Question

If $A+B+C=\pi$, then ${\sin }^{4}A+{\sin }^{4}B+{\sin }^{4}C=\frac{3}{2}+2\cos A\cos B\cos C+\frac{1}{2}\cos 2A\cos 2B\cos 2C$

Answer 1

We have,

$A+B+C=\pi$

We know that

${\sin }^{2}x=\left(\frac{1-\cos 2x}{2}\right)$

${\sin }^{4}x={\left(\frac{1-\cos 2x}{2}\right)}^2$

$\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C$ $........\left(1\right)$

${\cos }^{2}2A+{\cos }^{2}2B+{\cos }^{2}2C=1+2\cos 2A\cos 2B\cos 2C$ $........\left(2\right)$

Now,

L.H.S

$={\sin }^{4}A+{\sin }^{4}B+{\sin }^{4}C$

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

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

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

From equations $\left(1\right)$ and $\left(2\right)$

$=\frac{1}{2}[3+(1+\cos 2A\cos 2B\cos 2C)-2(-1-4\cos A\cos B\cos C\left)\right]$

$=\frac{1}{2}[4+\cos 2A\cos 2B\cos 2C+2+8\cos A\cos B\cos C]$

$=\frac{1}{2}[6+\cos 2A\cos 2B\cos 2C+8\cos A\cos B\cos C]$

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

R.H.S

Hence, proved.