Show that the general value of $\theta$ satisfying $\sin\theta=\sin\alpha$ and $\cos\theta = \cos\alpha$ is given by $\theta = 2n\pi + \alpha$

Question

The general value of $\theta$ simultaneously satisfying equations, $$\sin\theta = \sin\alpha \quad\text{and}\quad \cos\theta = \cos\alpha$$ is given by $$\theta = 2n\pi + \alpha \ \forall \ n \in \mathbb{Z} \\$$

My attempt:

Adding the two equations,

$$\sin\theta + \cos\theta = \sin\alpha + \cos\alpha$$ $$\sin\theta - \sin\alpha = \cos\alpha - \cos\theta$$ $$2\cos\left( \frac{\theta + \alpha}{2} \right)\sin\left( \frac{\theta - \alpha}{2} \right)= 2\sin\left( \frac{\theta + \alpha}{2} \right)\sin\left( \frac{\theta - \alpha}{2} \right)$$ $$\cos^2\left( \frac{\theta + \alpha}{2} \right) = \sin^2\left( \frac{\theta + \alpha}{2} \right)$$ $$\cos(\theta + \alpha) = 0$$ $$\therefore \theta + \alpha = (2n+1)\frac{\pi}{2},\ n \in \mathbb{Z}$$ $$\theta = (2n+1)\frac{\pi}{2} - \alpha, \ n \in \mathbb{Z} $$

Why doesn't my solution match with the correct solution?

Please help!

Answer 1

No matter how complicated expressions get, you still can't divide by zero, which you implicitly did from the third line.

In this case we might have that $\sin\frac{\theta-\alpha}2=0$, which reduces to $\theta=2n\pi+\alpha$ for some $n\in\mathbb Z$ as desired. If it is not zero, we can divide by it and get $$\cos\frac{\theta+\alpha}2=\sin\frac{\theta+\alpha}2$$ which reduces to $\theta=\frac\pi2+2n\pi-\alpha$, but this does not satisfy the original system.

Answer 2

One problem is you divided by $\sin\left(\frac{\theta - \alpha}{2}\right)$ when going from the third to fourth lines. However, with $\theta = 2n\pi + \alpha$, then $\frac{\theta - \alpha}{2} = n\pi$, but $\sin(n\pi) = 0$. When you divide by $0$, basically anything can occur from it, including incorrect results such as what you got.