Question

lf $\alpha +\beta +\gamma =2\theta$, then $\cos \theta +\cos (\theta -\alpha )+\cos (\theta -\beta )+\cos (\theta -\gamma )$ is equal to

• A. $4\sin \left(\frac{\alpha }{2}\right)\cos \left(\frac{\beta }{2}\right)\sin \left(\frac{\gamma }{2}\right)$
• B. $4\cos \left(\frac{\alpha }{2}\right)\cos \left(\frac{\beta }{2}\right)\cos \left(\frac{\gamma }{2}\right)$
• C. $4\sin \left(\frac{\alpha }{2}\right)\sin \left(\frac{\beta }{2}\right)\sin \left(\frac{\gamma }{2}\right)$
• D. $4\sin \alpha \cos \beta \sin \gamma$

Answer 1

The correct option is B $4\cos \left(\frac{\alpha }{2}\right)\cos \left(\frac{\beta }{2}\right)\cos \left(\frac{\gamma }{2}\right)$

Given $\alpha +\beta +\gamma =2\theta$

Consider $I=\cos \left(\theta \right)+\cos (\theta -\alpha )+\cos (\theta -\beta )+\cos (\theta -\gamma )$

$=2\cos \left(\frac{2\theta -\alpha }{2}\right)\cos \frac{\alpha }{2}+2\cos \left(\frac{2\theta -(\beta +\gamma )}{2}\right)\cos \left(\frac{\gamma -\beta }{2}\right)$

$\alpha +\beta +\gamma =2\theta \Rightarrow \beta +\gamma =2\theta -\alpha$

$I=2\cos \left(\frac{2\theta -\alpha }{2}\right)\cos \frac{\alpha }{2}+2\cos \left(\frac{2\theta -(2\theta +\alpha )}{2}\right)\cos \left(\frac{\gamma -\beta }{2}\right)$

$=2\cos \left(\frac{\beta +\gamma }{2}\right)\cos \frac{\alpha }{2}+2\cos \frac{\alpha }{2}\cos \left(\frac{\gamma -\beta }{2}\right)$

$=2\cos \frac{\alpha }{2}(\cos \left(\frac{\beta +\gamma }{2}\right)+\cos \left(\frac{\gamma -\beta }{2}\right))$

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