Question

Prove that $(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos \frac{5\pi }{8})(1+\cos \frac{7\pi }{8})=\frac{1}{8}$

Answer 1

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

$\Rightarrow LHS=(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos \frac{5\pi }{8})(1+\cos \frac{7\pi }{8})$

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

$=(1+\cos \frac{\pi }{8})(1+\cos (\pi -\frac{\pi }{8}))(1+\cos \frac{3\pi }{8})(1+\cos (\pi -\frac{3\pi }{8}))$

$=(1+\cos \frac{\pi }{8})(1-\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1-\cos \frac{3\pi }{8})$

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

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

Multiply and divide by $4,$ we get

$=\frac{1}{4}\left(2{\sin }^2\frac{\pi }{8}\right)\left(2{\sin }^2\frac{3\pi }{8}\right)$

$=\frac{1}{4}(1-\cos \frac{\pi }{4})(1+\cos \frac{3\pi }{4})$

$[\because 1-\cos 2\theta =2{\sin }^2\frac{2\theta }{2}]$

$[i.e.(1-\cos \theta )=2{\sin }^2\frac{\theta }{2}]$

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

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

$=\frac{1}{4}\times \frac{1}{2}$

$=\frac{1}{8}$

Hence, the answer is $L.H.S=R.H.S.$