Question

$\cos (\alpha +\beta +\gamma )+\cos (\alpha -\beta -\gamma )+\cos (\beta -\gamma -\alpha )+\cos (\gamma -\alpha -\beta )=$

• A. $2\cos \alpha \cos \beta \cos \gamma$
• B. $3\cos \alpha \cos \beta \cos \gamma$
  
• C. $4\cos \alpha \cos \beta \cos \gamma$
• D. $6\cos \alpha \cos \beta \cos \gamma$

Answer 1

The correct option is C $4\cos \alpha \cos \beta \cos \gamma$

Let $I=\cos (\alpha +\beta +\gamma )+\cos (\alpha -\beta -\gamma )+\cos (\beta -\gamma -\alpha )+\cos (\gamma -\alpha -\beta )$

$=\cos (\alpha +\beta +\gamma )+\cos (\gamma -\alpha -\beta )+\cos (\alpha -\beta -\gamma )+\cos (\beta -\gamma -\alpha )$

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

Using $\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

We get

$I=2\cos \gamma \cos (\alpha +\beta )+2\cos (\alpha -\beta )\cos \gamma$

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

Again using $\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$I=2\cos \gamma \left(\cos \right(\alpha +\beta )+\cos (\alpha -\beta \left)\right)$

$\Rightarrow I=2\cos \gamma \left(2\cos \alpha \cos \beta \right)=4\cos \alpha \cos \beta \cos \gamma$