Question 1: Express each of the following as a sum or difference of sine and cosines:

i) $2 \sin 3x \cos x$   ii) $2 \cos 3x \sin 2x$    iii) $2 \sin 4x \sin 3x$    iv) $2 \cos 7x \cos 3x$

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

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

iii) $2 \sin 4x \sin 3x = \cos ( 4x - 3x) - \cos (4x + 3x) = \cos x - \cos 7x$

iv) $2 \cos 7x \cos 3x = \cos ( 7x + 3x) + \cos (7x - 3x) = \cos 10 x + \cos 4x$

$\\$

Question 2: Prove that:

i) $2 \sin$ $\frac{5\pi}{12}$ $\sin$ $\frac{\pi}{12}$ $=$ $\frac{1}{2}$    ii) $2 \cos$ $\frac{5\pi}{12}$ $\cos$ $\frac{\pi}{12}$ $=$ $\frac{1}{2}$    iii) $2 \sin$ $\frac{5\pi}{12}$ $\cos$ $\frac{\pi}{12}$ $=$ $\frac{\sqrt{3}+2}{2}$

i) $2 \sin$ $\frac{5\pi}{12}$ $\sin$ $\frac{\pi}{12}$

$= \cos \Big($ $\frac{5\pi}{12}$ $-$ $\frac{\pi}{12}$ $\Big) - \cos \Big($ $\frac{5\pi}{12}$ $+$ $\frac{\pi}{12}$ $\Big)$

$= \cos$ $\frac{4\pi}{12}$ $- \cos$ $\frac{6\pi}{12}$

$= \cos$ $\frac{\pi}{3}$ $- \cos$ $\frac{\pi}{2}$

$=$ $\frac{1}{2}$ $- 0$

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

ii) $2 \cos$ $\frac{5\pi}{12}$ $\cos$ $\frac{\pi}{12}$

$= \cos \Big($ $\frac{5\pi}{12}$ $+$ $\frac{\pi}{12}$ $\Big) + \cos \Big($ $\frac{5\pi}{12}$ $-$ $\frac{\pi}{12}$ $\Big)$

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

$= 0 +$ $\frac{1}{2}$

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

iii) $2 \sin$ $\frac{5\pi}{12}$ $\cos$ $\frac{\pi}{12}$

$= \sin \Big($ $\frac{5\pi}{12}$ $+$ $\frac{\pi}{12}$ $\Big) + \sin \Big($ $\frac{5\pi}{12}$ $-$ $\frac{\pi}{12}$ $\Big)$

$= \sin$ $\frac{\pi}{2}$ $+ \sin$ $\frac{\pi}{3}$

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

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

$\\$

Question 3: Show that:

i) $\sin 50^o \cos 85^o =$ $\frac{1 - \sqrt{2} \sin 35^o}{2\sqrt{2}}$    ii) $\sin 25^o \cos 115^o =$ $\frac{1}{2}$ $( \sin 40^o - 1)$

i) LHS $= \sin 50^o \cos 85^o$

$=$ $\frac{1}{2}$ $\Big[ 2 \sin 50^o \cos 85^o \Big]$

$=$ $\frac{1}{2}$ $\Big[ \sin (50^o + 85^o) + \sin (50^o - 85^o) \Big]$

$=$ $\frac{1}{2}$ $\Big[ \sin 135^o + \sin ( - 35^o) \Big]$

$=$ $\frac{1}{2}$ $\Big[ \sin (90^o + 45^o) - \sin 35^o \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos 45^o - \sin 35^o \Big]$

$=$ $\frac{1}{2}$ $\Big[$ $\frac{1}{\sqrt{2}}$ $- \sin 35^o \Big]$

$=$ $\frac{1 - \sqrt{2} \sin 35^o}{2 \sqrt{2}}$ $=$ RHS. Hence Proved.

ii) LHS $= \sin 25^o \cos 115^o$

$=$ $\frac{1}{2}$ $\Big[ 2 \sin 25^o \cos 115^o \Big]$

$=$ $\frac{1}{2}$ $\Big[ \sin (25^o + 115^o) + \sin (25^o - 115^o) \Big]$

$=$ $\frac{1}{2}$ $\Big[ \sin 140^o + \sin ( - 90^o) \Big]$

$=$ $\frac{1}{2}$ $\Big[ \sin (90^o + 50^o) - \sin 90^o \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos 50^o - 1 \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos (90^o - 40^o) - 1 \Big]$

$=$ $\frac{1}{2}$ $\Big[ \sin 40^o - 1 \Big]$  RHS. Hence Proved.

