Question

If $A+B+C={180}^{\circ }$. Find ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$ =

• A. $-1-4\cos A\cos B\cos C$
• B. $1+2\cos A\cos B\cos C$
• C. $2+2\cos A\cos B\cos C$
• D. $4\sin A\sin B\sin C$

Answer 1

The correct option is C

$2+2\cos A\cos B\cos C$

${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$

We know conditional identities but all of them are in the form of degree one.Here, power of sine is 2.

First we should reduce this power to 1 and then simplify it

We know $\cos 2A=1-2{\sin }^{2}A$

$\Rightarrow {\sin }^{2}A=\frac{1-\cos 2A}{2}$

So, $\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+\frac{1-\cos 2C}{2}$

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

$\cos 2A+\cos 2B+\cos 2C$
$=2\cos \left(\frac{2A+2B}{2}\right)\cdot \cos \left(\frac{2A-2B}{2}\right)+\cos \left(2\right(\pi -(A+B)\left)\right)$
$=2\cos (A+B)\cdot \cos (A-B)+\cos (2\pi -2(A+B\left)\right)$
$=2\cos (A+B)\cdot \cos (A-B)+\cos 2(A+B)$
$=2\cos (A+B)\cdot \cos (A-B)+2{\cos }^2(A+B)-1$
$=2\cos (A+B)\left[\cos \right(A-B)+\cos (A+B\left)\right]-1$
$=2\cos (180-c)[2\cos \frac{(A-B+A+B)}{2}\cdot \cos \frac{(A-B-A-B)}{2}]-1$
$=-2\cos C\times [2\cos A\cdot \cos (-B\left)\right]-1$
$\left(\cos \right(-\theta )=\cos \theta )$
$\therefore$ given expression becomes $2+2\cos A\cos B\cos C$