Question

If $A+B+C=\pi$, then $\cos 2A+\cos 2B+\cos 2C$ is equal to
A. $1+4\cos A\cos B\cos C$
B. $-1+4\sin A\sin B\sin C$
C. $-1-4\cos A\cos B\cos C$
D. None of these

Answer 1

We first take the sum of angles for the trigonometric ratios. We also use the multiple angle formula of $\cos 2X=2{{\cos }^{2}}X-1$. We convert them to their multiple forms. We take $2\cos C$ common and find the required solution.

Complete step by step answer:
It is given that $A+B+C=\pi$. We get $A+B=\pi -C$ and $C=\pi -\left( A+B \right)$.
We are going to use the formulas of sum of angles for the trigonometric ratios. We have $\cos X+\cos Y=2\cos \left( \dfrac{X+Y}{2} \right)\cos \left( \dfrac{X-Y}{2} \right)$.
We use the representation of $X=2A;Y=2B$ and get
$\cos 2A+\cos 2B =2\cos \left( \dfrac{2A+2B}{2} \right)\sin \left( \dfrac{2A-2B}{2} \right) \\\ \Rightarrow \cos 2A+\cos 2B =2\cos \left( A+B \right)\cos \left( A-B \right) \\\$
We change the angle and get
$2\cos \left( A+B \right)\cos \left( A-B \right) =2\cos \left( \pi -C \right)\cos \left( A-B \right) \\\ \Rightarrow 2\cos \left( A+B \right)\cos \left( A-B \right) =-2\cos C\cos \left( A-B \right) \\\$
We also use the multiple angle formula of $\cos 2C=2{{\cos }^{2}}C-1$. Therefore,
$\cos 2A+\cos 2B+\cos 2C =-2\cos C\cos \left( A-B \right)+2{{\cos }^{2}}C-1$
We take $2\cos C$ common and get
$-2\cos C\cos \left( A-B \right)+2{{\cos }^{2}}C-1 =-1+2\cos C\left[ \cos C-\cos \left( A-B \right) \right] \\\$
We again change the $\cos C$ in the bracket.
$\cos C=\cos \left[ \pi -\left( A+B \right) \right]=-\cos \left( A+B \right)$.
$\Rightarrow \cos 2A+\cos 2B+\cos 2C =-1+2\cos C\left[ -\cos \left( A+B \right)-\cos \left( A-B \right) \right] \\\ \Rightarrow \cos 2A+\cos 2B+\cos 2C =-1-2\cos C\left[ \cos \left( A+B \right)+\cos \left( A-B \right) \right] \\\$
We represent $X=A+B;Y=A-B$ in $\cos X+\cos Y=2\cos \left( \dfrac{X+Y}{2} \right)\cos \left( \dfrac{X-Y}{2} \right)$ to get
$\cos X+\cos Y =2\cos \left( \dfrac{A+B+A-B}{2} \right)\cos \left( \dfrac{A+B-A+B}{2} \right) \\\ \Rightarrow \cos X+\cos Y =2\cos A\cos B \\\$
Therefore,
$\cos 2A+\cos 2B+\cos 2C =-1-2\cos C\left[ 2\cos A\cos B \right] \\\ \therefore \cos 2A+\cos 2B+\cos 2C =-1-4\cos A\cos B\cos C$

Hence, the correct option is C.

Note: Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty$. In that case we have to use the formula $x=n\pi \pm a$ for $\cos \left( x \right)=\cos a$ where $0\le a\le \pi$.