$\\$

Question 4: Prove that $4 \cos x \cos \Big($ $\frac{\pi}{3}$ $+x \Big) \cos \Big($ $\frac{\pi}{3}$ $-x \Big) = \cos 3x$

LHS $= 4 \cos x \cos \Big($ $\frac{\pi}{3}$ $+ x \Big) \cos \Big($ $\frac{\pi}{3}$ $- x \Big)$

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

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

$= 2 \cos x \Big[ \cos$ $\frac{2\pi}{3}$ $+ \cos 2x \Big]$

$= 2 \cos x \Big[ \cos \Big($ $\frac{\pi}{2}$ $+$ $\frac{\pi}{6}$ $\Big) + \cos 2x \Big]$

$= 2 \cos x \Big[ - \sin$ $\frac{\pi}{6}$ $+ \cos 2x \Big]$

$= 2 \cos x \Big[ -$ $\frac{1}{2}$ $+ \cos 2x \Big]$

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

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

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

$= \cos 3x =$ RHS

$\\$

Question 5: Prove that:

i) $\cos 10^o \cos 30^o \cos 50^o \cos 70^o =$ $\frac{3}{16}$   ii) $\cos 40^o \cos 80^o \cos 160^o = -$ $\frac{1}{8}$

iii) $\sin 20^o \sin 40^o \sin 80^o =$ $\frac{\sqrt{3}}{8}$   iv) $\cos 20^o \cos 40^o \cos 80^o =$ $\frac{1}{8}$

v) $\tan 20^o \tan 40^o \tan 60^o \tan 80^o = 3$    vi) $\tan 20^o \tan 30^o \tan 40^o \tan 80^o = 1$

vii) $\sin 10^o \sin 50^o \sin 60^o \sin 70^o =$ $\frac{\sqrt{3}}{16}$   viii) $\sin 20^o \sin 40^o \sin 60^o \sin 80^o =$ $\frac{3}{16}$

i) LHS $= \cos 10^o \cos 30^o \cos 50^o \cos 70^o$

$=$ $\frac{\sqrt{3}}{2}$ $\Big[ (\cos 10^o \cos 50^o) \cos 70^o \Big]$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ (2 \cos 50^o \cos 10^o) \cos 70^o \Big]$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ \Big( \cos (50^o+10^o) + \cos (50^o-10^o) \Big) \cos 70^o \Big]$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ ( \cos 60^o + \cos 40^o ) \cos 70^o \Big]$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ ( \frac{1}{2} + \cos 40^o ) \cos 70^o \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\cos 70^o +$ $\frac{\sqrt{3}}{4}$ $\cos 40^o \cos 70^o$

$=$ $\frac{\sqrt{3}}{8}$ $\cos 70^o +$ $\frac{\sqrt{3}}{8}$ $(2\cos 40^o \cos 70^o)$

$=$ $\frac{\sqrt{3}}{8}$ $\cos 70^o +$ $\frac{\sqrt{3}}{8}$ $( \cos (70^o+40^o) + \cos (70^o-40^o) )$

$=$ $\frac{\sqrt{3}}{8}$ $\cos 70^o +$ $\frac{\sqrt{3}}{8}$ $\cos 110 + \frac{\sqrt{3}}{8} \cos 30^o$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ \cos 70^o + \cos 110 + \cos 30 \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ \cos 70^o + \cos (180 - 70) + \cos 30 \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ \cos 70^o - \cos 70^o +$ $\frac{\sqrt{3}}{2}$ $\Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[$ $\frac{\sqrt{3}}{2}$ $\Big]$

$=$ $\frac{3}{16}$ $=$ RHS. Hence proved.

ii) LHS $= \cos 40^o \cos 80^o \cos 160^o$

$= \frac{1}{2}$ $\Big[ 2 \cos 40^o \cos 80^o \cos 160^o \Big]$

$= \frac{1}{2}$ $\Big[ \cos 80^o ( 2 \cos 160^o \cos 40^o) \Big]$

$= \frac{1}{2}$ $\Big[ \cos 80^o [ \cos ( 160^o+40^o) + \cos ( 160^o-40^o) ] \Big]$

