If $\sin a+\sin b=2$, then show that $\sin(a+b)=0$ If $\sin a+\sin b=2$, then show that $\sin(a+b)=0$.  
I have tried to solve this problem in the following way :   
\begin{align}&\sin a + \sin b=2 \\
\Rightarrow &2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)=2\\
\Rightarrow &\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)=1
\end{align}
What will be the next ?

HINT:
As $-1\le \sin x\le 1$ for real $x$
$\implies -2\le \sin a+\sin b\le 2$
The equality occurs if $\sin a=\sin b=1$
$a=2n\pi+\frac\pi2,  b=2m\pi+\frac\pi2,$ for some integer $m,n$
So, $a+b=2\pi(m+n)+\pi=\pi(2m+2n+1)$ and we know $\sin r\pi=0$ for integer $r$

Alternatively, if $\sin x=1,\cos x=\pm\sqrt{1-1^2}=0\implies $ here $\cos a=\cos b=0$
and we know $\sin(a+b)=\sin a\cos b+\cos a \sin b$
