How to determine the $dy$-term of the differential of $f(x,y,z)= \sin(x+y)\sin(y+z)$ to be $\sin(x+2y+z)$?

Question

So I want to know how I can simplify the $dy$-part of the differential of $f(x,y,z)= \sin(x+y)sin(y+z)$.

I already have the $dx$-term which is $[cos(x+y)sin(y+z)]dx$ and the $dz$-term which becomes $[\sin(x+y)\sin(y+z)]dz$.

So I already know that the differential of the entire thing is

$df(x,y,z)= [\cos(x+y)\sin(y+z)]dx+ \sin(x+2y+z)dy + [\sin(x+y)\cos(y+z)]dz$

The question actually is what is the $\dfrac{\partial f(x,y,z)}{\partial y}$? I don't know any trigonemetric identity to of 3 terms into 1 sin. Can someone explain this to me?

EDIT: I already have for the $dy$-term that $\cos(x+y)\sin(y+z)+ \cos(y+z)\sin(x+y)]dy$

Answer 1

Hint: $$\sin (a+b)=\cos(a)\sin(b)+ \cos(b)\sin(a)$$