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

Answer 1

The correct option is D

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

Using ${\sin }^{2}A=\frac{1-\cos 2A}{2}$$\therefore {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+\frac{1-\cos 2C}{2}$

$=\frac{3}{2}-\frac{1}{2}(\cos 2A+\cos 2B+\cos 2C)\cdots \left(1\right)$
$=\frac{1}{2}(3-\left[2\cos \right(A+B\left)\cos \right(A-B)+\cos (2C\left)\right])$
$=\frac{1}{2}(3-[(-2\cos C)\cos (A-B)+2{\cos }^{2}C-1]$
$=\frac{1}{2}(4-(2\cos C\{\cos C-\cos (A-B\left)\right\}\left)\right)$
$=\frac{1}{2}(4-2\cos C(-\cos (A+B)-\cos (A-B)\left)\right)$
$=\frac{1}{2}(4+2\cos C(\cos (A+B)+\cos (A-B)\left)\right)$
$=\frac{1}{2}(4+2\cos C\times 2\cos A\cos B)$
$=2+2\cos A\cos B\cos C$
Hence, The correct answer is option (c)