$= \frac{1}{2}$ $\Big[ \cos 80^o ( \cos 200^o + \cos 120^o ) \Big]$

$= \frac{1}{2}$ $\Big[ \cos 80^o [ \cos ( 180^o+20^o) + \cos ( 180^o-60^o) ] \Big]$

$= \frac{1}{2}$ $\Big[ \cos 80^o ( \cos 20^o + \cos 60^o ) \Big]$

$= -$ $\frac{1}{2}$ $\cos 80^o \cos 20^o +$ $\frac{1}{2}$ $\cos 80^o \cos 60^o$

$= -$ $\frac{1}{4}$ $[ 2 \cos 80^o \cos 20^o + 2 \cos 80^o \cos 60^o ]$

$= -$ $\frac{1}{4}$ $[ \cos (80^o+20^o) + \cos (80^o-20^o)+ 2 \cos 80^o \times$ $\frac{1}{2}$ $]$

$= -$ $\frac{1}{4}$ $[ \cos 100^o + \cos 60^o+ \cos 80^o ]$

$= -$ $\frac{1}{4}$ $[ \cos (180^o - 80^o) + \cos 60^o+ \cos 80^o ]$

$= -$ $\frac{1}{4}$ $[ -\cos 80^o + \cos 60^o+ \cos 80^o ]$

$= -$ $\frac{1}{4}$ $[ \cos 60^o ]$

$= -$ $\frac{1}{8}$ $=$ RHS. Hence proved.

iii) LHS $= \sin 20^o \sin 40^o \sin 80^o$

$=$ $\frac{1}{2}$ $\Big[ 2 \sin 20^o \sin 40^o \Big] \sin 80^o$

$=$ $\frac{1}{2}$ $\Big[ \cos (40^o-20^o) - \cos (40^o+20^o) \Big] \sin 80^o$

$=$ $\frac{1}{2}$ $\Big[ \cos 20^o - \cos 60^o \Big] \sin 80^o$

$=$ $\frac{1}{2}$ $\Big[ \cos 20^o -$ $\frac{1}{2}$ $\Big] \sin 80^o$

$=$ $\frac{1}{2}$ $\Big[ \cos 20^o \sin 80^o \Big] -$ $\frac{1}{4}$ $\sin 80^o$

$=$ $\frac{1}{4}$ $\Big[ 2 \cos 20^o \sin 80^o \Big] -$ $\frac{1}{4}$ $\sin 80^o$

$=$ $\frac{1}{4}$ $\Big[ \sin (80^o+20^o) + \sin (80^o-20^o) \Big] -$ $\frac{1}{4}$ $\sin 80^o$

$=$ $\frac{1}{4}$ $\Big[ \sin 100^o + \sin 60^o \Big] -$ $\frac{1}{4}$ $\sin 80^o$

$=$ $\frac{1}{4}$ $\Big[ \sin (180^o - 80^o) + \sin 60^o \Big] -$ $\frac{1}{4}$ $\sin 80^o$

$=$ $\frac{1}{4}$ $\sin 80^o +$ $\frac{\sqrt{3}}{8}$ $-$ $\frac{1}{4}$ $\sin 80^o$

$=$ $\frac{\sqrt{3}}{8}$ $=$ RHS. Hence proved.

iv) LHS $= \cos 20^o \cos 40^o \cos 80^o$

$= \frac{1}{2}$ $\Big[ 2 \cos 20^o \cos 40^o \Big] \cos 80^o$

$= \frac{1}{2}$ $\Big[ \cos ( 40^o+ 20^o) + \cos ( 40^o - 20^o) \Big] \cos 80^o$

$= \frac{1}{2}$ $\Big[ \cos 60^o + \cos 20^o \Big] \cos 80^o$

$= \frac{1}{2}$ $\Big[$ $\frac{1}{2}$ $+ \cos 20^o \Big] \cos 80^o$

$= \frac{1}{4}$ $\Big[ \cos 80^o + 2\cos 20^o \cos 80^o \Big]$

$= \frac{1}{4}$ $\Big[ \cos 80^o + \cos ( 80^o + 20^o) + \cos ( 80^o - 20^o) \Big]$

