Question

The value of the expression
$(1+cos\frac{\pi }{10})(1+cos\frac{3\pi }{10})(1+cos\frac{7\pi }{10})(1+cos\frac{9\pi }{10})$ is

Answer 1

$(1+cos\frac{\pi }{10})(1+cos\frac{3\pi }{10})(1+cos\frac{7\pi }{10})(1+cos\frac{9\pi }{10})$

$=(1+cos\frac{\pi }{10})(1+cos\frac{3\pi }{10})(1-cos\frac{3\pi }{10})(1-cos\frac{\pi }{10})$ $[cos\frac{k\pi }{10}=-cos(\frac{10-k\pi }{10}\left)\right]$

$=(1-co{s}^2\frac{\pi }{10})(1-co{s}^2\frac{3\pi }{10})$

$=si{n}^2\frac{\pi }{10}si{n}^2\frac{3\pi }{10}$

$=\frac{1}{4}(1-cos\frac{\pi }{5})(1-cos\frac{3\pi }{5})$

$=\frac{1}{4}(1-cos\frac{\pi }{5})(1-[\frac{1}{2}-cos\frac{\pi }{5}\left]\right)$ $[cos\frac{\pi }{5}+cos\frac{3\pi }{5}=\frac{1}{2}]$

$=\frac{1}{8}(1-cos\frac{\pi }{5})(1+2cos\frac{\pi }{5})$

$=\frac{1}{8}(1+cos\frac{\pi }{5}-2co{s}^2\frac{\pi }{5})$

$=\frac{1}{8}(cos\frac{\pi }{5}-cos\frac{2\pi }{5})$ $[2co{s}^2\frac{\pi }{5}-1=cos\frac{2\pi }{5}]$

$=\frac{1}{8}(cos\frac{\pi }{5}+cos\frac{3\pi }{5})$ $[cos\frac{k\pi }{5}=-cos(\frac{5-k\pi }{5}\left)\right]$

$=\frac{1}{8}\left(\frac{1}{2}\right)$ $[cos\frac{\pi }{5}+cos\frac{3\pi }{5}=\frac{1}{2}]$

$=\frac{1}{16}$