# Find maximum of $\sin{x}+\sin{y}-\sin{(x+y)}+\sqrt{3}\left(\cos{x}+\cos{y}+\cos{(x+y)}\right)$

The trick is that we can't use derivatives.

If we look at $$\cos{x}+\cos{y}+\cos{(x+y)}$$, the function cos attains its maximum of $$1$$ at $$0$$ radians, so $$x=y=0$$ and $$x+y=0$$ and the maximum of $$\cos{x}+\cos{y}+\cos{(x+y)}=1+1+1=3$$.

What about part of the function with $$\sin$$? If we use the same logic: $$\sin$$ attains its maximum of $$1$$ at $$\pi/2$$ radians, so $$x=y=\pi/2$$, so $$x+y=\pi$$, we get the maximum of $$\sin{x}+\sin{y}-\sin{(x+y)}=1+1-0=2$$, but I guess it's not true, cause Wolfram says the maximum is $$0$$.

Any hint would help a lot!! thank you!

• From a very casual look... Have you tried to use derivatives, if not for the "official" solution, then only to find the actual maximum ... which you can then try to justify by other means? Nov 12, 2021 at 16:57
• have you checked that all signs in particular are correct ? Nov 12, 2021 at 16:58

$$\sin{x}+\sin{y}-\sin{(x+y)}+\sqrt{3}[\cos{x}+\cos{y}+\cos{(x+y)}] \\=\sin x+\sin y -\sin(x+y)+\tan \frac\pi3[\cos x+\cos y+\cos(x+y)]\\ =\frac1{\cos\dfrac\pi3}[\sin(x+\frac\pi3)+\sin(y+\frac\pi3)+\sin(\frac\pi3-x-y)]\\ =\frac1{\cos\dfrac\pi3}[\sin(x+\frac\pi3)+\sin(y+\frac\pi3)+\sin[ \pi-(\frac{\pi}3+x)-(\frac\pi3+y)]\\$$ The expression in the square bracket is the sum of sines of 3 angles in a triangle. Its maximum is obtained when all are equal. This can be seen if we consider the sum of 2 angles in a triangle $$\alpha,\beta,\gamma$$. For 2 angles $$\alpha,\beta$$ and a fixed $$\gamma$$ the sum is: $$\sin \alpha+\sin\beta=2\sin\dfrac{\alpha+\beta}2 \cos\dfrac{\alpha-\beta}2$$. It is maximized when $$\alpha=\beta$$. Then the sum of each with a $$\gamma$$ will be maximized when they are equal as well and we conclude that all the 3 must be equal. It follows that: $$x+\dfrac\pi3=y+\dfrac\pi3= \pi-(\frac{\pi}3+x)+(\frac\pi3+y)$$ and therefore $$x=y=0$$ and the maximum value of the expression is: $$3\tan\dfrac\pi3=3\sqrt3$$

Let $$x+60^{\circ}=\alpha$$, $$y+60^{\circ}=\beta$$ and $$60^{\circ}-x-y=\gamma$$.

Thus, $$\alpha+\beta+\gamma=180^{\circ}$$ and $$\sin{x}+\sin{y}-\sin{(x+y)}+\sqrt{3}\left(\cos{x}+\cos{y}+\cos{(x+y)}\right)=$$ $$=2(\sin\alpha+\sin\beta+\sin(\alpha+\beta))=2(\sin\alpha+\sin\beta+\sin\gamma).$$ Since we need to find a maximal value of the last expression, we can assume $$\sin\alpha\geq0$$, $$\sin\beta\geq0$$, $$\sin\gamma\geq0$$ and from here we can assume $$\alpha\geq0$$, $$\beta\geq0$$ and $$\gamma\geq0$$.

Id est, by Jensen we obtain: $$2(\sin\alpha+\sin\beta+\sin\gamma)\leq6\sin\frac{\alpha+\beta+\gamma}{3}=3\sqrt3.$$ The equality occurs for $$\alpha=\beta=\gamma=60^{\circ},$$ which says that we got a maximal value.

The command of Mathematica

Maximize[{Sin[x] + Sin[y] - Sin[x + y] +
Sqrt[3]*(Cos[x] + Cos[y] + Cos[x + y]), x >= -Pi && x <= Pi,
y >= -Pi && y <= Pi}, {x, y}]


results in {3 Sqrt[3], {x -> 0, y -> 0}}. This is confirmed by NMaximize with DifferentialEvolution method.