Prove that:

$\displaystyle \text{Question 1: } \sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x$

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

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

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

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

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

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

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

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

$\displaystyle = 4 \sin x - 12 \sin^3 x + 8 \sin^5 x + \sin x - 8 \sin^3 x + 8 \sin^5 x$

$\displaystyle = 16 \sin^5 x - 20 \sin^3 x + 5 \sin x = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 2: } 4( \cos^3 10^{\circ} + \sin^3 20^{\circ}) = 3 ( \cos 10^{\circ} + \sin 20^{\circ})$

$\displaystyle \cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$

$\displaystyle 4 \cos^3 \theta = \cos 3\theta + 3 \cos \theta$

$\displaystyle \text{Taking } \theta = 10^{\circ}$

$\displaystyle 4 \cos^ 3 10^{\circ} = \cos 30^{\circ} + 3 \cos 10^{\circ}$ … … … … … i)

$\displaystyle \sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$

$\displaystyle 4 \sin^3 \theta = 3 \sin \theta - \sin 3 \theta$

$\displaystyle \text{Taking } \theta = 20^{\circ}$

$\displaystyle 4 \sin^3 20^{\circ} = 3 \sin 20^{\circ} - \sin 60^{\circ}$ … … … … … ii)

Adding ( i ) and ( ii ) Equation.

$\displaystyle 4( \cos^3 10^{\circ} + \sin^3 20^{\circ}) = 3( \cos 10^{\circ} + \sin 20^{\circ}) + \cos 30^{\circ} - \sin 60^{\circ}$

$\displaystyle 4( \cos^3 10^{\circ} + \sin^3 20^{\circ}) = 3( \cos 10^{\circ} + \sin 20^{\circ})$ [as we know that $\displaystyle \cos 30^{\circ} = \sin 60^{\circ}$ ]

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle = 2 \sin x - 2 \sin^3 x + \sin x - 2 \sin^3 x$

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

$\displaystyle \cos 3x = \cos ( 2x + x)$

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

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

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

$\displaystyle = 2 \cos^3 x - \cos x - 2 \cos x + 2 \cos^3 x$

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

$\displaystyle \text{LHS } = \cos^3 x \sin 3x + \sin^3 x \cos 3x$

$\displaystyle = \Big( \frac{\cos 3x + 3\cos x}{4} \Big) \sin 3x + \sin^3 x \Big( \frac{3 \sin x - \sin 3x}{4} \Big)$

$\displaystyle = \frac{1}{4} \Big[ \cos 3x \sin 3x + 3 \cos x \sin 3x + 3 \sin x \cos 3x - \sin 3x \cos x \Big]$

$\displaystyle = \frac{3}{4} ( \cos x \sin 3x + \sin x \cos 3x )$

$\displaystyle = \frac{3}{4} \sin 4x$

$\displaystyle \\$

Question 4:

$\displaystyle \tan x \tan \Big( x + \frac{\pi}{3} \Big) + \tan x \tan \Big( x - \frac{\pi}{3} \Big) + \tan \Big( x + \frac{\pi}{3} \Big) \tan \Big( x - \frac{\pi}{3} \Big) = -3$

$\displaystyle \text{LHS } = \tan x \tan ( x +60) + \tan x \tan (x-60) + \tan ( x +60) \tan (x-60)$

$\displaystyle = \tan x \Big[ \frac{\tan x + \sqrt{3}}{1 - \tan x \sqrt{3}} \Big] + \tan x \Big[ \frac{\tan x - \sqrt{3}}{1 + \tan x \sqrt{3}} \Big] + \frac{\tan x - \sqrt{3}}{1 + \tan x \sqrt{3}} \times \frac{\tan x + \sqrt{3}}{1 - \tan x \sqrt{3}}$

$\displaystyle = \tan x \Big[ \frac{\tan x + \sqrt{3}}{1 - \tan x \sqrt{3}} + \frac{\tan x - \sqrt{3}}{1 + \tan x \sqrt{3}} \Big] + \frac{\tan x - \sqrt{3}}{1 + \tan x \sqrt{3}} \times \frac{\tan x + \sqrt{3}}{1 - \tan x \sqrt{3}}$

