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Question: Answered & Verified by Expert
If $\alpha, \beta, \gamma$ are any three angles, then $\cos \alpha+\cos \beta-\cos \gamma-\cos (\alpha+\beta+\gamma)=$
MathematicsTrigonometric Ratios & IdentitiesAP EAMCETAP EAMCET 2018 (22 Apr Shift 2)
Options:
  • A $4 \cos \frac{\alpha+\beta}{2} \cos \frac{\beta+\gamma}{2} \cos \frac{\gamma+\alpha}{2}$
  • B $4 \cos \frac{\alpha+\beta}{2} \sin \frac{\beta+\gamma}{2} \sin \frac{\gamma+\alpha}{2}$
  • C $4 \cos \frac{\alpha+\beta}{2} \sin \frac{\beta-\gamma}{2} \sin \frac{\gamma-\alpha}{2}$
  • D $4 \sin \frac{\alpha+\beta}{2} \cos \frac{\beta+\gamma}{2} \cos \frac{\gamma+\alpha}{2}$
Solution:
1104 Upvotes Verified Answer
The correct answer is: $4 \cos \frac{\alpha+\beta}{2} \sin \frac{\beta+\gamma}{2} \sin \frac{\gamma+\alpha}{2}$
$\begin{aligned} & \cos \alpha+\cos \beta-\cos \gamma-\cos (\alpha+\beta+\gamma) \\ & =2 \sin \frac{2 \alpha+\beta+\gamma}{2} \sin \frac{\beta+\gamma}{2}+2 \sin \frac{\beta+\gamma}{2} \sin \left(\frac{\gamma-\beta}{2}\right) \\ & =2 \sin \frac{\beta+\gamma}{2}\left[\sin \frac{2 \alpha+\beta+\gamma}{2}+\sin \frac{\gamma-\beta}{2}\right] \\ & =4 \sin \frac{\beta+\gamma}{2} \sin \frac{\alpha+\gamma}{2} \cos \frac{\alpha+\beta}{2} \\ & =4 \cos \frac{\alpha+\beta}{2} \sin \frac{\beta+\gamma}{2} \sin \frac{\gamma+\alpha}{2} .\end{aligned}$

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