Question

Prove $\frac{b^2-c^2}{a^2}\sin 2A+\frac{c^2-a^2}{b^2}\sin 2B+\frac{a^2-b^2}{c^2}\sin 2C=0$

Answer 1

We know that in a triangle $a=k\sin A,b=k\sin B,c=k\sin C$ and $\cos A=\frac{b^2+c^2-a^2}{2bc}$ and similarly using it for $\cos Band\cos C$

So,$\frac{\sin 2A}{a^2}=\frac{2}{a^2}\sin A\cos A=\frac{2a}{k{a}^2}\frac{b^2+c^2-a^2}{2bc}=\frac{b^2+c^2-a^2}{2kabc}$

$\frac{\sin 2B}{b^2}=\frac{2}{b^2}\sin B\cos B=\frac{2b}{k{b}^2}\frac{c^2+a^2-b^2}{2ca}=\frac{c^2+a^2-b^2}{2kabc}$

$\frac{\sin 2C}{c^2}=\frac{2}{c^2}\sin C\cos C=\frac{2c}{k{c}^2}\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2-c^2}{2kabc}$

Thus,

$L.H.S=(b^2-c^2)\frac{\sin 2A}{a^2}+(c^2-a^2)\frac{\sin 2B}{b^2}+(a^2-b^2)\frac{\sin 2C}{c^2}$

$=(b^2-c^2)\frac{b^2+c^2-a^2}{2kabc}+(c^2-a^2)\frac{c^2+a^2-b^2}{2kabc}+(a^2-b^2)\frac{a^2+b^2-c^2}{2kabc}$

$=\frac{1}{2akbc}[(b^2-c^2)(b^2+c^2-a^2)+(c^2-a^2)(c^2+a^2-b^2)+(a^2-b^2)(a^2+b^2-c^2)]$

$=\frac{1}{2akbc}[b^4-c^4-a^2(b^2-c^2)+c^4-a^4-b^2(c^2-a^2)+a^4-b^4-c^2(a^2-b^2)]$

$=\frac{1}{2akbc}[-a^2{b}^2+a^2{c}^2-b^2{c}^2+b^2{a}^2-c^2{a}^2+c^2{b}^2]$

$=\frac{1}{2akbc}\left(0\right)=0=R.H.S$

Hence proved