Question

Prove that :

\[ \sin^2\frac{\pi}{18} + \sin^2\frac{\pi}{9} + \sin^2\frac{7\pi}{18} + \sin^2\frac{4\pi}{9} = 2 \]

Solution

\[ \frac{7\pi}{18} = \frac{\pi}{2}-\frac{\pi}{9} \]

\[ \sin^2\frac{7\pi}{18} = \cos^2\frac{\pi}{9} \]

Also,

\[ \frac{4\pi}{9} = \frac{\pi}{2}-\frac{\pi}{18} \]

\[ \sin^2\frac{4\pi}{9} = \cos^2\frac{\pi}{18} \]

Therefore,

\[ \begin{aligned} &\sin^2\frac{\pi}{18} +\sin^2\frac{\pi}{9} +\sin^2\frac{7\pi}{18} +\sin^2\frac{4\pi}{9} \\[8pt] =& \sin^2\frac{\pi}{18} +\cos^2\frac{\pi}{18} +\sin^2\frac{\pi}{9} +\cos^2\frac{\pi}{9} \\[8pt] =& 1+1 \\[8pt] =& 2 \end{aligned} \]

Hence Proved.