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: $$\sin\left(\frac{\pi}{2}-x\right)=\sin(x+y)=\sin\left(\frac{\pi}{2}-y\right).$$
On
$$f_x = \cos x - \sin(x+y) = 0\\ f_y = \cos y - \sin(x+y) = 0$$ 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 $\frac{\pi}{2}$
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.