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 $$\sin x+\sin y = \sqrt{3}\cos\frac{x-y}{2}=\sqrt3\cos\left(\frac\pi3-y\right)\ge\begin{cases}\sqrt3\cos\left(\frac\pi3-0\right)\\\sqrt3\cos\left(\frac\pi3-\frac{2\pi}3\right)\end{cases}=\frac{\sqrt3}2.$$

  • $\begingroup$ I am wondering why $-\sqrt{3}$ is wrong. What kind of thoughts should I have to find the new minimum and not answering $-\sqrt{3}$ ? $\endgroup$ – Trey Anupong Jun 15 at 14:16
  • 1
    $\begingroup$ 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 $\cos\pi/3=1/2>0$. $\endgroup$ – 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. $$\sin{x}+\sin{y}-\sin(x+y)=\sin{x}(1-\cos{y})+\sin{y}(1-\cos{x})\geq0.$$


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