Question

Show 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

$\begin{array}{c}\left(1+\cos \frac{\pi }{8}\right)\left(1+\cos \frac{3\pi }{8}\right)\left[1+\cos \left(\pi -\frac{3\pi }{8}\right)\right]\left[1+\cos \pi -\frac{\pi }{8}\right]\\ \Rightarrow \left(1+\cos \frac{\pi }{8}\right)\left(1+\cos \frac{3\pi }{8}\right)\left(1-\cos \frac{3\pi }{8}\right)\left(1-\cos \frac{\pi }{8}\right)\\ \Rightarrow \left(1+\cos \frac{\pi }{8}\right)\left(1-\cos \frac{\pi }{8}\right)\left(1+\cos \frac{\pi }{8}\right)\left(1-\cos \frac{3\pi }{8}\right)\\ \Rightarrow \left(1^2-{\cos }^2\frac{\pi }{8}\right)\left(1^2-{\cos }^2\frac{3\pi }{8}\right)\\ \Rightarrow {\sin }^2\frac{\pi }{8}\times {\sin }^2\frac{3\pi }{8}\Rightarrow {\sin }^2\frac{\pi }{8}\times {\sin }^2\left(\frac{\pi }{2}-\frac{\pi }{8}\right)\\ \Rightarrow \frac{4}{4}{\sin }^2\frac{\pi }{8}\times {\cos }^2\frac{\pi }{8}\\ \Rightarrow \frac{1}{4}4{\sin }^2\frac{\pi }{8}\times {\cos }^2\frac{\pi }{8}\\ \Rightarrow \frac{1}{4}\times {\left(2\sin \frac{\pi }{8}\cos \frac{\pi }{8}\right)}^2\\ \Rightarrow \frac{1}{4}{\left[{\sin }^2\frac{\pi }{8}\right]}^2\\ \Rightarrow \frac{1}{4}{\left[\sin {45}^{\circ }\right]}^2\Rightarrow \frac{1}{8}\end{array}$