# 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.

• Have you tried using $sin(x - y)$ decomposition on the rhs? Just to provide a potential starting point. – Nicg May 13 at 20:07
• Use that $\sin(x-y) = \cos(\frac{\pi}{2}-(x-y))$ – Vasya May 13 at 20:09

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.

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}

let $$u = x + \frac {\pi}{4}\\ v = \frac {\pi}{4} - y$$

$$u+v = \frac {\pi}{2} + x - y\\ u-v = x + y$$

$$\cos(u-v) + \sin(u+v - \frac{\pi}{2}) = 2\sin u\sin v\\ \sin(u+v - \frac{\pi}{2}) = - \sin \left(\frac{\pi}{2} - (u + v)\right) = -\cos (u+v)\\ \cos(u-v) - \cos(u+v) = 2\sin u\sin v\\ \cos u\cos v- \sin u\sin v -\cos u\cos v+\sin u \sin v = 2\sin u\sin v$$

The other solutions have chosen http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

$$2\sin A\sin B=\cos(A-B)-\cos(A+B)$$

Here $$A=x+\dfrac\pi4,B=\dfrac\pi4-y$$