# 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.

For the minimum, note that since $$x,y\ge0\implies y\le\dfrac{2\pi}3$$, we have $$\dfrac{x-y}2=\dfrac{x+y-2y}2=\dfrac{\dfrac{2\pi}3-2y}2=\frac\pi3-y$$ so $$\sin x+\sin y = \sqrt{3}\cos\frac{x-y}{2}=\sqrt3\cos\left(\frac\pi3-y\right)\ge\begin{cases}\sqrt3\cos\left(\frac\pi3-0\right)\\\sqrt3\cos\left(\frac\pi3-\frac{2\pi}3\right)\end{cases}=\frac{\sqrt3}2.$$

• I am wondering why $-\sqrt{3}$ is wrong. What kind of thoughts should I have to find the new minimum and not answering $-\sqrt{3}$ ? – Trey Anupong Jun 15 at 14:16
• Recall that $0\le y\le 2\pi/3$ and that $\cos(\pi/3-y)$ is symmetric about $\pi/3$! This means that you can never get a negative answer for the minimum in this case, since $\cos\pi/3=1/2>0$. – TheSimpliFire Jun 15 at 14:19

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. $$\sin{x}+\sin{y}-\sin(x+y)=\sin{x}(1-\cos{y})+\sin{y}(1-\cos{x})\geq0.$$