Question

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

Answer 1

We have $\cos \frac{7\pi }{8}=\cos (\pi -\frac{\pi }{8})=-\cos \frac{\pi }{8}$
$\cos \frac{5\pi }{8}=\cos (\pi -\frac{3\pi }{8})=-\cos \frac{3\pi }{8}$
L.H.S. $(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos \frac{5\pi }{8})(1+\cos \frac{7\pi }{8})=\frac{1}{b}$
$=(1-{\cos }^2\frac{\pi }{8})(1-{\cos }^2\frac{3\pi }{8})$
$={\sin }^2\frac{\pi }{8}{\sin }^2\frac{3\pi }{8}$
$=\frac{1}{4}\left(2{\sin }^2\frac{\pi }{8}\right)\left(2{\sin }^2\frac{3\pi }{8}\right)$
$=\frac{1}{4}\left[(1-\cos \frac{\pi }{4})(1-\cos \frac{3\pi }{4})\right]$ $\because [1-\cos \theta =2{\sin }^2\frac{\theta }{2}]$
$=\frac{1}{4}\left[(1-\frac{1}{\sqrt{2}})(1+\frac{1}{\sqrt{2}})\right]=\frac{1}{4}(1-\frac{1}{2})=\frac{1}{8}=$ R.H.S.
Ans: 8