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 $\displaystyle (1-25)$

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

Answer:

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

$\displaystyle \\$

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

Answer:

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

$\displaystyle \\$

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

Answer:

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

$\displaystyle = 2 \cos x = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

Answer:

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

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

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

$\displaystyle = \tan x = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle = \tan x = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

$\displaystyle = 2 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle = 2 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

Answer:

$\displaystyle \text{LHS } = (\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2$

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{LHS } = 1 + \cos^2 2x$

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

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{RHS } = 4 ( \cos^6 x - \sin^6 x)$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

$\displaystyle \text{Question 15: } (\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0$

Answer:

$\displaystyle \text{LHS } = (\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x$

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

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

$\displaystyle = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

Answer:

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

$\displaystyle \text{Since } \cos 2x = \cos^2 x - \sin^2 x$

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

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

$\displaystyle = \sin 2x = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 17: } \cos 4x = 1 - 8 \cos^2 x + 8 \cos ^4 x$

Answer:

$\displaystyle \text{LHS } = \cos 4x$

$\displaystyle = \cos 2 \times 2x$

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

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

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

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

$\displaystyle = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 18: } \sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x$

Answer:

$\displaystyle \text{LHS } = \sin 4x$

$\displaystyle = \sin 2 .2x$

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

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

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

$\displaystyle = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

$\displaystyle = 7 + 6 = 13 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{LHS } = \cos^6 x - \sin^6 x$

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

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

$\displaystyle = 2 \sec 2x = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

Answer:

$\displaystyle \text{LHS } = \cot^2 x - \tan^2 x$

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{LHS } = \cos 4x - \cos 4 \alpha$

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

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

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

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

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

$\displaystyle = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

Answer:

$\displaystyle \text{LHS } = \sin 3x + \sin 2x - \sin x$

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

Answer:

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

$\displaystyle \text{We know } \cos 2x = \cos^2 x - \sin^2 x$

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

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

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

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

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

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

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

Similarly,

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

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

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

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

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

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

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

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

Given that $\displaystyle x$ lies in III Quadrant, $\displaystyle \sin x \text{ is negative. }$

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

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

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

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

$\displaystyle \text{Since } x \text{ lies in II Quadrant, } \sin x$ is positive

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

$\displaystyle \text{We know } \sin 2x = 2 \sin x \cos x$

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

$\displaystyle \text{Since } x$ lies in II Quadrant , $\displaystyle \cos x$ is positive

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 30:

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

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

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

Answer:

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

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

$\displaystyle \text{Since } x$ lie in II Quadrant, $\displaystyle \cos x \text{ is negative. }$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \text{Since } x$ likes in I Quadrant, $\displaystyle 2x$ lies in II Quadrant $\displaystyle \therefore \cos 2x \text{ is negative. }$

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{LHS } = \cos 7^o \cos 14^o \cos 28^o \cos 56^o$

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

$\displaystyle \text{LHS } = \tan (\alpha - \beta)$

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

Answer:

$\displaystyle \text{i) } \text{Given } \sin \alpha + \sin \beta = a$

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

$\displaystyle \text{Given } \cos \alpha + \cos \beta = b$

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

Dividing i) by ii) we get

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

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

ii) Squaring and adding

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

Applying Componendo and Dividendo

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

Squaring and adding the two

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

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

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

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

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

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

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

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

Answer:

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

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

$\displaystyle \text{Also } \text{Given } \cos \beta = \frac{5}{13}$

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

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

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

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

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

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

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

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

Answer:

$\displaystyle \text{i) } \text{Given } a \cos 2x + b \sin 2x = c$ … … … … … i)

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

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

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

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

$\displaystyle \text{If } \alpha \text{ and } \beta$ are the roots of i) , then $\displaystyle \tan \alpha \text{ and } \tan \beta$ are the roots of ii)

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

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

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

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$\displaystyle \text{Question 45: } \text{If } \cos \alpha + \cos \beta = 0 = \sin \alpha + \sin \beta , \text{ then prove that } \cos 2\alpha + \cos 2\beta= -2 \cos (\alpha + \beta)$