Relationship between $\beta, \theta$ when $\sin\beta = \sin\theta, \beta \neq \theta$ As in the title, if we have $\sin\theta = \sin\beta$ and $\beta \neq \theta$ isn't it true that $\beta = \theta + \frac{\pi}{2}.$ If $\beta = \theta + \pi$ we should have $\sin\beta = -\sin\theta$, no?
I can't see how to show this formally is my issue here. 
 A: Draw the graph and see where a horizontal line intersects it. That is the answer.
Thus you have either $\theta = \beta + (\text{a multiple of } 2\pi)$ or $\beta = \pi -\theta + (\text{a multiple of } 2\pi).$
So, for example, $\sin88^\circ = \sin 92^\circ,$ since those are $(90^\circ\pm2^\circ) = (90^\circ \pm \text{something}).,$ i.e. either of them is $180^\circ$ minus the other.
A: We have 
$$\sin\theta=\sin\beta\iff\begin{cases}\theta\equiv \beta\mod2\pi\\
\quad\text{or}\\\theta\equiv\pi-\beta\mod 2\pi\end{cases}.$$
A: From the sum formula : $\quad\sin(\beta)-\sin(\theta)=2\sin(\frac{\beta-\theta}2)\cos(\frac{\beta+\theta}2)$
We have equality when either the sinus or the cosinus annulates.
$\begin{cases}
\frac{\beta-\theta}2=k\pi\iff \beta=\theta+2k\pi\\\\
\frac{\beta+\theta}2=\frac{\pi}2+k\pi\iff \beta=(\pi-\theta)+2k\pi\\
\end{cases}$
But you can also visualize this easily graphically.

Regarding your other question in Michael Hardy's comment section
If you want $|\theta-\beta|\le \frac{\pi}{2}$ this means you get $(\pi-\beta)-\beta\le \frac{\pi}2\iff \beta\ge\frac{\pi}4$
Substituted in the sinus, this gives : $\theta>\beta \mid \sin(\theta)=\sin(\beta)\ge\frac{\sqrt{2}}2$.