$= \frac{1}{4}$ $\Big[ \cos 80^o + \cos 100^o + \cos 60^o \Big]$

$= \frac{1}{4}$ $\Big[ \cos 80^o + \cos (180^o - 80^o) + \cos 60^o \Big]$

$= \frac{1}{4}$ $\Big[ \cos 80^o - \cos 80^o +$ $\frac{1}{2}$ $\Big]$

$= \frac{1}{8}$ $=$ RHS. Hence Proved.

v) LHS $= \tan 20^o \tan 40^o \tan 60^o \tan 80^o$

$= \sqrt{3} [ \tan 20^o \tan 40^o \tan 80^o ]$

$= \sqrt{3} \Big[$ $\frac{\sin 20^o \sin 40^o \sin 80^o }{\cos 20^o \cos 40^o \cos 80^o }$ $\Big]$

$= \sqrt{3} \Big[$ $\frac{ [ \cos ( 40^o-20^o) - \cos ( 40^o + 20^o) ] \sin 80^o }{ [ \cos ( 20^o+40^o) + \cos ( 40^o - 20^o) ] \cos 80^o }$ $\Big]$

$= \sqrt{3} \Big[$ $\frac{ [ \cos 20^o - \cos 60^o ] \sin 80^o }{ [ \cos 60^o + \cos 20^o ] \cos 80^o }$ $\Big]$

$= \sqrt{3} \Big[$ $\frac{ [ 2 \cos 20^o - 1 ] \sin 80^o }{ [ 1 + 2 \cos 20^o ] \cos 80^o }$ $\Big]$

$= \sqrt{3} \Big[$ $\frac{ 2 \cos 20^o \sin 80^o - \sin 80^o }{ \cos 80^o + 2 \cos 20^o \cos 80^o }$ $\Big]$

$= \sqrt{3} \Big[$ $\frac{ \sin (80+20) + \sin (80^o-20^o) - \sin 80^o }{ \cos 80^o + \cos (80^o+20^o) + \cos (80^o-20^o) }$ $\Big]$

$= \sqrt{3} \Big[$ $\frac{ \sin 100^o + \sin 60^o - \sin 80^o }{ \cos 80^o + \cos 100^o + \cos 60^o }$ $\Big]$

$= \sqrt{3} \Big[$ $\frac{ \sin (180^o-80^o) + \sin 60^o - \sin 80^o }{ \cos 80^o + \cos (180^o-80^o) + \cos 60^o }$ $\Big]$

$= \sqrt{3} \Big[$ $\frac{ \sin 80^o + \sin 60^o - \sin 80^o }{ \cos 80^o - \cos 80^o + \cos 60^o }$ $\Big]$

$= \sqrt{3} \tan 60^o$

$= \sqrt{3} \times \sqrt{3} = 3 =$ RHS. Hence proved.

vi) LHS $= \tan 20^o \tan 30^o \tan 40^o \tan 80^o$

$=$ $\frac{1}{\sqrt{3}}$ $[ \tan 20^o \tan 40^o \tan 80^o ]$

$=$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{\sin 20^o \sin 40^o \sin 80^o }{\cos 20^o \cos 40^o \cos 80^o }$ $\Big]$

$=$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{ (2 \sin 20^o \sin 40^o) \sin 80^o }{ (2 \cos 20^o \cos 40^o) \cos 80^o }$ $\Big]$

$=$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{ [ \cos ( 40^o-20^o) - \cos ( 40^o + 20^o) ] \sin 80^o }{ [ \cos ( 20^o+40^o) + \cos ( 40^o - 20^o) ] \cos 80^o }$ $\Big]$

$=$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{ [ \cos 20^o - \cos 60^o ] \sin 80^o }{ [ \cos 60^o + \cos 20^o ] \cos 80^o }$ $\Big]$

$=$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{ [ 2 \cos 20^o - 1 ] \sin 80^o }{ [ 1 + 2 \cos 20^o ] \cos 80^o }$ $\Big]$

$=$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{ 2 \cos 20^o \sin 80^o - \sin 80^o }{ \cos 80^o + 2 \cos 20^o \cos 80^o }$ $\Big]$

$=$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{ \sin (80+20) + \sin (80^o-20^o) - \sin 80^o }{ \cos 80^o + \cos (80^o+20^o) + \cos (80^o-20^o) }$ $\Big]$

