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Trigonometrical Equations
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The roots of the equation $1 - \cos \theta = \sin \theta .\sin \frac{\theta }{2}$ is
A
$k\pi ,k \in I$
B
$2k\pi ,k \in I$
C
$k\frac{\pi }{2},k \in I$
D
None of these
Solution
(b) We have, $1 – \cos \theta = \sin \theta \,.\,\sin \frac{\theta }{2}$
==> $2{\sin ^2}\frac{\theta }{2} = 2\sin \frac{\theta }{2}\,.\,\cos \frac{\theta }{2}\,.\,\sin \frac{\theta }{2}$
==> $2{\sin ^2}\frac{\theta }{2}\,\left[ {1 – \cos \frac{\theta }{2}} \right] = 0$
==> $\sin \frac{\theta }{2} = 0$ or $2{\sin ^2}\frac{\theta }{4} = 0$
==> $\sin \frac{\theta }{2} = 0$ or $\sin \frac{\theta }{4} = 0$
==> $\frac{\theta }{2} = k\pi $ or $\frac{\theta }{4} = k\pi $.
Hence, $\theta = 2k\pi $ or $\theta = 4k\pi $, $k \in I$.
Standard 11
Mathematics
