Question

If $\cos A+\cos B+\cos C=0$, prove that $\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C$.

Answer 1

$\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C$

$=(4{\cos }^{3}A-3\cos A)+(4{\cos }^{3}B-3\cos B)+(4{\cos }^{3}C-3\cos C)$

$=4({\cos }^{3}A+{\cos }^{3}B+{\cos }^{3}C)-3(\cos A+\cos B+\cos C)$

$=4[3\cos A+\cos B+\cos C]-3\left(0\right)$ [$\because$ If $a+b+c=0$,then $a^3+b^3+c^3=3abc$]

$=12\cos A+\cos B+\cos C$