$\displaystyle = \tan x \Big[ \frac{(\tan x + \sqrt{3}) (1 + \tan x \sqrt{3}) +(\tan x - \sqrt{3}) (1 - \tan x \sqrt{3}) }{(1 - \tan x \sqrt{3})(1 + \tan x \sqrt{3} ) } \Big] + \frac{\tan x - \sqrt{3}}{1 + \tan x \sqrt{3}} \times \frac{\tan x + \sqrt{3}}{1 - \tan x \sqrt{3}}$

$\displaystyle = \tan x \Big[ \frac{ \tan x +\sqrt{3} + \sqrt{3} \tan^2 x + 3 \tan x + \tan x -\sqrt{3} - \sqrt{3} \tan^2 x + 3 \tan x }{ 1 - 3 \tan^2 x } \Big] + \frac{\tan^2 x - 3 }{1 - 3 \tan^2 x }$

$\displaystyle = \frac{8 \tan x}{1 - 3 \tan^2 x} + \frac{\tan^2 x - 3}{1 - 3 \tan^2 x}$

$\displaystyle = \frac{9\tan^2 x - 3}{1 - 3 \tan^2 x}$

$\displaystyle = -3 \Big( \frac{1 - 3 \tan^2 x}{1 - 3 \tan^2 x} \Big)$

$\displaystyle = - 3 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 5: } \tan x + \tan \Big( \frac{\pi}{3} + x \Big) - \tan \Big( \frac{\pi}{3} - x \Big) = 3 \tan 3x$

$\displaystyle \text{LHS } = \tan x + \tan ( 60^{\circ}+x) + \tan ( 60^{\circ}-x)$

$\displaystyle = \tan x + \frac{\tan 60^{\circ} + \tan x }{1 - \tan 60^{\circ} \tan x } + \frac{\tan 60^{\circ} - \tan x }{1 + \tan 60^{\circ} \tan x }$

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

$\displaystyle = \tan x + \frac{(\sqrt{3} + \tan x)(1 + \sqrt{3} \tan x) + (\sqrt{3} - \tan x)(1 - \sqrt{3} \tan x)}{(1 - \sqrt{3} \tan x)(1 + \sqrt{3} \tan x)}$

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

$\displaystyle = \tan x + \frac{8 \tan x}{1 - 3 \tan^2 x}$

$\displaystyle = \frac{\tan x - 3 \tan^3 x + 8 \tan x }{1 - 3 \tan^2 x}$

$\displaystyle = \frac{9\tan x - 3 \tan^3 x }{1 - 3 \tan^2 x}$

$\displaystyle = 3 \times \frac{3\tan x - \tan^3 x }{1 - 3 \tan^2 x}$

$\displaystyle = 3 \tan 3x = \text{ RHS. Hence proved. }$

Since

$\displaystyle \tan 3 x = \tan ( 2x + x)$

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

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

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

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

$\displaystyle = \frac{3 \tan x - \tan^3 x }{1-3\tan^2 x }$

$\displaystyle \\$

$\displaystyle \text{Question 6: } \cot x + \cot \Big( \frac{\pi}{3} + x \Big) - \cot \Big( \frac{\pi}{3} - x \Big) = 3 \cot 3x$

$\displaystyle \text{LHS } = \cot x + \cot ( 60^{\circ} + x) - \cot ( 60^{\circ} - x)$

$\displaystyle = \frac{1}{\tan x } + \frac{1}{\tan (60^{\circ} + x)} - \frac{1}{\tan (60^{\circ} - x)}$

$\displaystyle = \frac{1}{\tan x } + \frac{1 - \tan 60^{\circ} \tan x }{\tan 60^{\circ} + \tan x} - \frac{1 + \tan 60^{\circ} \tan x }{\tan 60^{\circ} - \tan x}$

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

$\displaystyle = \frac{1}{\tan x } + \frac{(1 - \sqrt{3} \tan x)(\sqrt{3} - \tan x) - (1 + \sqrt{3} \tan x)(\sqrt{3} + \tan x) }{(\sqrt{3} + \tan x) (\sqrt{3} - \tan x)}$

