Question

Solve ${\sin }^{2}n\theta -{\sin }^2(n-1)\theta -{\sin }^2\theta$.

Answer 1

$\because {\sin }^{2}A-{\sin }^{2}B=\sin (A+B)\sin (A-B)$
we have $\sin (2n-1)\theta \cdot \sin \theta -{\sin }^2\theta =0$
$\sin \theta \left[\sin \right(2n-1)\theta -\sin \theta ]=0$
$\sin \theta =0$ $\therefore \theta =r\pi$
$\sin (2n-1)\theta =\sin \theta$
$\therefore (2n-1)\theta =r\pi +(-1{)}^r\theta$
r even $=2k,(2n-2)\theta =2k\pi$ $\therefore \theta =\frac{k\pi }{n-1}$
r odd $=2k+1,2n\theta =(2k+1)\pi$
$\therefore \theta =(k+\frac{1}{2})\frac{\pi }{n}$.