Question

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

Answer 1

$A+B+C=\pi$

${\mathrm{Sin}}^2\theta =\frac{1-\cos 2\theta }{2}$$\Rightarrow {\mathrm{Sin}}^4\theta ={\left(\frac{1-\cos 2\theta }{2}\right)}^2$

$\cos 2A+\cos 2B+\cos 2B=-1-4\cos A\cos B\cos C$

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

LHS=$si{n}^{4}A+si{n}^{4}B+si{n}^{4}C$

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

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

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

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

$=\frac{1}{4}\left[6+8\cos A\cos B\cos C+2\cos 2A\cos 2B\cos 2C\right]$

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

Lhs= RHS

Hence proof