Question

If $A$ is not an integeral multiple of $\pi$, find k such that $\cos A.\cos 2A.\cos 4A.\cos 8A=\frac{\sin 16A}{k\sin A}$.

Answer 1

We have,

$\cos A.\cos 2A.\cos 4A.\cos 8A$

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

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

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

$=\frac{1}{4\sin A}\left(\sin 4A\cos 4A\cos 8A\right)$

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

$=\frac{1}{8\sin A}\left(\sin 8A\cos 8A\right)$

$=\frac{1}{16\sin A}\left(2\sin 8A\cos 8A\right)$

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

$=\frac{\sin 16A}{16\sin A}$

So, the value of $k$ is $16$.

Hence, this is the answer.