<s>[INST] Find maximum and minimum of $\sin x + \sin y$ I am working on my scholarship exam practice but I am stuck on finding the minimum. Pre-university maths background is assumed.

When $x + y = \frac{2\pi}{3}, x\geq0, y\geq0$, the maximum of 
  $\sin x+\sin y$ is ....., and the minimum of that is .....

Let me walk you through what I have got.
$\sin x+\sin y = 2\sin (\frac{x+y}{2})\cos (\frac{x-y}{2})$
By substituting $x + y = \frac{2\pi}{3}$ into the sine function, we have
$\sin x+\sin y = 2\sin (\frac{2\pi}{3\cdot2})\cos (\frac{x-y}{2})$
$\sin x+\sin y = \sqrt{3}\cos (\frac{x-y}{2})$
To find the maximum and minimum, we know that
$-1 \leq\cos (\frac{x-y}{2})\leq1$
$-\sqrt{3} \leq\sqrt{3}\cos (\frac{x-y}{2})\leq\sqrt{3}$
Hence, the maximum is $\sqrt{3}$ which is correct and in accordance with the answer key.
However, it seems that the minimum equals to $-\sqrt{3}$ is incorrect. The answer key provided is  $\frac {\sqrt{3}}{2}$. Could you please elucidate how I can get to this answer? My guess is something to do with the condition $x\geq0$ and $y\geq0$ given by the question.
 [/INST] Since $f(x)=\sin{x}$ is a concave function on $\left[0,\frac{2\pi}{3}\right]$ and $\left(\frac{2\pi}{3},0\right)\succ(x,y),$ where $x\geq y,$ 
by Karamata we obtain:
$$\sin{x}+\sin{y}\geq\sin(x+y)+\sin0=\frac{\sqrt3}{2}.$$
The equality occurs for $x=\frac{2\pi}{3}$ and $y=0$, which says that we got a minimal value.
The maximal value we can get by Jensen:
$$\sin{x}+\sin{y}\leq2\sin\frac{x+y}{2}=\sqrt3,$$
where the equality accurs for $x=y$.
The first inequality we can prove also by the following way.
$$\sin{x}+\sin{y}-\sin(x+y)=\sin{x}(1-\cos{y})+\sin{y}(1-\cos{x})\geq0.$$
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