Prove that \sin x+\cos x=\sqrt2 \sin (x+\frac{\pi}{4}) \sqrt2(x+\frac{\pi}{4})=\sqrt2(\sin x \cos \frac{\pi}{4}+\cos

From
${\displaystyle \sin (x+y)=\sin x\cos y+\cos x\sin y}$ (1)
and
${\displaystyle \sin (-y)=-\sin y}$ (2)
(both of which identities I assume as known and will not prove here), we obtain
$\sin (x-y)=\sin x\cos y-\cos x\sin y$ (3)
and by adding (1)+(3),
${\displaystyle \sin (x+y)+\sin (x-y)=2\sin x\cos y}$
Substitue ${\displaystyle x+y\leftarrow u}$ and ${\displaystyle x-y\leftarrow v}$, i.e., ${\displaystyle x\leftarrow \frac{u+v}{2}}$ and ${\displaystyle y\leftarrow \frac{u-v}{2}}$ to arrive at
${\displaystyle \sin u+\sin v=2\sin \frac{u+v}{2}\cos \frac{u-v}{2}}$

Using the formula $\sin (x+y)=\sin x\cos y+\cos x\sin y$
See for first line we know $\cos (\frac{\pi }{4})and\sin (\frac{\pi }{4})$ is equal to $\frac{1}{\sqrt{2}}$
So the expression reduces to
$\sqrt{2}\sin (x+\frac{\pi }{4})=\sqrt{2}(\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}})$
Now I hope you can reduce it to second line.
For the second line use the formula I gave in starting of answer.