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.

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}
