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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?

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  • 2
    $\begingroup$ If $\alpha=\beta=0$ and $\gamma=\pi/2$, on the left hand side I get $1+0+0=1$, but on the right I get $4*(1/\sqrt2)*0*1=0$, so this proposition might be difficult to prove. $\endgroup$ – user5713492 Dec 8 '17 at 15:54
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use that $$\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\left(\left(\sin(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)\right)$$

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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?

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  • $\begingroup$ @BiliDebili, Have you meant this? $\endgroup$ – lab bhattacharjee Dec 12 '17 at 15:37

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