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 1

LHS = $${\sin }^2\frac{\pi }{18}+{\sin }^2\frac{\pi }{9}+{\sin }^2\frac{7\pi }{18}+{\sin }^2\frac{4\pi }{9}$$
$$={\sin }^2\frac{\pi }{18}+{\sin }^2\frac{2\pi }{18}+{\sin }^2\frac{7\pi }{18}+{\sin }^2\frac{8\pi }{18}$$
$$={\sin }^2\frac{\pi }{18}+{\sin }^2\frac{2\pi }{18}+{\sin }^2\left(\frac{7\pi }{18}\right)+{\sin }^2\left(\frac{8\pi }{18}\right)$$
$$={\sin }^2\frac{\pi }{18}+{\sin }^2\frac{2\pi }{18}+{\sin }^2(\frac{\pi }{2}-\frac{2\pi }{18})+{\sin }^2(\frac{\pi }{2}-\frac{\pi }{18})$$
$$={\sin }^2\frac{\pi }{18}+{\sin }^2\frac{2\pi }{18}+{\cos }^2\frac{2\pi }{18}+{\cos }^2\frac{\pi }{18}$$
$$={\sin }^2\frac{\pi }{18}+{\cos }^2\frac{\pi }{18}+{\sin }^2\frac{2\pi }{18}+{\cos }^2\frac{2\pi }{18}$$
$$=1+1$$
$$=2$$
 = RHS
Hence proved.