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

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

Vedclass · Accepted Answer

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

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(B) Given $\alpha+\beta+\gamma=2 	heta$,so $	heta = \frac{\alpha+\beta+\gamma}{2}$.Let $S = \cos 	heta + \cos (	heta-\alpha) + \cos (	heta-\beta) + \cos (	heta-\gamma)$.Using the formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$,we group the terms:$S = [\cos 	heta + \cos (	heta-\alpha)] + [\cos (	heta-\beta) + \cos (	heta-\gamma)]$$S = 2 \cos \frac{2	heta-\alpha}{2} \cos \frac{\alpha}{2} + 2 \cos \frac{2	heta-\beta-\gamma}{2} \cos \frac{\beta-\gamma}{2}$Since $2	heta = \alpha+\beta+\gamma$,we have $2	heta-\beta-\gamma = \alpha$.$S = 2 \cos \frac{\beta+\gamma}{2} \cos \frac{\alpha}{2} + 2 \cos \frac{\alpha}{2} \cos \frac{\beta-\gamma}{2}$$S = 2 \cos \frac{\alpha}{2} [\cos \frac{\beta+\gamma}{2} + \cos \frac{\beta-\gamma}{2}]$Using $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$:$S = 2 \cos \frac{\alpha}{2} [2 \cos \frac{\beta}{2} \cos \frac{\gamma}{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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