Prove that:

Question 1: $\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x$

LHS $= \sin 4x \cos x + \cos 4x \sin x$

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

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

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

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

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

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

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

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

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

$\\$

Question 2: $4( \cos^3 10^o + \sin^3 20^o) = 3 ( \cos 10^o + \sin 20^o)$

$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$
$4 \cos^3 \theta = \cos 3\theta + 3 \cos \theta$
Taking $\theta = 10^o$
$4 \cos^ 3 10^o = \cos 30^o + 3 \cos 10^o$     … … … … … i)

$\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$
$4 \sin^3 \theta = 3 \sin \theta - \sin 3 \theta$
Taking $\theta = 20^o$
$4 \sin^3 20^o = 3 \sin 20^o - \sin 60^o$    … … … … … ii)

Adding ( i ) and ( ii ) Equation.
$4( \cos^3 10^o + \sin^3 20^o) = 3( \cos 10^o + \sin 20^o) + \cos 30^o - \sin 60^o$
$4( \cos^3 10^o + \sin^3 20^o) = 3( \cos 10^o + \sin 20^o)$ [as we know that $\cos 30^o = \sin 60^o$ ]

$\\$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

LHS $= \cos^3 x \sin 3x + \sin^3 x \cos 3x$

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

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

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

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

$\\$

Question 4:

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

LHS $= \tan x \tan ( x +60) + \tan x \tan (x-60) + \tan ( x +60) \tan (x-60)$

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

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

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

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

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

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

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

$= - 3 =$ RHS. Hence proved.

$\\$

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

LHS $= \tan x + \tan ( 60^o+x) + \tan ( 60^o-x)$

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

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

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

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

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

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

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

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

$= 3 \tan 3x =$ RHS. Hence proved.

Since

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

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

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

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

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

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

$\\$

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

LHS $= \cot x + \cot ( 60^o + x) - \cot ( 60^o - x)$

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

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

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

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

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

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

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

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

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

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

$= 3 \cot 3x =$ RHS. Hence proved.

$\\$

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

LHS $= \cot x + \cot ( 60^o + x) + \cot ( 120^o + x)$

$= \cot x + \cot ( 60^o + x) - \cot [ 180 - ( 120^o + x) ]$

$= \cot x + \cot ( 60^o + x) - \cot ( 60^o - x)$

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

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

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

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

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

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

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

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

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

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

$= 3 \cot 3x =$ RHS. Hence proved.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\\$

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

LHS $= \sin^3 x + \sin^3 (120^o+x) + \sin^3 (240^o+x)$

$= \sin^3 x + [\sin (120^o+x)]^3 + [\sin (240^o+x) ]^3$

$= \sin^3 x + [\sin 120^o \cos x + \cos 120^o \sin x]^3 + [\sin 240^o \cos x + \cos 240^o \sin x ]^3$

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

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

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

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

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

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

Hence proven.

$\\$

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

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

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

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

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

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

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

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

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

Hence proven.