Question

If $\cos \alpha +\cos \beta +\cos \gamma =0=\sin \alpha +\sin \beta +\sin \gamma$ prove that $\cos 3\alpha +\cos 3\beta +\cos 3\gamma =3\cos (\alpha +\beta +\gamma )$ and $\sin 3\alpha +\sin 3\beta +\sin 3\gamma =3\sin (\alpha +\beta +\gamma )$

Answer 1

Given $\cos \alpha +\cos \beta +\cos \gamma =\sin \alpha +\sin \beta +\sin \gamma =0$

Consider $(\cos \alpha +i\sin \alpha )+(\cos \beta +i\sin \beta )+(\cos \gamma +i\sin \gamma )=(\cos \alpha +\cos \beta +\cos \gamma )+i(\sin \alpha +\sin \beta +\sin \gamma )=0$

$\Rightarrow$ $e^{i\alpha }+e^{i\beta }+e^{i\gamma }=0$

We know that when $a+b+c=0$ then $a^3+b^3+c^3=3abc$

Therefore we get $e^{i3\alpha }+e^{i3\beta }+e^{i3\gamma }=3e^{i\alpha }{e}^{i\beta }{e}^{i\gamma }=3e^{i(\alpha +\beta +\gamma )}$

$\Rightarrow$ $\cos 3\alpha +\cos 3\beta +\cos 3\gamma =3\cos (\alpha +\beta +\gamma )$ and $\sin 3\alpha +\sin 3\beta +\sin 3\gamma =3\sin (\alpha +\beta +\gamma )$