Proof of $ \cos \alpha + \sin \beta + \cos \gamma = 4* \sin( \frac\alpha2 + 45°)* \sin \frac\beta2 * \sin (\frac\gamma2 + 45°) $ Proof of 
$$ \cos \alpha + \sin \beta + \cos \gamma = 4* \sin( \frac\alpha2 + 45°)* \sin \frac\beta2 * \sin (\frac\gamma2 + 45°) $$ 
if $ \alpha + \beta + \gamma = \fracπ2 $
I tried to simplify right side:
$4(-\frac12(\cos(\alpha/2+\gamma/2+π/2)-\cos(\alpha/2-\gamma/2))*\sin(\beta/2)=$
$=-2*\sin(\beta)*\cos(\alpha/2+\gamma/2+π/2)+2\cos(\alpha/2-\gamma/2)*\sin(\beta/2)= $
$=-\sin(\beta/2+\alpha/2+\gamma/2+π/2)-\sin(\beta/2-\alpha/2-\gamma/2-π/2)+\sin(\beta/2+\alpha/2-\gamma/2)+\sin(\beta/2-\alpha/2+\gamma/2) $
Is it possible to simplify this to the left hand side?

use that $$\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\left(\left(\sin(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)\right)$$
