Question

Prove that : $\cos a\cos 2a\cdot \cos 4a\cos 8a=\frac{\sin 16a}{16\sin a}$

Answer 1

LHS :

$\cos a\cos 2a.\cos 4a\cos 8a$

$=\frac{1}{2\sin a}\left[\right(2\sin a\cos a\left)\cos 2a\cos 4a\cos 8a\right]$

$=\frac{1}{2\sin a}\left[\sin 2a\cos 2a\cos 4a\cos 8a\right]$

$=\frac{1}{4\sin a}\left[\right(2\sin 2a\cos 2a\left)\cos 4a\cos 8a\right]$

$=\frac{1}{4\sin a}\left[\sin 4a\cos 4a\cos 8a\right]$

$=\frac{1}{8\sin a}\left[\right(2\sin 4a\cos 4a\left)\cos 8a\right]$

$=\frac{1}{8\sin a}\left[\sin 8a\cos 8a\right]$

$=\frac{1}{16\sin a}\left[2\sin 8a\cos 8a\right]$

$=\frac{1}{16\sin a}\left(\sin 16a\right)$

$=\frac{\sin 16a}{16\sin a}$ = RHS