minimum and maximum of $f(x,y)=\sin(x)+\sin(y)-\sin(x+y)$ we are asked to find the minimum and maximum of the function$f:A \to A$ $f(x,y)=\sin(x)+\sin(y)-\sin(x+y)$
Where $A$ is the triangle bound by $x=0$,$y=0$ and $y=-x+2\pi$
I'd like someone to review my answer.
What I did:
$A$ is a closed and bounded set, $f(x,y)$ is continuous, so according to Weierstrass theorem, $f$ receives maximal / minimal values, either on the boundary, or an internal point $(a,b)$ where $\triangle f(a,b)=(0,0)$,($\triangle$ represents gradient.)
First let's find points where the gradient is zero:
$\triangle f(x,y)=(\cos(x)-\cos(x+y),\cos(y)-\cos(x+y))=0$
this implies $\cos(x)=\cos(y)=\cos(x+y)$.
On the triangle we were given, this can only happen at $(0,2\pi)$ or $(2\pi,0)$, otherwise we are outside the boundaries of the triangle. and $f(0,2\pi)=f(2\pi,0)=0$.
Let's see what happens on the boundary, assume first $x=0$:
$f(0,y)=\sin(0)+\sin(y)-\sin(0+y)=0$
Same thing happens when $y=0$.
Now let's see $y=-x+2\pi$:
$f(x,-x+2\pi)=\sin(x)+\sin(-x+2\pi)-\sin(x+x-2\pi)=\sin(x)+\sin(-x)-\sin(2x)=\sin(x)-\sin(x)-\sin(2x)=-\sin(2x)$
Let's denote $g(x)=-\sin(2x)$, then $g'(x)=-2\cos(2x)$. $g'(x)=0$ implies $\cos(2x)=0$, which implies $2x=\frac{\pi}{2}+k\pi$, then $x=\frac{\pi}{4}+k\frac{\pi}{2}$
The only such valid point on our triangle would be the point $(\frac{\pi}{4},\frac{7\pi}{4})$. and at that point: $f(\frac{\pi}{4},\frac{7\pi}{4})=\sin(\frac{\pi}{4})+\sin(\frac{7\pi}{4})-\sin(2\pi)=0$
At all the potentially extreme points, we got $f=0$. this makes me believe that $f(A)=\{0\}$. There is no point on the triangle where $f$ is not zero. Is this true? Is it possible to verify this result with trigonometric identities? to simplify $\sin(x)+\sin(y)-\sin(x+y)$ and eventually hope to reach zero?

You are correct in saying that if $f$ attains a minimum/maximum at $(x,y)$ in the interior of $A$, then $(x,y)$ must satisfy $\cos x = \cos y = \cos(x+y)$. 
However, $(0,2\pi)$ and $(2\pi,0)$ are not the only points in $A$ at which this condition can be met. 
Suppose $(x,y) \in A$ satisfies $\cos x = \cos y = \cos(x+y)$. 
Then, $0 = \cos x - \cos(x+y) = 2\sin \dfrac{y}{2} \sin\left(x + \dfrac{y}{2}\right)$. 
Hence, $\sin \dfrac{y}{2} = 0$ or $\sin\left(x + \dfrac{y}{2}\right) = 0$.
Thus, we must have either $y = 2\pi m$, or $2x+y = 2\pi m$ for some integer $m$. 
Similarly, we must have either $x = 2\pi n$, or $x+2y = 2\pi n$ for some integer $n$. 
With a bit of casework, we can see that the only $(x,y)$ in the interior of $A$ is $(x,y) = \left(\dfrac{2\pi}{3},\dfrac{2\pi}{3}\right)$ (the solution to $2x+y = x+2y = 2\pi$). 
So you need to test this point as well. (As it turns out, this ends up being the maximum). 
