Question

In $\Delta ABC$ prove: $\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C$

Answer 1

In the answer we want $-1$ and as such, we write $\cos 2A$ as $2{\cos }^{2}A-1$ and combine the other two terms.

L.H.S$=(2{\cos }^{2}A-1)+2\cos (B+C)\cos (B-C)$

$=-1+2{\cos }^{2}A-2\cos A\cos (B-C)$

$=-1+2\cos A[\cos A-\cos (B-C\left)\right]$

$=-1+2\cos A[-cos(B+C)-\cos (B-C\left)\right]$

$=-1-2\cos A\left(2\cos B\cos C\right)$ ....... $[2\cos A\cos B=\cos (A+B)+\cos (A-B\left)\right]$

$=-1-4\cos A\cos B\cos C$.