# 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.