How to prove $\cos(x+y)+\sin(x-y)=2 \sin(x+\frac{\pi}{4}) \sin(\frac{\pi}{4}-y)$ I spend some time trying to figure out how to prove the following identity:
$$ \cos(x+y)+\sin(x-y)=2 \sin\left(x+\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}-y\right) $$
I tried to use the following identities:
$$ \cos(x+y)=\cos(x) \sin(y) - \sin(x) \sin(y) $$
and
$$ \sin(x-y)=\sin(x) \cos(y) - \cos(x) \sin(y). $$
After that, I wanted to use 
$$ \cos(x) = \sin\left(x+ \frac{\pi}{2}\right). $$
Unfortunately I can't reach the correct identity. Is there another way of doing it ? Thank you.
 A: Note that
$$\sin(x+y)+\sin(x-y)=2\sin x \cos y$$
i.e.
$$\sin(A)+\sin(B)=2\sin \frac{A+B}{2} \cos \frac{A-B}{2}$$
Then you can say
\begin{align}
\cos(x+y)+\sin(x-y)
&= \sin(x+y+\frac{\pi}{2})+\sin(x-y)\\
&= 2 \sin(x+\frac{\pi}{4})\cos(y+\frac{\pi}{4})\\
&= 2 \sin(x+\frac{\pi}{4})\sin(\frac{\pi}{4}-y)
\end{align}
A: Use
$$
\cos p-\cos q=-2\sin\frac{p+q}{2}\sin\frac{p-q}{2}
$$
and
$$
\sin\alpha=\cos\Bigl(\frac{\pi}{2}-\alpha\Bigr)
$$
Alternatively, expand both sides using the addition formulas.
A: let $u = x + \frac {\pi}{4}\\
v = \frac {\pi}{4} - y$
$u+v = \frac {\pi}{2} + x - y\\
u-v = x + y$
$\cos(u-v) + \sin(u+v - \frac{\pi}{2}) = 2\sin u\sin v\\
\sin(u+v - \frac{\pi}{2}) = - \sin \left(\frac{\pi}{2} - (u + v)\right) = -\cos (u+v)\\
\cos(u-v) - \cos(u+v) = 2\sin u\sin v\\
\cos u\cos v- \sin u\sin v -\cos u\cos v+\sin u \sin v = 2\sin u\sin v$
A: The other solutions have chosen http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
I shall use http://mathworld.wolfram.com/WernerFormulas.html
$$2\sin A\sin B=\cos(A-B)-\cos(A+B)$$
Here $A=x+\dfrac\pi4,B=\dfrac\pi4-y$
