Question

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

is equal to

(a) \(1\)
(b) \(4\)
(c) \(2\)
(d) \(0\)

Solution

Using identity

\[ \sin^2\theta+\sin^2\left(\frac{\pi}{2}-\theta\right)=1 \]

\[ \sin^2\frac{\pi}{18} + \sin^2\frac{4\pi}{9} =1 \]

because

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

\[ \sin^2\frac{2\pi}{9} + \sin^2\frac{7\pi}{18} =1 \]

because

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

Therefore,

\[ 1+1=2 \]

Answer

\[ \boxed{2} \]

Correct Option: (c)