If alpha+beta+gamma=2 theta,then cos theta+co | vedclass

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(B) Given that $\alpha+\beta+\gamma=2 	heta$,so $	heta = \frac{\alpha+\beta+\gamma}{2}$.Let $S = \cos 	heta + \cos (	heta-\alpha) + \cos (	heta-\

Vedclass · Accepted Answer

$4 \cos \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \cdot \cos \frac{\gamma}{2}$

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(B) Given that $\alpha+\beta+\gamma=2 	heta$,so $	heta = \frac{\alpha+\beta+\gamma}{2}$.Let $S = \cos 	heta + \cos (	heta-\alpha) + \cos (	heta-\beta) + \cos (	heta-\gamma)$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$:$S = [\cos 	heta + \cos (	heta-\gamma)] + [\cos (	heta-\alpha) + \cos (	heta-\beta)]$$S = 2 \cos \frac{2	heta-\gamma}{2} \cos \frac{\gamma}{2} + 2 \cos \frac{2	heta-\alpha-\beta}{2} \cos \frac{\beta-\alpha}{2}$Since $2	heta = \alpha+\beta+\gamma$,we have $2	heta-\gamma = \alpha+\beta$ and $2	heta-\alpha-\beta = \gamma$.$S = 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\gamma}{2} + 2 \cos \frac{\gamma}{2} \cos \frac{\beta-\alpha}{2}$$S = 2 \cos \frac{\gamma}{2} [\cos \frac{\alpha+\beta}{2} + \cos \frac{\beta-\alpha}{2}]$Using $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ again:$S = 2 \cos \frac{\gamma}{2} [2 \cos \frac{\beta}{2} \cos \frac{\alpha}{2}]$$S = 4 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}$.

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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