Question

$A+B+C={180}^0$. Find the value of ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C$

• A. $1+2\cos A\cos B\cos C$
• B. $1+2\sin A\sin B\sin C$
• C. $1-2\cos A\cos B\cos C$
• D. $1-2\sin A\sin B\sin C$

Answer 1

Answer: (C)

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

Using $\cos 2x=2{\cos }^{2}x-1\Rightarrow {\cos }^{2}x=\frac{\cos 2x+1}{2}$

$I={\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C$
$=\frac{1}{2}(\cos 2A+\cos 2B+\cos 2C+3)$

Now, from $A+B+C=180\Rightarrow 2C=360-2(A+B)$

We get

$I=\frac{1}{2}(2\cos (A+B)\cos (A-B)+\cos (360-2(A+B))+3)=\frac{1}{2}(2\cos (A+B)\cos (A-B)+\cos (2(A+B))+3)=\frac{1}{2}(2\cos (A+B)\cos (A-B)+2{\cos }^2(A+B)-1+3)=\frac{1}{2}\left[2\cos \right(180-C\left)\right(\cos (A+B)+\cos (A-B))+2]=\frac{1}{2}[-2\cos C\left(2\cos A\cos B\right)+2]$

$=1-2\cos A\cos B\cos C$