If $\alpha+\beta+\gamma=2 \theta$, then $\cos \theta+\cos (\theta-\alpha)$ $+\cos (\theta-\beta)+\cos…

If $\alpha+\beta+\gamma=2 \theta$, then $\cos \theta+\cos (\theta-\alpha)$ $+\cos (\theta-\beta)+\cos (\theta-\gamma)$ is equal to
  1. $4 \sin \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}$
  2. $4 \cos \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \cdot \cos \frac{\gamma}{2}$
  3. $4 \sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}$
  4. $4 \sin \alpha \cdot \sin \beta \cdot \sin \gamma$

Solution

Now, $ \begin{aligned} & \cos (\theta-\alpha)+\cos (\theta-\beta)+\cos \theta+\cos (\theta-\gamma) \\ &= 2 \cos \left(\theta-\left(\frac{\alpha+\beta}{2}\right)\right) \cos \left(\frac{\beta-\alpha}{2}\right) \\ &+2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\theta-\frac{\gamma}{2}\right) \\ &=2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\frac{\beta-\alpha}{2}\right) \\ &+2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right) \\ &=2 \cos \left(\frac{\gamma}{2}\right)\left[\cos \left(\frac{\beta-\alpha}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)\right] \\ &=2 \cos \left(\frac{\gamma}{2}\right) 2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \\ &= 4 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2} \end{aligned} $

Asked in: AP EAMCET 2008

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