Question

sin² π/18 + sin² π/9 + sin² 7π/18 + sin² 4π/9 =
(a) 1
(b) 4
(c) 2
(d) 0

Answer 1

(c) 2

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ {\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{\mathrm{\pi }}{9}+{\sin }^2\frac{7\mathrm{\pi }}{18}+{\sin }^2\frac{4\mathrm{\pi }}{9} \\ ={\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{2\mathrm{\pi }}{18}+{\sin }^2\frac{7\mathrm{\pi }}{18}+{\sin }^2\frac{8\mathrm{\pi }}{18} \\ ={\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{2\mathrm{\pi }}{18}+{\sin }^2\left({\displaystyle \frac{7\mathrm{\pi }}{18}}\right)+{\sin }^2\left(\frac{8\mathrm{\pi }}{18}\right) \\ ={\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{2\mathrm{\pi }}{18}+{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{2}}-{\displaystyle \frac{2\mathrm{\pi }}{18}}\right)+{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{2}}-{\displaystyle \frac{\mathrm{\pi }}{18}}\right) \\ ={\sin }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{2\mathrm{\pi }}{18}+{\cos }^2\frac{2\mathrm{\pi }}{18}+{\cos }^2\frac{\mathrm{\pi }}{18} \\ ={\sin }^2\frac{\mathrm{\pi }}{18}+{\cos }^2\frac{\mathrm{\pi }}{18}+{\sin }^2\frac{2\mathrm{\pi }}{18}+{\cos }^2\frac{2\mathrm{\pi }}{18} \\ =1+1 \\ =2 \end{gathered}$$