Question 1: $\sin^2$ $\frac{2\pi}{5}$ $- \sin^2$ $\frac{\pi}{3}$ $=$ $\frac{\sqrt{5}-1}{8}$

LHS $= \sin^2 72^o - \sin^2 60^o$

$= \sin^2 ( 90^o-18^o) - \Big($ $\frac{\sqrt{3}}{2}$ $\Big)^2$

$= \cos^2 18^o -$ $\frac{3}{4}$

$\because \cos 18^o =$ $\frac{\sqrt{10+2\sqrt{5}}}{4}$

$= \Big($ $\frac{\sqrt{10+2\sqrt{5}}}{4}$ $\Big)^2 -$ $\frac{3}{4}$

$=$ $\frac{10+2\sqrt{5}}{16}$ $-$ $\frac{3}{4}$

$=$ $\frac{10 + 2 \sqrt{5}-12}{16}$ $=$ $\frac{2\sqrt{5} - 2 }{16}$ $=$ $\frac{\sqrt{5} - 1}{8}$ $=$ RHS. Hence proved.

$\\$

Question 2: $\sin^2 24^o - \sin^2 6^o =$ $\frac{\sqrt{5}-1}{8}$

LHS $= \sin^2 24^o - \sin^2 6^o$

$\because \sin ( A+B) \sin ( A - B) = \sin^2 A - \sin^2 B$

$= \sin ( 24^o +6^o) \sin (24^o-6^o)$

$= \sin 30^o \sin 18^o$

$\because \sin 18^o =$ $\frac{\sqrt{5}-1}{4}$

$=$ $\frac{1}{2}$ $\Big($ $\frac{\sqrt{5}-1}{4}$ $\Big)$

$=$ $\frac{\sqrt{5}-1}{8}$ $=$ RHS. Hence proved.

$\\$

Question 3: $\sin^2 42^o - \cos^2 78^o =$ $\frac{\sqrt{5}+1}{8}$

LHS $= \sin^2 42^o - \cos^2 78^o$

$= \sin^2 (90^o-48^o) - \cos^2 (90^o-12^o)$

$= \cos^2 48^o - \sin^2 12^o$

$= \cos ( 48^o + 12^o) \cos ( 48^o - 12^o)$

$= \cos 60^o \cos 36^o$

$\because \cos 36^o =$ $\frac{\sqrt{5}+1}{4}$

$=$ $\frac{1}{2}$ $\times \frac{\sqrt{5}+1}{4}$

$=$ $\frac{\sqrt{5}+1}{8}$ $=$ RHS. Hence proved.

$\\$

Question 4: $\cos 78^o \cos 42^o \cos 36^o =$ $\frac{1}{8}$

LHS $=$ $\cos 78^o \cos 42^o \cos 36^o$

$=$ $\frac{1}{2}$ $[ 2 \cos 78^o \cos 42^o ] \cos 36^o$

$=$ $\frac{1}{2}$ $[ \cos 120^o + \cos 36^o ] \cos 36^o$

$=$ $\frac{1}{2}$ $\Big[ -$ $\frac{1}{2}$ $+$ $\frac{\sqrt{5}+ 1}{4}$ $\Big] \Big($ $\frac{\sqrt{5}+ 1}{4}$ $\Big)$

$=$ $\frac{1}{2}$ $\Big[$ $\frac{\sqrt{5}- 1}{4}$ $\Big] \Big($ $\frac{\sqrt{5}+ 1}{4}$ $\Big)$

$=$ $\frac{1}{2}$ $\times$ $\frac{5-1}{16}$ $=$ $\frac{1}{8}$ $=$ RHS

$\\$

Question 5: $\cos$ $\frac{\pi}{15}$ $\cos$ $\frac{2\pi}{15}$ $\cos$ $\frac{4\pi}{15}$ $\cos$ $\frac{7\pi}{15}$ $=$ $\frac{1}{16}$

LHS $= \cos$ $\frac{\pi}{15}$ $\cos$ $\frac{2\pi}{15}$ $\cos$ $\frac{4\pi}{15}$ $\cos$ $\frac{7\pi}{15}$

$=$ $\frac{1}{2\sin \frac{\pi}{15}}$ $\Big[ 2 \sin$ $\frac{\pi}{15}$ $\cos$ $\frac{\pi}{15}$ $\Big] \cos$ $\frac{2\pi}{15}$ $\cos$ $\frac{4\pi}{15}$ $\cos$ $\frac{7\pi}{15}$

