Help me prove that
$\cos(x / 2) + \cos(y / 2) - \cos(z / 2) = 4 \sin((\pi - x) / 4) \sin((\pi - y) / 4) \sin ((\pi + z) / 4$
where $x + y + z = \pi$
I've reached $2 \sin((x + z) / 4) (\cos((x + z) / 4) - \sin((x - z) / 4))$, but I'm not sure how to continue.
Using Prosthaphaeresis Formula,
$$\cos\frac y2-\cos\frac z2=-2\sin\frac{y+z}4\sin\frac{y-z}4$$
Now, $\displaystyle\cos\frac x2=\cos\frac{\pi-(y+z)}2=\sin\frac{y+z}2=2\sin\frac{y+z}4\cos\frac{y+z}4$ (Double Angle formula $\sin2A$)
$$\implies \cos\frac x2+\cos\frac y2-\cos\frac z2=2\sin\frac{y+z}4\cos\frac{y+z}4-2\sin\frac{y+z}4\sin\frac{y-z}4$$
$$=2\sin\frac{y+z}4\left(\cos\frac{y+z}4-\sin\frac{y-z}4\right)$$
$\displaystyle\sin\frac{y-z}4=\cos\left(\frac\pi2-\frac{y-z}4\right)=\cos\left(\frac{x+y+z}2-\frac{y-z}4\right)$
Again apply Prosthaphaeresis formula $\displaystyle\cos C-\cos D$