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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.

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

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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}

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

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

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

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

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