$=$ $\frac{1}{4\sin \frac{\pi}{15}}$ $\Big[ 2 \sin$ $\frac{2\pi}{15}$ $\cos$ $\frac{2\pi}{15}$ $\Big] \cos$ $\frac{4\pi}{15}$ $\cos$ $\frac{7\pi}{15}$

$=$ $\frac{1}{8\sin \frac{\pi}{15}}$ $\Big[ 2 \sin$ $\frac{4\pi}{15}$ $\cos$ $\frac{4\pi}{15}$ $\Big] \cos$ $\frac{7\pi}{15}$

$=$ $\frac{1}{16\sin \frac{\pi}{15}}$ $\Big[ 2 \sin$ $\frac{8\pi}{15}$ $\cos$ $\frac{7\pi}{15}$ $\Big]$

$=$ $\frac{1}{16\sin \frac{\pi}{15}}$ $\Big[ \sin \big($ $\frac{8\pi}{15}$ $+$ $\frac{7\pi}{15}$ $\big) + \sin \big($ $\frac{8\pi}{15}$ $-$ $\frac{7\pi}{15}$ $\big) \Big]$

$=$ $\frac{1}{16\sin \frac{\pi}{15}}$ $\Big[ \sin \pi + \sin$ $\frac{\pi}{15}$ $\Big]$

$=$ $\frac{1}{16\sin \frac{\pi}{15}}$ $\Big[ \sin$ $\frac{\pi}{15}$ $\Big]$

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

$\\$

Question 6: $\cos 6^o \cos 42^o \cos 66^o \cos 78^o =$ $\frac{1}{16}$

LHS $= \cos 6^o \cos 42^o \cos 66^o \cos 78^o$

$=$ $\frac{1}{4}$ $( 2 \cos 6^o \cos 66^o)( 2 \cos 42^o \cos 78^o)$

$=$ $\frac{1}{4}$ $( \cos 72^o + \cos 60^o) ( \cos 120^o + \cos 36^o)$

$=$ $\frac{1}{4}$ $\big(\sin 18^o +$ $\frac{1}{2}$ $\big) \big( -$ $\frac{1}{2}$ $+$ $\frac{\sqrt{5}+ 1}{4}$ $\big )$

$=$ $\frac{1}{4}$ $\big ($ $\frac{\sqrt{5}-1}{4}$ $+$ $\frac{1}{2}$ $\big) \big( +$ $\frac{\sqrt{5}+ 1}{4}$ $-$ $\frac{1}{2}$ $\big )$

$=$ $\frac{1}{4}$ $\big ($ $\frac{\sqrt{5}+1}{4}$ $\big) \big($ $\frac{\sqrt{5}-1}{4}$ $\big )$

$=$ $\frac{1}{4}$ $\times$ $\frac{4}{16}$ $=$ $\frac{1}{16}$ $=$ RHS. Hence proved.

$\\$

Question 7: $\sin 6^o \sin 42^o \sin 66^o \sin 78^o =$ $\frac{1}{16}$

LHS $= \sin 6^o \sin 42^o \sin 66^o \sin 78^o$

$=$ $\frac{1}{4}$ $( 2 \sin 6^o \sin 66^o)( 2 \sin 42^o \sin 78^o)$

$=$ $\frac{1}{4}$ $( \cos 60^o - \cos 72^o) ( \cos 36^o - \cos 120^o)$

$=$ $\frac{1}{4}$ $\big($ $\frac{1}{2}$ $- \sin 18^o \big) \big($ $\frac{\sqrt{5}+ 1}{4}$ $+$ $\frac{1}{2}$ $\big )$

$=$ $\frac{1}{4}$ $\big($ $\frac{1}{2}$ $-$ $\frac{\sqrt{5}- 1}{4}$ $\big) \big($ $\frac{\sqrt{5}+ 1}{4}$ $+$ $\frac{1}{2}$ $\big )$

$=$ $\frac{1}{4}$ $\big ($ $\frac{3 - \sqrt{5}}{4}$ $\big) \big($ $\frac{3 + \sqrt{5}}{4}$ $\big )$

$=$ $\frac{1}{4}$ $\times$ $\frac{4}{16}$ $=$ $\frac{1}{16}$ $=$ RHS. Hence proved.

