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?
$\endgroup$ 24 Answers
$\begingroup$Let's do the (harder) method attempted by the OP, "but it didn't work". $$ \sin(x-y)\cos(y)+\cos(x-y)\sin(y) \\ = \big[\sin(x)\cos(y)-\cos(x)\sin(y)\big]\cos(y)+\big[\cos(x)\cos(y)+\sin(x)\sin(y)\big]\sin(y) \\ = \sin(x)\cos(y)\cos(y)-\cos(x)\sin(y)\cos(y)+\cos(x)\cos(y)\sin(y)+\sin(x)\sin(y)\sin(y) \\= \sin(x)\cos^2(y)+\sin(x)\sin^2(y) \\= \sin(x)\big[\cos^2(y)+\sin^2(y)\big] \\= \sin(x)\big[ 1 \big] \\=\sin(x) $$
$\endgroup$ $\begingroup$HINT:
Recall that $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$. Now, let $a=x-y$ and $b=y$.
$\endgroup$ $\begingroup$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 $y=0$:
$$\begin{align} &\sin(x - 0)\cos(0)+\cos(x-0)\sin(0)\\ &=\sin(x)\end{align}$$
$\endgroup$ 2 $\begingroup$by using well known identity we get
$\endgroup$$$\\ \sin { x=\sin { \left( x-y+y \right) = } } \sin (x-y)\cos (y)+\cos (x-y)\sin (y)$$
