If $\alpha, \beta, \gamma$ are any three angles, then $\cos \alpha+\cos \beta-\cos \gamma-\cos (\alpha+\beta+\gamma)=$

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Aicharya · Accepted Answer

$\begin{aligned} & \cos \alpha+\cos \beta-\cos \gamma-\cos (\alpha+\beta+\gamma) \ & =2 \sin \frac{2 \alpha+\beta+\gamma}{2} \sin \frac{\beta+\gamma}{2}+2 \sin \frac{\beta+\gamma}{2} \sin \left(\frac{\gamma-\beta}{2}ight) \ & =2 \sin \frac{\beta+\gamma}{2}\left[\sin \frac{2 \alpha+\beta+\gamma}{2}+\sin \frac{\gamma-\beta}{2}ight] \ & =4 \sin \frac{\beta+\gamma}{2} \sin \frac{\alpha+\gamma}{2} \cos \frac{\alpha+\beta}{2} \ & =4 \cos \frac{\alpha+\beta}{2} \sin \frac{\beta+\gamma}{2} \sin \frac{\gamma+\alpha}{2} .\end{aligned}$

If $\alpha, \beta, \gamma$ are any three angles, then $\cos \alpha+\cos \beta-\cos \gamma-\cos…

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