How to simplify $\sin(x-y)\cos(y)+\cos(x-y)\sin(y)$ the question
How to simplify $\sin(x-y)\cos(y)+\cos(x-y)\sin(y)$
my steps
I tried to use trig identities on the $\sin(x-y)$ and $\cos(x-y)$ and tried to distribute the others in but it didn't work. Any ideas?
 A: Let's do the (harder) method attempted by the OP, "but it didn't work".
$$
\sin(x-y)\cos(y)+\cos(x-y)\sin(y)
\\ =
\big[\sin(x)\cos(y)-\cos(x)\sin(y)\big]\cos(y)+\big[\cos(x)\cos(y)+\sin(x)\sin(y)\big]\sin(y)
\\ =
\sin(x)\cos(y)\cos(y)-\cos(x)\sin(y)\cos(y)+\cos(x)\cos(y)\sin(y)+\sin(x)\sin(y)\sin(y)
\\=
\sin(x)\cos^2(y)+\sin(x)\sin^2(y)
\\=
\sin(x)\big[\cos^2(y)+\sin^2(y)\big]
\\= \sin(x)\big[ 1 \big]
\\=\sin(x)
$$
A: HINT:
Recall that $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$.  Now, let $a=x-y$ and $b=y$.
A: Knowing the solution (using methods posted by others) suggests an alternate solution. Apply $\frac{\partial}{\partial y}$ to your expression and you get:
$$-\cos(x-y)\cos(y)-\sin(x-y)\sin(y)+\sin(x-y)\sin(y)+\cos(x-y)\cos(y)=0$$
So the expression is actually constant in $y$. It takes on the same value for all values of $y$ as for when $y=0$:
$$\begin{align}
&\sin(x - 0)\cos(0)+\cos(x-0)\sin(0)\\
&=\sin(x)\end{align}$$
A: by using well known identity we get

$$\\ \sin { x=\sin { \left( x-y+y \right) = }  } \sin  (x-y)\cos  (y)+\cos  (x-y)\sin  (y)$$