$\\$

Question 8: $\cos 36^o \cos 42^o \cos 60^o \cos 78^o =$ $\frac{1}{16}$

LHS $= \cos 36^o \cos 42^o \cos 60^o \cos 78^o$

$=$ $\frac{1}{2}$ $\cos 36^o \cos 60^o ( 2 \cos 42^o \cos 78^o)$

$=$ $\frac{1}{2}$ $\Big($ $\frac{\sqrt{5}+1}{4}$ $\Big) \times$ $\frac{1}{2}$ $( \cos 120^o + \cos 36^o)$

$= \Big($ $\frac{\sqrt{5}+1}{16}$ $\Big) \Big( -$ $\frac{1}{2}$ $+$ $\frac{\sqrt{5}+1}{4}$ $\Big)$

$= \Big($ $\frac{\sqrt{5}+1}{16}$ $\Big) \Big($ $\frac{\sqrt{5}-1}{4}$ $\Big)$

$=$ $\frac{4}{64}$ $=$ $\frac{1}{16}$ $=$ RHS. Hence proved.

$\\$

Question 9: $\sin$ $\frac{\pi}{5}$ $\sin$ $\frac{2\pi}{5}$ $\sin$ $\frac{3\pi}{5}$ $\sin$ $\frac{4\pi}{5}$ $=$ $\frac{5}{16}$

LHS $= \sin 36^o \sin 72^o \sin 108^o \sin 144^o$

$= \sin 36^o \sin 72^o \sin (180^o - 72^o) \sin ( 180^o - 36^o)$

$= \sin 36^o \sin 72^o \sin 72^o \sin 36^o$

$= ( 2 \sin 36^o \sin 72^o )^2$

$= ( 2 \sin 36^o \cos 18^o )^2$

$= \Big($ $\frac{\sqrt{10 - 2 \sqrt{5}}}{4}$ $\times$ $\frac{\sqrt{10 + 2 \sqrt{5}}}{4}$ $\Big)$

$=$ $\frac{100 - 20}{16 \times 16}$ $=$ $\frac{80}{16 \times 16}$ $=$ $\frac{5}{16}$ $=$ RHS. Hence proved.

$\\$

Question 10: $\cos$ $\frac{\pi}{15}$ $\cos$ $\frac{2\pi}{15}$ $\cos$ $\frac{3\pi}{15}$ $\cos$ $\frac{4\pi}{15}$ $\cos$ $\frac{5\pi}{15}$ $\cos$ $\frac{6\pi}{15}$ $\cos$ $\frac{7\pi}{15}$$=$ $\frac{1}{128}$

LHS $=$ $\cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$

$=$ $\frac{1}{2 \sin \frac{\pi}{15}}$ $\Big[ 2 \sin \frac{\pi}{15} \cos \frac{\pi}{15} \Big] \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$

$=$ $\frac{1}{4 \sin \frac{\pi}{15}}$ $\Big[ 2 \sin \frac{2\pi}{15} \cos \frac{2\pi}{15} \Big] \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$

$=$ $\frac{1}{8 \sin \frac{\pi}{15}}$ $\Big[ 2 \sin \frac{4\pi}{15} \cos \frac{4\pi}{15} \Big] \cos \frac{3\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$

$=$ $\frac{1}{16 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}}$ $\Big[ 2 \sin \frac{3\pi}{15} \cos \frac{3\pi}{15} \Big] \sin \frac{8\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$

$=$ $\frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}}$ $\Big[ 2 \sin \frac{6\pi}{15} \cos \frac{6\pi}{15} \Big] \sin \frac{8\pi}{15} \cos \frac{7\pi}{15}$

$=$ $\frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}}$ $\Big[ \sin \frac{12\pi}{15} \sin \frac{8\pi}{15} \cos \frac{7\pi}{15} \Big]$

$=$ $\frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}}$ $\Big[ \sin ( \pi - \frac{3\pi}{15}) \sin (\pi - \frac{7\pi}{15} ) \cos \frac{7\pi}{15} \Big]$

$=$ $\frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}}$ $\Big[ \sin \frac{3\pi}{15} \sin \frac{7\pi}{15} \cos \frac{7\pi}{15} \Big]$

$=$ $\frac{1}{64 \sin \frac{\pi}{15}}$ $\Big[ 2 \sin \frac{7\pi}{15} \cos \frac{7\pi}{15} \Big]$

$=$ $\frac{1}{64 \sin \frac{\pi}{15}}$ $\Big[ \sin \frac{14\pi}{15} \Big]$

$=$ $\frac{1}{64 \sin \frac{\pi}{15}}$ $\Big[ \sin ( \pi - \frac{\pi}{15} ) \Big]$

$=$ $\frac{1}{64 \sin \frac{\pi}{15}}$ $\Big[ \sin \frac{\pi}{15} \Big]$

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