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

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

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

$=(1+\cos \left(\frac{\pi }{8}\right))(1+\sin \left(\frac{\pi }{8}\right))(1-\sin \left(\frac{\pi }{8}\right))(1-\cos \left(\frac{\pi }{8}\right))$

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

$={\sin }^2\left(\frac{\pi }{8}\right){\cos }^2\left(\frac{\pi }{8}\right)$

$=\frac{1}{4}\times {\left(2\sin \left(\frac{\pi }{8}\right)\cos \left(\frac{\pi }{8}\right)\right)}^2\{\because \sin 2x=2sinxcosx\}$

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

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

$=\frac{1}{4}\times {\left(\frac{1}{\sqrt{2}}\right)}^2$

$=\frac{1}{8}$

Hence proved