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Find the critical points of function$$ f(x,y)=\sin x + \sin y + \cos(x+y),$$ where $0<x<\dfrac{\pi}{2}$, $0<y<\dfrac{\pi}{2}$.

What I have done: $$f_{x}=\cos(x)-\sin(x+y),\\ f_{y}=\cos(y)-\sin(x+y).$$

From $f_{x}=0$, $\cos(x)=\sin(x+y)$. From $f_{x}=0$, $\cos(y)=\sin(x+y)$. I do not know where to go from here.

My attemps: $$\sin\left(\frac{\pi}{2}-x\right)=\sin(x+y)=\sin\left(\frac{\pi}{2}-y\right).$$

Ѕᴀᴀᴅ
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Jack Sparrow
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2 Answers2

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$$f_x = \cos x - \sin(x+y) = 0\\ f_y = \cos y - \sin(x+y) = 0$$ Subtracting one from the other we get $\cos x = \cos y$ and with the restrictions of $x,y$ to $(0,\frac {\pi}{2})$ we can say $x= y$ and $\cos x - \sin 2x = 0$

$$\cos x(1-\sin x) = 0\\x = \frac {\pi}{6}$$

$\frac {\pi}{2}$ would also solve the equation but the domain says strictly less than $\frac{\pi}{2}$

Doug M
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Hint:

$$\frac{df}{dx}=\cos x - \sin(x+y) = 0$$

$$\frac{df}{dy}=\cos y - \sin(x+y) = 0$$

Exploding the $\sin$s

$$\cos x - \sin x \cos y - \cos x \sin y = 0$$

$$\cos y - \sin x \cos y - \cos x \sin y = 0$$

Dividing the first with $\cos x$, the second with $\cos y$:

$$1 - \tan x \cos y - \sin y = 0$$

$$1 - \sin x - \cos x \tan y = 0$$

Expressing $\tan x$ from the first:

$$\tan x = \frac{1- \sin y}{\cos y}$$

...and now substitute it into the second equation. You have to express $\sin x$ with $\tan x$, there is a trigonometrical identity for that, but unfortunately this small margin is too narrow here to contain. ;-)

But the result will be a second degree equation and you will simply solve it.

peterh
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  • can't you just do $\sin(\frac{\pi }{2}-x)=\sin(x+y)$? – Jack Sparrow Jun 26 '18 at 01:44
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    @peterh This is not a hint. In fact it is the solution. – tien lee Jun 26 '18 at 01:45
  • @JackSparrow Why would it be true? It looks to me non-sense (if you don't have some special circumstance). – peterh Jun 26 '18 at 01:46
  • @tienlee He still has to dig out the identity to express sinus in the term of tangent, then solve the resulting equation, and substitute the result back into the original equation. This is actually bigger work as here I gave. Thus, here I only shown the direction. – peterh Jun 26 '18 at 01:48
  • @peterh there's no way it will take that long, it's supposed to be exam problem which takes 5-6 mins each question. – Jack Sparrow Jun 26 '18 at 02:07
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    @JackSparrow It is right, the other answer is better. Btw, the intention of the site is not to solve exam problems. – peterh Jun 26 '18 at 02:10
  • Shouldn’t the second equation be: $1-sin(x)-cos(x)tan(y) = 0$ -after dividing by cos(y)? – elson1608 Mar 23 '23 at 09:24
  • @elson1608 Thanks, I fixed it :-) – peterh Mar 23 '23 at 18:55