<s>[INST] 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 ?
 [/INST] The maximum value of the $sin$ function is 1, at $\pi / 2 + 2k\pi$, for any integer $k$.
So $a = \pi / 2 + 2k\pi$, and $b = \pi / 2 + 2l\pi$, for any integers $k$, $l$.
Therefore, $\sin(a + b) = \sin(\pi + 2(k + l)\pi) = 0$.
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