$=$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{ \sin 100^o + \sin 60^o - \sin 80^o }{ \cos 80^o + \cos 100^o + \cos 60^o }$ $\Big]$

$=$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{ \sin (180^o-80^o) + \sin 60^o - \sin 80^o }{ \cos 80^o + \cos (180^o-80^o) + \cos 60^o }$ $\Big]$

$=$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{ \sin 80^o + \sin 60^o - \sin 80^o }{ \cos 80^o - \cos 80^o + \cos 60^o }$ $\Big]$

$=$ $\frac{1}{\sqrt{3}}$ $\tan 60^o$

$=$ $\frac{1}{\sqrt{3}}$ $\times \sqrt{3} = 1 =$ RHS. Hence proved.

vii) LHS $= \sin 10^o \sin 50^o \sin 60^o \sin 70^o$

$=$ $\frac{\sqrt{3}}{2}$ $\Big[ \sin 10^o \sin 50^o \sin 70^o \Big]$

$=$ $\frac{\sqrt{3}}{2}$ $\Big[ \sin (90^o - 80^o) \sin (90^o - 40^o) \sin (90^o - 20^o) \Big]$

$=$ $\frac{\sqrt{3}}{2}$ $\Big[ \cos 80^o \cos 40^o \cos 20^o \Big]$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ \cos 80^o (2 \cos 40^o \cos 20^o) \Big]$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ \cos (40^o + 20^o) + \cos (40^o - 20^o) \Big] \cos 80^o$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ \cos 60^o+ \cos 20^o \Big] \cos 80^o$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ \cos 80^o+ 2 \cos 20^o \cos 80^o \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ \cos 80^o+ \cos ( 80^o + 20^o) + \cos (80^o - 20^o) \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ \cos 80^o+ \cos 100^o + \cos 60^o \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ \cos 80^o+ \cos (180 - 80^o) + \cos 60^o \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ \cos 80^o - \cos 80^o + \cos 60^o \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\cos 60^o$

$=$ $\frac{\sqrt{3}}{16}$ $=$ RHS. Hence proved.

viii) LHS $= \sin 20^o \sin 40^o \sin 60^o \sin 80^o$

$=$ $\frac{\sqrt{3}}{2}$ $\Big[ \sin 20^o \sin 40^o \sin 80^o \Big]$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ (2 \sin 20^o \sin 40^o) \sin 80^o \Big]$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ \cos (40^o - 20^o) - \cos (40^o + 20^o) \Big] \sin 80^o$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ - \cos 60^o+ \cos 20^o \Big] \sin 80^o$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ - \frac{1}{2}+ \cos 20^o \Big] \sin 80^o$

$=$ $\frac{\sqrt{3}}{4}$ $\Big[ - \frac{1}{2} \sin 80^o + \cos 20^o \sin 80^o \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ - \sin 80^o + 2 \cos 20^o \sin 80^o \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ - \sin 80^o + \sin (80^o + 20^o) + \sin ( 80^o - 20^o) \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ - \sin 80^o + \sin 100^o + \sin 60^o \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ - \sin 80^o + \sin (180^o - 80^o) + \sin 60^o \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\Big[ - \sin 80^o + \sin 80^o + \sin 60^o \Big]$

$=$ $\frac{\sqrt{3}}{8}$ $\sin 60^o$

$=$ $\frac{\sqrt{3}}{16}$ $=$ RHS. Hence proved.

$\\$

Question 6: Show that:

i) $\sin A \sin (B-C) + \sin B \sin (C-A) + \sin C \sin (A-B) = 0$

ii) $\sin (B-C) \cos (A - D) + \sin (C-A) \cos (B-D) + \sin (A-B) \cos (C-D) = 0$

i) LHS $= \sin A \sin (B-C) + \sin B \sin (C-A) + \sin C \sin (A-B)$

$= \frac{1}{2} \Big[ 2\sin A \sin (B-C) + 2\sin B \sin (C-A) + 2\sin C \sin (A-B) \Big]$

$= \frac{1}{2} \Big[ \cos ( A - B + C) - \cos ( A + B - C) + \cos ( B - C + A ) - \cos ( B + C - A ) + \cos ( C - A + B ) - \cos ( C + A - B) \Big]$

$= \frac{1}{2} \Big[ \cos ( A - B + C) - \cos ( A - B + C) - \cos ( A + B - C ) + \cos ( A + B - C ) - \cos ( B + C - A ) + \cos ( B + C - A ) \Big]$

