Question

If $A+B+C=\pi$ , then prove that
${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2+2\cos A\cdot \cos B\cdot \cos C$ .

Answer 1

$A+BC=\pi$ (given)

$\Rightarrow ,C=\pi -(A+B)-\left(i\right)$

Now,

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

[As $\cos 2x=1-2{\sin }^{2}x$ ]

$=\frac{3}{2}-\frac{1}{2}[\cos 2A+\cos 2B+\cos 2C]$

$=[Using\cos C+\cos 0=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}]$

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

$=\frac{3}{2}-\frac{1}{2}\left[2\cos \right(\pi -C\left)\cos \right(A-B)+\cos 2C]$ (From eq (i))

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

$=\frac{3}{2}+\cos C\cos (A-B)-{\cos }^{2}C+\frac{1}{2}$ [multiplying $\frac{1}{2}$ inside bracket]

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

$=2+\cos C\left[\cos \right(A-B)-\cos (\pi -(A+B)\left)\right]$ [From eq (i)]

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

$=2+\cos C\times \left[2\cos A\cos B\right]$

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