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How to find the stationary point(s) of $f(x,y)=\sin x \sin y \sin (x+y)$

With $x,y\in(0,\pi)$

So far I have found $$\nabla f =(\color{red}{\sin x\cos (x+y)+\cos x\sin (x+y)\sin y},\color{blue}{\sin y\cos (x+y)+\cos y\sin (x+y)\sin x)}$$

So we need

$$(\sin x \cos (x+y)+\cos x \sin (x+y)) \sin y=0 \\ (\sin y \cos (x+y)+\cos y\sin (x+y))\sin x=0$$

Problem is I don't know how to solve this.

Tim D
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  • $\sin x \cos (x+y)+\cos x \sin (x+y)=\sin(x+(x+y))=\sin(2x+y)$ (classical formula $\sin a \cos b+ \sin b \cos a=\sin(a+b)$ and the same kind of transformation for the other parenthesis. – Jean Marie Mar 25 '16 at 17:52
  • Okay so we get $\sin (2x+y)\sin y=0$ and $\sin (x+2y)\sin x=0$ but $\sin x$ and $\sin y$ are never $0$ on the region we care about so $\sin (2x+y)=0=\sin(x+2y) \implies (x+2y)=(2x+y) \implies x=y$ so we need to search for in the range such that $3x=\pi$ so we have $x=\pi /3$ so there is one stationary point at $(\pi /3 , \pi /3)$ is that correct? – Tim D Mar 25 '16 at 18:00

1 Answers1

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$\left(\sin x \cos(x+y) + \cos x\sin(x+y)\right) \sin y = 0$ when $\sin x \cos(x+y) + \cos x\sin(x+y) = 0$ or when $\sin y = 0$. I trust that you know when $\sin y = 0$. What about that first one?

\begin{align} \sin x \cos(x+y) + \cos x\sin(x+y) &= 0 \\ \sin x \cos(x+y) &= -\cos x\sin(x+y)\\ -\tan x &= \tan(x+y)\\ \tan(-x) &= \tan(x+y) \end{align}

Because tangent is periodic with period $\pi$, then this equation is satisfied when $x+y = -x + n\pi$, where $n$ is any integer (provided there are no domain restriction violations). This is equivalent to $y = -2x + n\pi$.

The second equation you mentioned is nearly identical.