpha}{2}$ . Hence proved.

$\\$

Question 42: $\cos \alpha + \cos \beta =$ $\frac{1}{3}$ and $\sin \alpha + \sin \beta =$ $\frac{1}{4}$ , prove that $\cos$ $\frac{\alpha - \beta}{2}$ $= \pm$ $\frac{5}{24}$

Answer:

Given $\cos \alpha + \cos \beta =$ $\frac{1}{3}$ and $\sin \alpha + \sin \beta =$ $\frac{1}{4}$

Squaring and adding the two

$(\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 =$ $\frac{1}{9}$ $+$ $\frac{1}{16}$

$\Rightarrow \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta + \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta =$ $\frac{1}{9}$ $+$ $\frac{1}{16}$

$\Rightarrow 2 + 2 \cos \alpha \cos \beta + 2 \sin \alpha \sin \beta =$ $\frac{25}{144}$

$\Rightarrow 2 + 2 \cos( \alpha - \beta) =$ $\frac{25}{144}$

$\Rightarrow \cos( \alpha - \beta) = -$ $\frac{263}{288}$

Since $\cos^2$ $\big( \frac{\alpha - \beta}{2} \big)$ $=$ $\frac{1+ \cos ( \alpha - \beta)}{2}$ $=$ $\frac{1- \frac{263}{288}}{2}$ $=$ $\frac{25}{576}$

$\Rightarrow \cos \big($ $\frac{\alpha - \beta}{2}$ $\big) = \pm$ $\frac{5}{24}$

$\\$

Question 43: If $\sin \alpha =$ $\frac{4}{5}$ and $\cos \beta =$ $\frac{5}{13}$ , prove that $\cos$ $\frac{\alpha - \beta}{2}$ $=$ $\frac{8}{\sqrt{65}}$

Answer:

Given $\sin \alpha =$ $\frac{4}{5}$

$\Rightarrow \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$

Also given $\cos \beta =$ $\frac{5}{13}$

$\Rightarrow \sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \frac{25}{169}} = \frac{12}{13}$

$\cos( \alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

$\Rightarrow \cos ( \alpha - \beta) =$ $\frac{3}{5}$ $\times$ $\frac{5}{13}$ $+$ $\frac{4}{5}$ $\times$ $\frac{12}{13}$ $=$ $\frac{63}{65}$

Therefore $\cos$ $\frac{ \alpha - \beta}{2}$ $=$ $\sqrt{\frac{1+ \cos ( \alpha - \beta)}{2} }$ $=$ $\sqrt{\frac{1+\frac{63}{65}}{2}}$ $= \pm$ $\frac{8}{\sqrt{65}}$

$\\$

Question 44: If $a \cos 2x + b \sin 2x = c$ has $\alpha$ and $\beta$ as its roots, then prove that:

i) $\tan \alpha + \tan \beta =$ $\frac{2b}{a+c}$

ii) $\tan \alpha \tan \beta =$ $\frac{c-a}{c+a}$

iii) $\tan ( \alpha + \beta) =$ $\frac{b}{a}$

Answer:

i) Given $a \cos 2x + b \sin 2x = c$ … … … … … i)

$\Rightarrow a \Big($ $\frac{1 - \tan^2 x}{1 + \tan^2 x}$ $\Big) + b \Big($ $\frac{2 \tan x}{1 + \tan^2 x}$ $\Big) = c$

$\Rightarrow a ( 1 - \tan^2 x ) + 2 b \tan x = c ( 1 + \tan^2 x)$

$\Rightarrow a - a \tan^2 x + 2 b \tan x = c + c \tan^2 x$

$\Rightarrow (c+a) \tan^2 x - 2 b \tan x + ( c-a) = 0$ … … … … … ii)

if $\alpha$ and $\beta$ are the roots of i) , then $\tan \alpha$ and $\tan \beta$ are the roots of ii)

$\therefore \tan \alpha + \tan \beta =$ $\frac{-(-2b)}{a+c}$ $=$ $\frac{2b}{a+c}$

ii) Similarly $\tan \alpha \tan \beta =$ $\frac{c-a}{c+a}$

iii) $\tan ( \alpha - \beta ) =$ $\frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$ $=$ $\frac{\frac{2b}{a+c}}{1 - \frac{c-a}{c+a} }$ $=$ $\frac{2b}{a+c-c+a}$ $=$ $\frac{b}{a}$