What is the General Solution of \(sin \theta\) = \(sin \alpha\) ?

Solution :

The general solution of \(sin \theta\) = \(sin \alpha\) is given by \(\theta\) = \(n\pi + (-1)^n \alpha\),  n \(\in\) Z.

Proof :

We have,  \(sin \theta\) = \(sin \alpha\)

\(\implies\)  \(sin \theta\) – \(sin \alpha\) = 0

\(\implies\)   \(2 sin ({\theta – \alpha\over 2}) cos({\theta + \alpha\over 2})\) = 0

\(\implies\)  \(sin ({\theta – \alpha\over 2})\) = 0   or,   \(cos({\theta + \alpha\over 2})\) = 0

\(\implies\)  \({\theta – \alpha\over 2}\) = \(m\pi\)   or  \({\theta + \alpha\over 2}\) = \((2m + 1){\pi\over 2}\),  m \(\in\) Z

\(\implies\)   \(\theta\)  =  \(2m\pi + \alpha\), \(\in\)  Z   or,  \(\theta\)  =  \((2m + 1)\pi – \alpha\),  m \(\in\) Z.

\(\implies\)   \(\theta\) = (any even multiple of \(\pi\)) + \(\alpha\)  or,  \(\theta\) = (any odd multiple of \(\pi\)) – \(\alpha\)

\(\implies\)    \(\theta\) = \(n\pi + (-1)^n \alpha\),  where n \(\in\) Z.

Remark : The equation \(cosec \theta\) = \(cosec \alpha\) is equivalent to \(sin \theta\) = \(sin \alpha\). Thus, \(cosec \theta\) = \(cosec \alpha\) and \(sin \theta\) = \(sin \alpha\) have the same general solution.

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