$x+y = 120^\circ, x\geq 0, y\geq 0.$ max and min of $\sin x + \sin y$

Question

$x+y = 120^\circ, y = (120^\circ - x)$

$\sin x + \sin (120^\circ - x) = f(x) $

How to find max and min?

Answer 1

You don't have to differentiate whatever, just use a basic factorisation formula in trigonometry: $$\sin x+\sin y=2\sin \frac{x+y}2\,\cos\frac{x-y}2=2\sin 60\,\cos\frac{x-y}2=\sqrt 3\,\cos\frac{x-y}2.$$ Now as $0\le x,y\le 120$, we have $-120\le x-y\le 120$, so that $\cos 60\le\cos\frac{x-y}2\le \cos 0$, and eventually $$\frac{\sqrt 3}2\le \sin x+\sin y\le\sqrt 3.$$

Answer 2

Hint: $$\sin(120-x)=\sin(120)\cos(x)-\cos(120)\sin(x)=\frac 12\cos x-\frac{\sqrt 3}{2}\sin x$$ May help

Answer 3

Note that $f(x)=\sqrt{3}\cos(60^\circ-x)\le\sqrt{3}$, with equality at $x=60^\circ$.

Answer 4

Use angle sum formula, simplify, recast it into a single trigonometric ratio:

$\sin x + \sin(120^{\circ}-x) = \frac 32 \sin x + \frac{\sqrt 3}{2}\cos x = \sqrt 3 \cos(x-60^{\circ})$

Which has an unrestricted maximum value of $\sqrt 3$ when $x=60^{\circ} + 360^{\circ}m$ and an unrestricted minimum value of $-\sqrt 3$ when $x=240^{\circ} + 360^{\circ}n$ where $m,n$ range over the integers.

With the added constraints, it's clear that the maximum value of $\sqrt 3$ occurs at $x=y=60^{\circ}$ and the minimum value of $\frac{\sqrt 3}2$ occurs symmetrically at $x=0^{\circ},y=120^{\circ}$ and $x=120^{\circ},y=0^{\circ}$.

To understand the restricted minimum, observe that $x$ and $y$ are jointly non-negative for $0^{\circ} \leq x \leq 120^{\circ}$ and that $\cos(x-60^{\circ})$ is increasing and then decreasing for equal intervals on either side of the maximum at the midpoint of that range. So the minimum is reached at the extremes of the range.