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If $\alpha+\beta+\gamma=2 \theta$, then $\cos \theta+\cos (\theta-\alpha)$ $+\cos (\theta-\beta)+\cos (\theta-\gamma)$ is equal to
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Verified Answer
The correct answer is:
$4 \cos \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \cdot \cos \frac{\gamma}{2}$
Now,
$$
\begin{aligned}
& \cos (\theta-\alpha)+\cos (\theta-\beta)+\cos \theta+\cos (\theta-\gamma) \\
&= 2 \cos \left(\theta-\left(\frac{\alpha+\beta}{2}\right)\right) \cos \left(\frac{\beta-\alpha}{2}\right) \\
&+2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\theta-\frac{\gamma}{2}\right) \\
&=2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\frac{\beta-\alpha}{2}\right) \\
&+2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right) \\
&=2 \cos \left(\frac{\gamma}{2}\right)\left[\cos \left(\frac{\beta-\alpha}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)\right] \\
&=2 \cos \left(\frac{\gamma}{2}\right) 2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \\
&= 4 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \cos (\theta-\alpha)+\cos (\theta-\beta)+\cos \theta+\cos (\theta-\gamma) \\
&= 2 \cos \left(\theta-\left(\frac{\alpha+\beta}{2}\right)\right) \cos \left(\frac{\beta-\alpha}{2}\right) \\
&+2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\theta-\frac{\gamma}{2}\right) \\
&=2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\frac{\beta-\alpha}{2}\right) \\
&+2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right) \\
&=2 \cos \left(\frac{\gamma}{2}\right)\left[\cos \left(\frac{\beta-\alpha}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)\right] \\
&=2 \cos \left(\frac{\gamma}{2}\right) 2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \\
&= 4 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}
\end{aligned}
$$
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