If $\alpha=\beta+\omega$, then $1 + \cos\alpha+\cos\beta+\cos\omega = 4 \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\omega}{2}$

Question

How can I prove the following? The RHS is not about SIN

If $\alpha = \beta + \omega$, then $$1 + \cos\alpha+\cos\beta+\cos\omega = 4 \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\omega}{2}$$

I've already tried that one: $$\cos (\alpha+\omega) + \cos\omega +\cos\alpha + \cos 2(\alpha+\omega)$$ What more can I do?

See https://math.stackexchange.com/questions/176892/prove-trigonometry-identity-for-cos-a-cos-b-cos-c — lab bhattacharjee, May 02 '18 at 10:09
@Hollycow Thanks for accepting my answer. Please also consider upvoting it. — J.G., May 02 '18 at 11:23

score 0 · Accepted Answer · answered May 02 '18 at 10:35

If you don't mind I'll write $x:=\beta,\,y:=\omega$. We have $$1+\cos x+\cos y-4\cos\frac{x+y}{2}\cos\frac{x}{2}\cos\frac{y}{2}=1+2\cos\frac{x+y}{2}(\cos\frac{x-y}{2}-2\cos\frac{x}{2}\cos\frac{y}{2}).$$The right-hand side simplifies to $1-2\cos^2\frac{x+y}{2}=-\cos(x+y)$.

If $\alpha=\beta+\omega$, then $1 + \cos\alpha+\cos\beta+\cos\omega = 4 \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\omega}{2}$

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