Question

The value of $(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos \frac{5\pi }{8})(1+\cos \frac{7\pi }{8})=\frac{a}{256}$. Find a.

Answer 1

$(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos \frac{5\pi }{8})(1+\cos \frac{7\pi }{8})=\frac{a}{256}$.

$(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos (\pi -\frac{3\pi }{8}))(1+\cos (\pi -\frac{\pi }{8}\left)\right)=\frac{a}{256}$.

$(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1-\cos \frac{3\pi }{8})(1-\cos (\frac{\pi }{8}\left)\right)=\frac{a}{256}$.

$(1-{\cos }^2\frac{\pi }{8})$ $(1-{\cos }^2\frac{3\pi }{8})=\frac{a}{256}$

${\sin }^2\frac{\pi }{8}{\sin }^2\frac{3\pi }{8}=\frac{a}{256}$

${\sin }^2\frac{\pi }{8}{\sin }^2(\frac{\pi }{4}-\frac{\pi }{8})=\frac{a}{256}$

${\sin }^2\frac{\pi }{8}{\cos }^2\frac{\pi }{8}=\frac{a}{256}$

$4{\sin }^2\frac{\pi }{8}{\cos }^2\frac{\pi }{8}=\frac{4a}{256}$

${\sin }^2\frac{\pi }{4}=\frac{a}{64}$

(As $2\sin A\cos A=\sin 2A$)

$\frac{1}{2}=\frac{a}{64}$

$a=32$

Hence the answer is $a=32$