Question

In a $\Delta$ ABC, prove that ${\sin }^{4}A+{\sin }^{4}B+{\sin }^{4}C=\frac{3}{2}+2\cos A\cos B\cos C+\frac{1}{2}\cos 2A\cos 2B\cos 2C$.

Answer 1

L.H.S $={\left(\frac{1-\cos 2A}{2}\right)}^2+{\left(\frac{1-\cos 2B}{2}\right)}^2+{\left(\frac{1-\cos 2C}{2}\right)}^2$
$=\frac{3}{4}+\frac{1}{4}({\cos }^{2}2A+{\cos }^{2}2B+{\cos }^{2}2C)-\frac{1}{2}(\cos 2A+\cos 2B+\cos 2C)$
$=\frac{3}{4}+\frac{1}{4}\left\{\frac{1}{2}\right(1+\cos 4A)+\frac{1}{2}(1+\cos 4B)+\frac{1}{2}(1+\cos 4C\left)\right\}-\frac{1}{2}(\cos 2A+\cos 2B+\cos 2C)$
$=\frac{3}{4}+\frac{3}{8}+\frac{1}{8}(\cos 4A+\cos 4B+\cos 4C)-\frac{1}{2}(\cos 2A+\cos 2B+\cos 2C)$
$=\frac{9}{8}+\frac{1}{8}\left[2\cos \right(2A+2B\left)\cos \right(2A-2B)+2{\cos }^{2}2C-1]-\frac{1}{2}\left[2\cos \right(A+B\left)\cos \right(A-B)+2{\cos }^{2}C-1]$
$=\frac{3}{2}+\frac{1}{4}\left[\cos \right(2\pi -2C\left)\cos \right[2A-2B)+{\cos }^{2}2C]-\left[\cos \right(\pi -C\left)\cos \right(A-B)+{\cos }^{2}C]$
$\because \frac{9}{8}-\frac{1}{8}+\frac{1}{2}=\frac{3}{2}$
$[\because A+B+C=\pi$ gives $2A+2B=2\pi -2C$ and $A+B=\pi -C]$
$=\frac{3}{2}+\frac{1}{4}\cos 2C\left[\cos \right(2A-2B)+\cos (2A-2B\left)\right]-\cos C[-\cos (A-B)-\cos (A+B\left)\right]$
$=\frac{3}{2}+\frac{1}{4}\cos 2C\cdot 2\cos 2A\cos 2B+\cos C\cdot 2\cos A\cos B$
$=\frac{3}{2}+2\cos A\cos B\cos C+\frac{1}{2}\cos 2A\cos 2B\cos 2C$.