2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.