Question

If $\alpha +\beta +\gamma =2\theta $ , then $\cos \theta +\cos \left( \theta -\alpha \right)+\cos \left( \theta -\beta \right)+\cos \left( \theta -\gamma \right)$ is equal to:
1) $4\sin \left( \dfrac{\alpha }{2} \right)\cos \left( \dfrac{\beta }{2} \right)\sin \left( \dfrac{\gamma }{2} \right)$
2) $4\cos \left( \dfrac{\alpha }{2} \right)\cos \left( \dfrac{\beta }{2} \right)\cos \left( \dfrac{\gamma }{2} \right)$
3) $4\sin \left( \dfrac{\alpha }{2} \right)\sin \left( \dfrac{\beta }{2} \right)\sin \left( \dfrac{\gamma }{2} \right)$
4) $4\sin \alpha \sin \beta \sin \gamma $

Answer 1

Hint: Here in this question it is given that $\alpha +\beta +\gamma =2\theta $ we have been asked to find the simplified value of the given expression $\cos \theta +\cos \left( \theta -\alpha \right)+\cos \left( \theta -\beta \right)+\cos \left( \theta -\gamma \right)$ . For answering this question we will use the trigonometric identity given as $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ .

Complete step-by-step solution:
Now considering from the question it is given that $\alpha +\beta +\gamma =2\theta $ we have been asked to find the simplified value of the given expression $\cos \theta +\cos \left( \theta -\alpha \right)+\cos \left( \theta -\beta \right)+\cos \left( \theta -\gamma \right)$ .
From the basic concepts of trigonometry we know the trigonometric identity given as $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ .
Now by using this formula and simplifying the given expression we will have $\Rightarrow 2\cos \left( \dfrac{2\theta -\alpha }{2} \right)\cos \left( \dfrac{\alpha }{2} \right)+\cos \left( \theta -\beta \right)+\cos \left( \theta -\gamma \right)$ .
Now by again using the same formula in the above expression we will have $\Rightarrow 2\cos \left( \dfrac{2\theta -\alpha }{2} \right)\cos \left( \dfrac{\alpha }{2} \right)+2\cos \left( \dfrac{2\theta -\beta -\gamma }{2} \right)\cos \left( \dfrac{\gamma -\beta }{2} \right)$.
As it is given that $\alpha +\beta +\gamma =2\theta $ we will have $\Rightarrow 2\theta -\beta -\gamma =\alpha $ by using this in the above expression we will have $\Rightarrow 2\cos \left( \dfrac{2\theta -\alpha }{2} \right)\cos \left( \dfrac{\alpha }{2} \right)+2\cos \left( \dfrac{\alpha }{2} \right)\cos \left( \dfrac{\gamma -\beta }{2} \right)$ .
Now by taking out $2\cos \left( \dfrac{\alpha }{2} \right)$ we will have $\Rightarrow 2\cos \left( \dfrac{\alpha }{2} \right)\left( \cos \left( \dfrac{2\theta -\alpha }{2} \right)+\cos \left( \dfrac{\gamma -\beta }{2} \right) \right)$ .
Now again using the trigonometric identity that we have discussed above in the above expression we will have $\Rightarrow 2\cos \left( \dfrac{\alpha }{2} \right)\left( 2\cos \left( \dfrac{\left( \dfrac{2\theta -\alpha +\gamma -\beta }{2} \right)}{2} \right)\cos \left( \dfrac{\left( \dfrac{2\theta -\alpha +\beta -\gamma }{2} \right)}{2} \right) \right)$ .
As it is given that $\alpha +\beta +\gamma =2\theta $ we will have $\Rightarrow 2\theta -\alpha -\beta =\gamma $ by using this in the above expression we will have
$\begin{align}
  & \Rightarrow 2\cos \left( \dfrac{\alpha }{2} \right)\left( 2\cos \left( \dfrac{\left( \dfrac{\gamma +\gamma }{2} \right)}{2} \right)\cos \left( \dfrac{\left( \dfrac{2\theta -\alpha +\beta -\gamma }{2} \right)}{2} \right) \right) \\
 & \Rightarrow 2\cos \left( \dfrac{\alpha }{2} \right)\left( 2\cos \left( \dfrac{\gamma }{2} \right)\cos \left( \dfrac{\left( \dfrac{2\theta -\alpha +\beta -\gamma }{2} \right)}{2} \right) \right) \\
\end{align}$ .
As it is given that $\alpha +\beta +\gamma =2\theta $ we will have $\Rightarrow 2\theta -\alpha -\gamma =\beta $ by using this in the above expression we will have
$\begin{align}
  & \Rightarrow 2\cos \left( \dfrac{\alpha }{2} \right)\left( 2\cos \left( \dfrac{\gamma }{2} \right)\cos \left( \dfrac{\left( \dfrac{\beta +\beta }{2} \right)}{2} \right) \right) \\
 & \Rightarrow 2\cos \left( \dfrac{\alpha }{2} \right)\left( 2\cos \left( \dfrac{\gamma }{2} \right)\cos \left( \dfrac{\beta }{2} \right) \right) \\
\end{align}$ .
Therefore we can conclude that when it is given that $\alpha +\beta +\gamma =2\theta $ we have been asked to find the simplified value of the given expression $\cos \theta +\cos \left( \theta -\alpha \right)+\cos \left( \theta -\beta \right)+\cos \left( \theta -\gamma \right)=4\cos \left( \dfrac{\alpha }{2} \right)\cos \left( \dfrac{\beta }{2} \right)\cos \left( \dfrac{\gamma }{2} \right)$ .
Hence we will mark the option “2” as correct.

Note: During the process of answering questions of this type we should be sure with the formula that we are going to use in between the steps of simplifying the expression. This is a very easy question and can be answered in a short span of time just by using a single formula only. Similarly we have another formula given as $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ .