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maxima and minima of f(x,y)=sinxsinysin(x+y)

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minimum and maximum of f(x,y)=sin(x)+sin(y)−sin(x+y)
calculus multivariable-calculus derivatives proof-verification
we are asked to find the minimum and maximum of the functionf:A→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+2π
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 △f(a,b)=(0,0),(△ represents gradient.)

First let's find points where the gradient is zero:

△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 (0,2π) or (2π,0), otherwise we are outside the boundaries of the triangle. and f(0,2π)=f(2π,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+2π:

f(x,−x+2π)=sin(x)+sin(−x+2π)−sin(x+x−2π)=sin(x)+sin(−x)−sin(2x)=sin(x)−sin(x)−sin(2x)=−sin(2x)
Let's denote g(x)=−sin(2x), then g′(x)=−2cos(2x). g′(x)=0 implies cos(2x)=0, which implies 2x=π2+kπ, then x=π4+kπ2
The only such valid point on our triangle would be the point (π4,7π4). and at that point: f(π4,7π4)=sin(π4)+sin(7π4)−sin(2π)=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