$a\sin x = b\sin \left( {x + \frac{{2\pi }}{3}} \right) = c\sin \left( {x + \frac{{4\pi }}{3}} \right)$

If $a\sin x = b\sin \left( {x + \frac{{2\pi }}{3}} \right) = c\sin \left( {x + \frac{{4\pi }}{3}} \right) \neq 0$, prove that $(a+b+c)^2=a^2+b^2+c^2$.

Solution

Dividing by $abc$ ($\because a,b,c \neq 0$),

$\frac{{\sin x}}{{bc}} = \frac{{\sin \left( {x + \frac{{2\pi }}{3}} \right)}}{{ca}} = \frac{{\sin \left( {x + \frac{{4\pi }}{3}} \right)}}{{ab}} = \frac{{\sin x + \sin \left( {x + \frac{{2\pi }}{3}} \right) + \sin \left( {x + \frac{{4\pi }}{3}} \right)}}{{bc + ca + ab}}$

$\therefore (ab + bc + ca)\sin x = bc\left\{ {\sin x + \underbrace {\sin \left( {x + \frac{{2\pi }}{3}} \right) + \sin \left( {x + \frac{{4\pi }}{3}} \right)}_{}} \right\}$

$ \Rightarrow (ab + bc + ca)\sin x = bc\left\{ {\sin x + 2\sin \left( {x + \pi } \right)\cos \left( {\frac{\pi }{3}} \right)} \right\}$

$ \Rightarrow (ab + bc + ca)\sin x = bc\left\{ {\sin x - 2\sin x.\frac{1}{2}} \right\} = 0$

$\therefore ab + bc + ca = 0 \because \sin x \ne 0$

Now, ${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2(ab + bc + ca) = {a^2} + {b^2} + {c^2}$