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