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

Answer

$\text{L.H.S}=\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}+\sin^2\frac{4\pi}{9}$ $=\sin^2\frac{\pi}{18}+\sin^2\frac{4\pi}{9}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}$ $=\sin^2\Big(\frac{\pi}{2}-\frac{4\pi}{9}\Big)+\sin^2\frac{4\pi}{9}+\sin^2\frac{\pi}{9}+\sin^2\Big(\frac{\pi}{2}-\frac{\pi}{9}\Big)$ $\big(\because\frac{\pi}{18}=\frac{\pi}{2}-\frac{4\pi}{9}\text{ and }\frac{7\pi}{18}=\frac{\pi}{2}-\frac{\pi}{9}\big)$ $=\cos^2\frac{4\pi}{9}+\sin^2\frac{4\pi}{9}+\sin^2\frac{\pi}{9}+\cos^2\frac{\pi}{9}$ $=1+1$ $(\because\sin^2\theta+\cos^2\theta=1)$ $= 2$ $= \text{R.H.S}$ $\text{proved}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free