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

$\displaystyle = \frac{1}{\tan x } - \frac{8 \tan x}{3 - \tan^2 x}$

$\displaystyle = \frac{3 - \tan^2 x - 8 \tan^2 x }{\tan x (3 - \tan^2 x)}$

$\displaystyle = \frac{3 - 9\tan^2 x }{3\tan x - \tan^3 x}$

$\displaystyle = 3 \times \frac{1 - 3\tan^2 x }{3\tan x - \tan^3 x}$

$\displaystyle = \frac{3}{\tan 3x}$

$\displaystyle = 3 \cot 3x = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 7: } \cot x + \cot \Big( \frac{\pi}{3} + x \Big) + \cot \Big( \frac{2\pi}{3} + x \Big) = 3 \cot 3x$

$\displaystyle \text{LHS } = \cot x + \cot ( 60^{\circ} + x) + \cot ( 120^{\circ} + x)$

$\displaystyle = \cot x + \cot ( 60^{\circ} + x) - \cot [ 180 - ( 120^{\circ} + x) ]$

$\displaystyle = \cot x + \cot ( 60^{\circ} + x) - \cot ( 60^{\circ} - x)$

$\displaystyle = \frac{1}{\tan x } + \frac{1}{\tan (60^{\circ} + x)} - \frac{1}{\tan (60^{\circ} - x)}$

$\displaystyle = \frac{1}{\tan x } + \frac{1 - \tan 60^{\circ} \tan x }{\tan 60^{\circ} + \tan x} - \frac{1 + \tan 60^{\circ} \tan x }{\tan 60^{\circ} - \tan x}$

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

$\displaystyle = \frac{1}{\tan x } + \frac{(1 - \sqrt{3} \tan x)(\sqrt{3} - \tan x) - (1 + \sqrt{3} \tan x)(\sqrt{3} + \tan x) }{(\sqrt{3} + \tan x) (\sqrt{3} - \tan x)}$

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

$\displaystyle = \frac{1}{\tan x } - \frac{8 \tan x}{3 - \tan^2 x}$

$\displaystyle = \frac{3 - \tan^2 x - 8 \tan^2 x }{\tan x (3 - \tan^2 x)}$

$\displaystyle = \frac{3 - 9\tan^2 x }{3\tan x - \tan^3 x}$

$\displaystyle = 3 \times \frac{1 - 3\tan^2 x }{3\tan x - \tan^3 x}$

$\displaystyle = \frac{3}{\tan 3x}$

$\displaystyle = 3 \cot 3x = \text{ RHS. Hence proved. }$

$\displaystyle \text{Question 8: } \sin 5x = 5 \cos^4 x \sin x - 10 \cos^2 x \sin^3 x + \sin^5 x$

$\displaystyle \sin 5x = \sin (3x + 2x) = \sin 3x \cos 2x + \cos 3x \sin 2x$

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

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

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

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

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

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

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

$\displaystyle = 3 \sin x \cos^4 x - 3 \sin^3 x \cos^2 x - \sin^3 x \cos^2 x + \sin^5 x$

$\displaystyle = 3 \sin x \cos^4 x - 4 \sin^3 x \cos^2 x + \sin^5 x$

$\displaystyle \cos 3x = \cos ( 2x + x)$

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

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

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

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

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

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

$\displaystyle \therefore \sin 5x = 3 \sin x \cos^4 x - 4 \sin^3 x \cos^2 x + \sin^5 x + 2\cos^4 x \sin x - 6 \sin^3 x \cos^2 x$

$\displaystyle = 5 \cos^4 x \sin x - 10 \cos^2 x \sin^3 x + \sin^5 x = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 9: } \sin^3 x + \sin^3 \Big( \frac{2\pi}{3} + x \Big) + \sin^3 \Big( \frac{4\pi}{3} + x \Big) = - \frac{3}{4} \sin 3x$

$\displaystyle \text{LHS } = \sin^3 x + \sin^3 (120^{\circ}+x) + \sin^3 (240^{\circ}+x)$

$\displaystyle = \sin^3 x + [\sin (120^{\circ}+x)]^3 + [\sin (240^{\circ}+x) ]^3$

$\displaystyle = \sin^3 x + [\sin 120^{\circ} \cos x + \cos 120^{\circ} \sin x]^3 + [\sin 240^{\circ} \cos x + \cos 240^{\circ} \sin x ]^3$

