Question

${\sin }^2\left(\frac{\pi }{18}\right)+{\sin }^2\left(\frac{\pi }{9}\right)+{\sin }^2\left(\frac{7\pi }{18}\right)+{\sin }^2\left(\frac{4\pi }{9}\right)=$

(a) 1
(b) 2
(c) 4
(d) none of these.

Answer 1

(b) 2

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