Question

lf $A+B+C=\pi$, then $\cos 2A+\cos 2B+\cos 2C+4\cos A\cos B\cos C$ is equal to

• A. $0$
• B. $-1$
• C. $-2$
• D. $-3$

Answer 1

The correct option is B $-1$

$A+B+C=180\Rightarrow C=180-(A+B)$

Using $A+B+C=180\Rightarrow C=180-(A+B)$

We get

$\cos 2A+\cos 2B+\cos 2C=2\cos (A+B)\cos (A-B)+\cos 2(180-(A+B\left)\right)$

$=2\cos (A+B)\cos (A-B)+\cos (360-2(A+B\left)\right)$

$=2\cos (A+B)\cos (A-B)+\cos 2(A+B)$

Now from $\cos 2\theta =2co{s}^2\theta -1$

$2\cos (A+B)\cos (A-B)+\cos 2(A+B)$

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

$=2\cos (A+B)(\cos (A-B)+\cos (A+B))-1$

From substituting value of $A+B=180-C$ and $\cos C+\cos D=2\cos C+D)\cos (D-C)$

We get, $2\cos (A+B)(\cos (A-B)-\cos (A+B))-1$

$=2\cos (180-C)\left(2\cos A\cos B\right)-1$

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

$=-4\sin A\sin B\sin C-1$

Therefore

$\cos 2A+\cos 2B+\cos 2C+4\cos A\cos B\cos C-1$

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