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

$\displaystyle = \sin^3 x + \frac{1}{8} (\sqrt{3} \cos x - \sin x)^3 - \frac{1}{8} (\sqrt{3} \cos x + \sin x)^3$

$\displaystyle = \sin^3 x + \frac{1}{8} \Big[ (\sqrt{3} \cos x - \sin x)^3 - (\sqrt{3} \cos x + \sin x)^3 \Big]$

$\displaystyle \because a^3 - b^3 = ( a-b)(a^2 + ab + b^2)$

$\displaystyle = \sin^3 x + \frac{1}{8} \Big[ (\sqrt{3} \cos x - \sin x - \sqrt{3} \cos x - \sin x ) \big( (\sqrt{3} \cos x - \sin x)^2 +(\sqrt{3} \cos x - \sin x)(\sqrt{3} \cos x + \sin x) + (\sqrt{3} \cos x + \sin x)^2 \big) \Big]$

$\displaystyle = \sin^3 x + \frac{1}{8} \Big[ (-2 \sin x) (3 \cos^2 x + \sin^2 x - 2 \sqrt{3} \cos x \sin x + 3 \cos^2 x - \sqrt{3} \sin x \cos x + \sqrt{3} \sin x \cos x - \sin^2 x + 3 \cos^2 x + \sin^2 x + 2 \sqrt{3} \cos x \sin x ) \Big]$

$\displaystyle = \sin^3 x + \frac{1}{8} \Big[ (-2 \sin x) ( 9 \cos^2 x + \sin^2 x) \Big]$

$\displaystyle = \sin^3 x - \frac{1}{4} \sin x ( 9 \cos^2 x + \sin^2 x)$

$\displaystyle = \frac{1}{4} \Big[ 4 \sin^3 x - 9 \sin x \cos^2 x - \sin^2 x \Big]$

$\displaystyle = \frac{1}{4} ( 3 \sin^2 x - 9 \sin x \cos^2 x)$

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

$\displaystyle \\$

$\displaystyle \text{Question 10: } \Big| \sin x \sin \Big( \frac{\pi}{3} - x \Big) \sin \Big( \frac{\pi}{3} + x \Big) \Big| \leq \frac{1}{4} \text{ for all values of } x$

$\displaystyle \text{LHS } = \Big| \sin x \sin ( 60- x) \sin ( 60 + x) \Big|$

$\displaystyle = \Big| \sin x ( \sin 60 \cos x - \cos 60 \sin x) ( \sin 60 \cos x + \cos 60 \sin x) \Big|$

$\displaystyle = \Big| \sin x ( \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x) ( \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x) \Big|$

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

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

$\displaystyle = \Big| \frac{1}{4} \sin x ( 3 - 4 \sin^2 x ) \Big|$

$\displaystyle = \Big| \frac{1}{4} ( 3 \sin x - 4 \sin^3 x ) \Big|$

$\displaystyle = \Big| \frac{1}{4} \sin 3x \Big| \leq \frac{1}{4}$ Since $\displaystyle \Big| \sin 3x \Big| \leq 1$

Hence proven.

$\displaystyle \\$

$\displaystyle \text{Question 11: } \Big| \cos x \cos \Big( \frac{\pi}{3} - x \Big) \cos \Big( \frac{\pi}{3} + x \Big) \Big| \leq \frac{1}{4} \text{ for all values of } x$

$\displaystyle \text{LHS } = \Big| \cos x \cos ( 60- x) \cos ( 60 + x) \Big|$

$\displaystyle = \Big| \cos x ( \cos 60 \cos x + \sin 60 \sin x) ( \cos 60 \cos x - \sin 60 \sin x) \Big|$

$\displaystyle = \Big| \cos x ( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x) ( \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x) \Big|$

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

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

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

$\displaystyle = \Big| \frac{1}{4} ( 4 \cos^3 x - 3 \cos x ) \Big|$

$\displaystyle = \Big| \frac{1}{4} \cos 3x \Big| \leq \frac{1}{4}$ Since $\displaystyle \Big| \cos 3x \Big| \leq 1$

Hence proven.