# 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

Prove that $\mathrm{sin}x+\mathrm{cos}x=\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)$
$\sqrt{2}\left(x+\frac{\pi }{4}\right)=\sqrt{2}\left(\mathrm{sin}x\mathrm{cos}\frac{\pi }{4}+\mathrm{cos}x\mathrm{sin}\frac{\pi }{4}\right)=\mathrm{sin}x+\mathrm{cos}x$
Could you solve it from opposite?
$\mathrm{sin}x+\mathrm{cos}x=\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)$
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Paul Mitchell
$\mathrm{sin}x+\mathrm{cos}x=\sqrt{2}\frac{1}{\sqrt{2}}\left(\mathrm{sin}x+\mathrm{cos}x\right)$
$=\sqrt{2}\left(\frac{1}{\sqrt{2}}\mathrm{sin}x+\frac{1}{\sqrt{2}}\mathrm{cos}x\right)$
$=\sqrt{2}\left(\mathrm{sin}x\mathrm{cos}\left(\frac{\pi }{4}\right)+\mathrm{sin}\left(\frac{\pi }{4}\right)\mathrm{cos}x\right)$
Hint: $\mathrm{sin}A\mathrm{cos}B+\mathrm{sin}B\mathrm{cos}A=\mathrm{sin}\left(A+B\right)$
$=\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)$
###### Not exactly what you’re looking for?
yotaniwc

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

###### Not exactly what you’re looking for?
nick1337

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