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?

I think the second angle should be $\dfrac\beta2 + 45^\circ$
Let  $\displaystyle\frac\alpha2 + 45^\circ=x, \frac\beta2 + 45^\circ=y, \frac\gamma2 + 45°=z$
$90^\circ=\alpha+\beta+\gamma=2x-90^\circ+2y+2z-90^\circ\iff x+y+z)=180^\circ$
$\implies \sin(x+y)=\sin z,\cos(x+y)=-\cos z$
So, the left hand side reduces to $$\sin2x+\sin2y+\sin2z$$
$$=2\sin(x+y)\cos(x-y)+2\sin z\cos z$$
$$=2\sin(z)\cos(x-y)+2\sin z\{-\cos(x+y)\}$$
Can you take it from here?
