Question

If $\cos \alpha +\cos \beta +\cos \gamma =0=\sin \alpha +\sin \beta +\sin \gamma$, Prove that
$\cos 3\alpha +\cos 3\beta +\cos 3\gamma =3\cos (\alpha +\beta +\gamma )$

Answer 1

Given that cosα + cosβ + cosγ = 0.....Equation 1and sinα +sinβ + sinγ = 0.....Equation 2
Multiply Equation 2 by i and add to Equation 1 ,
cosα + cosβ + cosγ +i sinα +i sinβ + i sinγ = 0 ,

z1 + z2 + z3=0 (z1 = cos α + i sin α) , z2 = cos β + i sin β and z3= cos γ + i sin γ
z1+z2+z3 =0 => z1 ³ + z2 ³ +z3 ³ = 3 z1 z2 z3
(cos α + i sin α)³ +(cos β+ i sin β)³ + (cos γ + i sin γ)³ = 3 (cos α + i sin α) (cos β+ i sin β) (cos γ + i sin γ)

i.e (cos 3α + i sin 3α) +(cos 3β+ i sin 3β ) + (cos 3γ + i sin 3γ) = 3 (cos (α+β+γ) + i 3 sin (α+β+γ)
i.e cos 3α +cos 3β+cos 3γ +i( sin 3α + sin 3β+sin 3γ)=3 (cos (α+β+γ) + i 3 sin (α+β+γ)

Equate real and imaginary terms , we get cos3α + cos3β + cos3γ = 3cos( α+β+γ) and sin3α +sin3β + sin3γ = 3sin( α+β+γ)