Question

Find the value of b if $\cos \alpha +\cos \beta +\cos \gamma +\cos \left( \alpha +\beta +\gamma \right)=b\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\beta +\gamma }{2} \right)\cos \left( \dfrac{\gamma +\alpha }{2} \right).$

Answer 1

We know the formula for the extension of addition of two cosines that converts it into the multiplication form. According to this formula, $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ . We need to use this multiple times, and exploit the fact that $\cos \left( -\theta \right)=\cos \theta$ , to find the value of b.

Complete step by step solution:
We know that to convert the addition of cosines into multiplication of cosines, we have the following formula
$\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)...\left( i \right)$
We are given that $\cos \alpha +\cos \beta +\cos \gamma +\cos \left( \alpha +\beta +\gamma \right)=b\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\beta +\gamma }{2} \right)\cos \left( \dfrac{\gamma +\alpha }{2} \right)...\left( ii \right)$
Let us use the LHS part of this equation.
$LHS=\cos \alpha +\cos \beta +\cos \gamma +\cos \left( \alpha +\beta +\gamma \right)$
Let us use equation (i),
$LHS=\left[ \cos \alpha +\cos \beta \right]+\left[ \cos \gamma +\cos \left( \alpha +\beta +\gamma \right) \right]$
$\Rightarrow LHS=\left[ 2\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right) \right]+\left[ 2\cos \left( \dfrac{\gamma +\left( \alpha +\beta +\gamma \right)}{2} \right)\cos \left( \dfrac{\gamma -\left( \alpha +\beta +\gamma \right)}{2} \right) \right]$
Simplifying the above equation, we get
$LHS=\left[ 2\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right) \right]+\left[ 2\cos \left( \dfrac{\alpha +\beta +2\gamma }{2} \right)\cos \left( \dfrac{-\alpha -\beta }{2} \right) \right]$
We are very well aware of the fact that $\cos \left( -\theta \right)=\cos \theta$ . Hence, $\cos \left( \dfrac{-\alpha -\beta }{2} \right)=\cos \left( \dfrac{\alpha +\beta }{2} \right)$ .
Thus, we can write,
$LHS=\left[ 2\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right) \right]+\left[ 2\cos \left( \dfrac{\alpha +\beta +2\gamma }{2} \right)\cos \left( \dfrac{\alpha +\beta }{2} \right) \right]$
We can further simplify the above equation as
$LHS=2\cos \left( \dfrac{\alpha +\beta }{2} \right)\left[ \cos \left( \dfrac{\alpha -\beta }{2} \right)+\cos \left( \dfrac{\alpha +\beta +2\gamma }{2} \right) \right]$
We can again use equation (i) for the addition of cosines $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
Thus, we get
$LHS=2\cos \left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\cos \left( \dfrac{\dfrac{\alpha -\beta }{2}+\dfrac{\alpha +\beta +2\gamma }{2}}{2} \right)\cos \left( \dfrac{\dfrac{\alpha -\beta }{2}-\dfrac{\alpha +\beta +2\gamma }{2}}{2} \right) \right]$
We can simplify the above equation as
$LHS=2\cos \left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\cos \left( \dfrac{\dfrac{\alpha -\beta +\alpha +\beta +2\gamma }{2}}{2} \right)\cos \left( \dfrac{\dfrac{\alpha -\beta -\alpha -\beta -2\gamma }{2}}{2} \right) \right]$
We can cancel few terms and we will get
$LHS=2\cos \left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\cos \left( \dfrac{\dfrac{2\alpha +2\gamma }{2}}{2} \right)\cos \left( \dfrac{\dfrac{-2\beta -2\gamma }{2}}{2} \right) \right]$
$\Rightarrow LHS=2\cos \left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\cos \left( \dfrac{\alpha +\gamma }{2} \right)\cos \left( \dfrac{-\beta -\gamma }{2} \right) \right]$
We can, now, again use the formula $\cos \left( -\theta \right)=\cos \theta$ . Hence, we can also say that $\cos \left( \dfrac{-\beta -\gamma }{2} \right)=\cos \left( \dfrac{\beta +\gamma }{2} \right)$ .
Thus, our equation is simplified as
$\Rightarrow LHS=2\cos \left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\cos \left( \dfrac{\alpha +\gamma }{2} \right)\cos \left( \dfrac{\beta +\gamma }{2} \right) \right]$
Or we can write it as
$\Rightarrow LHS=4\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\beta +\gamma }{2} \right)\cos \left( \dfrac{\gamma +\alpha }{2} \right)$
We can now compare this LHS part with the RHS part of equation (ii). So, now we have
$4\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\beta +\gamma }{2} \right)\cos \left( \dfrac{\gamma +\alpha }{2} \right)=b\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\beta +\gamma }{2} \right)\cos \left( \dfrac{\gamma +\alpha }{2} \right)$
Hence, by comparing, we can say that $b=4$ .
Thus, the value of b is 4.

Note: This problem requires a good amount of calculation. So, we must be very careful, and try not to make mistakes while calculating. Some of the students, in hurry, may consider $\cos \left( \dfrac{\dfrac{2\alpha +2\gamma }{2}}{2} \right)\text{ as }\cos \left( \dfrac{2\alpha +2\gamma }{\dfrac{2}{2}} \right).$We must be clear that \cos \left( \dfrac{\dfrac{2\alpha +2\gamma }{2}}{2} \right)=\cos \left( \dfrac{\left\\{ \dfrac{2\alpha +2\gamma }{2} \right\\}}{2} \right) and not \cos \left( \dfrac{2\alpha +2\gamma }{\left\\{ \dfrac{2}{2} \right\\}} \right).