Question

Prove that: ${\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{\mathrm{\pi }}{9}+{\sin }^2\frac{7\mathrm{\pi }}{18}+{\sin }^2\frac{4\mathrm{\pi }}{9}=2$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{\mathrm{\pi }}{9}+{\sin }^2\frac{7\mathrm{\pi }}{18}+{\sin }^2\frac{4\mathrm{\pi }}{9} \\ ={\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{2\mathrm{\pi }}{18}+{\sin }^2\frac{7\mathrm{\pi }}{18}+{\sin }^2\frac{8\mathrm{\pi }}{18} \\ ={\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{2\mathrm{\pi }}{18}+{\sin }^2\left({\displaystyle \frac{7\mathrm{\pi }}{18}}\right)+{\sin }^2\left(\frac{8\mathrm{\pi }}{18}\right) \\ ={\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{2\mathrm{\pi }}{18}+{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{2}}-{\displaystyle \frac{2\mathrm{\pi }}{18}}\right)+{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{2}}-{\displaystyle \frac{\mathrm{\pi }}{18}}\right) \\ ={\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{2\mathrm{\pi }}{18}+{\cos }^2\frac{2\mathrm{\pi }}{18}+{\cos }^2\frac{\mathrm{\pi }}{18} \\ ={\sin }^2\frac{\mathrm{\pi }}{18}+{\cos }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{2\mathrm{\pi }}{18}+{\cos }^2\frac{2\mathrm{\pi }}{18} \\ =1+1 \\ =2 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$