If $x$, $y$ and $z$ are real numbers with the property $x+y+z= \pi$, then the maximum of $\sin x+\sin y+\sin z$ is $3\sqrt{3}/2$.
Now, if $x+y+z=0$ then is the maximum of $|\sin x| + |\sin y| + |\sin z|$ again $3\sqrt{3}/2$?
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We know that $|\sin x + \sin y + \sin z| \leq |\sin x| + |\sin y| + |\sin z|$, so the maximum has to be at least that much. – Ben Grossmann Dec 02 '13 at 20:51
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@Omnomnomnom Not quite obvious, since the condition on $x+y+z$ is different. – Zubin Mukerjee Dec 02 '13 at 20:58
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At most that much. – Emily Dec 02 '13 at 20:58
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@ZubinMukerjee I missed that, good catch – Ben Grossmann Dec 02 '13 at 21:23
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If $x,y,z$ maximize $|\sin x|+|\sin y|+|\sin z|$ subject to $x+y+z=\pi$, then $x,y,z'=z-\pi$ maximize $|\sin x|+|\sin y|+|\sin z'|=|\sin x|+|\sin y|+|\sin z|$ subject to $x+y+z'=0$ and vice versa.
Hagen von Eitzen
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$$f(x,y):=|\sin x|+|\sin y|+|\sin(-x-y)|=|\sin x|+|\sin y|+|\sin(x+y)|$$
In order to maximize $f$, let's consider $$\nabla f=\left\langle\frac{\sin x}{|\sin x|}\cos x+\frac{\sin (x+y)}{|\sin(x+y)|}\cos(x+y),\frac{\sin y}{|\sin y|}\cos y+\frac{\sin (x+y)}{|\sin(x+y)|}\cos(x+y)\right\rangle$$
If we require $\cos x-\cos(x+y)=0$, then $y=2k\pi-2x$ for some $k\in \Bbb Z$. If their sum is equal to $0$, $y=(2k+1)\pi-2x$.
Then $$ \begin{align} f(x)&=\left\{\begin{array}{l}|\sin x|+|\sin(2k\pi-2x)|+|\sin(2k\pi-x)|\\|\sin x|+|\sin((2k+1)\pi-2x)|+|\sin((2k+1)\pi-x)|\end{array}\right.\\ &=2|\sin x|+|\sin 2x| \end{align}$$ Now, when is $f(x)$ maximized? Well, $f'(x)=2\left(\frac{\sin x}{|\sin x|}\cos x+\frac{\sin 2x}{|\sin 2x|}\cos 2x\right)$. So we either want $\cos x+\cos 2x=0$ or $\cos x-\cos 2x=0$ This gets $\cos x=\pm1,\pm\frac12$. When this is true, we get:
$$f(x)=0,\frac{3\sqrt3}2$$
Tim Ratigan
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