The roots of the equation 1 - cos theta = sin | vedclass

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Question

(b) We have, $1 - \cos 	heta = \sin 	heta \,.\,\sin \frac{	heta }{2}$

==> $2{\sin ^2}\frac{	heta }{2} = 2\sin \frac{	heta }{2}\,.\,\cos \fr

Vedclass · Accepted Answer

$2k\pi ,k \in I$

Vedclass · Answer

(b) We have, $1 - \cos 	heta = \sin 	heta \,.\,\sin \frac{	heta }{2}$

==> $2{\sin ^2}\frac{	heta }{2} = 2\sin \frac{	heta }{2}\,.\,\cos \frac{	heta }{2}\,.\,\sin \frac{	heta }{2}$

==> $2{\sin ^2}\frac{	heta }{2}\,\left[ {1 - \cos \frac{	heta }{2}} ight] = 0$

==> $\sin \frac{	heta }{2} = 0$ or $2{\sin ^2}\frac{	heta }{4} = 0$

==> $\sin \frac{	heta }{2} = 0$ or $\sin \frac{	heta }{4} = 0$

==> $\frac{	heta }{2} = k\pi $ or $\frac{	heta }{4} = k\pi $.

Hence, $	heta = 2k\pi $ or $	heta = 4k\pi $, $k \in I$.

The roots of the equation $1 - \cos \theta = \sin \theta .\sin \frac{\theta }{2}$ is

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