Question

Find the value of $(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos \frac{5\pi }{8})(1+\cos \frac{7\pi }{8})$

Answer 1

Given

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

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

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

$=(1-co{s}^2\frac{\pi }{8})\left(\right(1-co{s}^2\frac{3\pi }{8}\left)\right)$ __ (1)

Now $co{s}^2\frac{\pi }{8}=\frac{1+cos\frac{\pi }{4}}{2}=\frac{1+\frac{1}{\sqrt{2}}}{2}=\frac{\sqrt{2}+1}{2\sqrt{2}}$

$co{s}^2\frac{3\pi }{8}=\frac{1+cos\frac{3\pi }{4}}{2}=\frac{1-\frac{1}{\sqrt{2}}}{2}=\frac{\sqrt{2}-1}{2\sqrt{2}}$

using in (1)

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

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

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

$=\frac{(2-1)}{2\times \sqrt{2}\times 2\sqrt{2}}=\frac{1}{2\sqrt{2}\times 2\sqrt{2}}=\frac{1}{8}$