Question

If the value of $\sin \theta +\sin 3\theta +\sin 5\theta +...+\sin (2n-1)\theta =\frac{{\sin }^{a}n\theta }{\sin \theta }$ Find $a.$

Answer 1

We have, $\sin \theta +\sin 3\theta +\sin 5\theta +...+\sin (2n-1)\theta =\frac{\sin \left[n\left(\frac{2\theta }{2}\right)\right]}{\sin \left(\frac{2\theta }{2}\right)}\sin \left(\frac{\theta +(2n-1)\theta }{2}\right)$

$\sin \theta +\sin 3\theta +\sin 5\theta +....+\sin (2n-1)\theta =\frac{{\sin }^{2}n\theta }{\sin \theta }$
Ans: a = 2