Question

Prove that:

${\sin }^2\frac{\pi }{8}+{\sin }^2\frac{3\pi }{8}+{\sin }^2\frac{5\pi }{8}+{\sin }^2\frac{7\pi }{8}=2$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sin }^2\frac{\mathrm{\pi }}{8}+{\sin }^2\frac{3\mathrm{\pi }}{8}+{\sin }^2\frac{5\mathrm{\pi }}{8}+{\sin }^2\frac{7\mathrm{\pi }}{8} \\ ={\sin }^2\left(\frac{\mathrm{\pi }}{2}-\frac{3\mathrm{\pi }}{8}\right)+{\sin }^2\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{8}\right)+{\sin }^2\frac{5\mathrm{\pi }}{8}+{\sin }^2\frac{7\mathrm{\pi }}{8} \\ ={\cos }^2\frac{3\mathrm{\pi }}{8}+{\sin }^2\frac{\mathrm{\pi }}{8}+{\sin }^2\left(\mathrm{\pi }-\frac{3\mathrm{\pi }}{8}\right)+{\sin }^2\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{8}\right) \end{gathered}$$

$$\begin{gathered} ={\cos }^2\frac{3\mathrm{\pi }}{8}+{\sin }^2\frac{\mathrm{\pi }}{8}+{\sin }^2\frac{3\mathrm{\pi }}{8}+{\cos }^2\frac{\mathrm{\pi }}{8} \\ =\left({\cos }^2\frac{\mathrm{\pi }}{8}+{\sin }^2\frac{\mathrm{\pi }}{8}\right)+\left({\cos }^2\frac{3\mathrm{\pi }}{8}+{\sin }^2\frac{3\mathrm{\pi }}{8}\right) \\ =1+1=2=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$