$= \frac{1}{2} \Big[ 0 \Big] = 0 =$ RHS. Hence proved.

ii) LHS $= \sin (B-C) \cos (A - D) + \sin (C-A) \cos (B-D) + \sin (A-B) \cos (C-D)$

$= \frac{1}{2} \Big[ 2\sin (B-C) \cos (A - D) + 2\sin (C-A) \cos (B-D) + 2\sin (A-B) \cos (C-D) \Big]$

$= \frac{1}{2} \Big[ \sin ( B - C + A - D) + \sin ( B - C - A + D) + \sin ( C - A + B - D) + \sin ( C - A - B + D) + \sin ( A - B + C - D) + \sin ( A - B - C + D) \Big]$

$= \frac{1}{2} \Big[ \sin ( A + B -C-D) + \sin ( B + D - C - A) + \sin ( -( A + D - B - C) ) + \sin (- ( A + B - C - D) ) + \sin ( -( B + D - A - C) ) + \sin ( A + D - B - C) \Big]$

$= \frac{1}{2} \Big[ \sin ( A + B -C-D) + \sin ( B + D - C - A) - \sin ( A + D - B - C) - \sin ( A + B - C - D) - \sin ( B + D - A - C) + \sin ( A + D - B - C) \Big]$

$= \frac{1}{2} \Big[ 0 \Big] = 0 =$ RHS. Hence proved.

$\\$

Question 7: Prove that: $\tan x \tan \Big($ $\frac{\pi}{3}$ $- x \Big) \tan \Big($ $\frac{\pi}{3}$ $+ x \Big) = \tan 3x$

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

$=$ $\frac{2 \sin x \sin ( 60-x) \sin ( 60+x)}{2 \cos x \cos ( 60-x) \cos ( 60+x)}$

$=$ $\frac{\sin x }{\cos x} \Big[ \frac{2 \sin ( 60-x) \sin ( 60+x)}{2 \cos ( 60-x) \cos ( 60+x)}$ $\Big]$

$=$ $\frac{\sin x }{\cos x} \Big[ \frac{\cos ( 60-x- 60 - x) - \cos ( 60 - x + 60 + x) }{\cos ( 60 - x + 60 + x) + \cos ( 60 - x - 60 - x)}$ $\Big]$

$=$ $\frac{\sin x }{\cos x} \Big[ \frac{\cos ( -2x) - \cos 120^o }{\cos 120^o + \cos ( -2x)}$ $\Big]$

$=$ $\frac{\sin x }{\cos x} \Big[ \frac{\cos 2x - \cos 120^o }{\cos 120^o + \cos 2x }$ $\Big]$

$=$ $\frac{\sin x }{\cos x} \Big[ \frac{\cos 2x - \cos (90^o + 30^o) }{\cos (90^o + 30^o) + \cos 2x }$ $\Big]$

$=$ $\frac{\sin x }{\cos x} \Big[ \frac{\cos 2x + \sin 30^o }{- \sin 30^o + \cos 2x }$ $\Big]$

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

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

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

$=$ $\frac{\sin 3x - \sin x + \sin x}{\cos 3x + \cos x - \cos x}$

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

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

$\\$

Question 8: If $\alpha + \beta =$ $\frac{\pi}{2}$, show that the maximum value of $\cos \alpha \cos \beta$ is $\frac{1}{2}$

Let $x = \cos \alpha \cos \beta$

$\Rightarrow x =$ $\frac{1}{2}$ $\Big[ 2 \cos \alpha \cos \beta \Big]$

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

$\Rightarrow x =$ $\frac{1}{2}$ $\Big[ \cos 90^o + \cos (\alpha - \beta) \Big]$

$\Rightarrow x =$ $\frac{1}{2}$ $\Big[ 0+ \cos (\alpha - \beta) \Big]$

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

Now we know$-1 \leq \cos (\alpha - \beta) \leq 1$

$\Rightarrow -$ $\frac{1}{2}$ $\leq$ $\frac{1}{2}$ $\cos (\alpha - \beta) \leq$ $\frac{1}{2}$

$\Rightarrow -$ $\frac{1}{2}$ $\leq x \leq$ $\frac{1}{2}$

Therefore the max value of $x$ is $\frac{1}{2}$