Question

If cos α + cos β + cos γ = sin α + sin β + sin γ = 0. then show that sin 3α + sin 3β + sin 3γ = 3 sin(α + β + γ)

Answer 1

Let a = cos α + i sin α = e^iα

b = cos β + i sin β = e^iβ

c = cos γ + i sin γ = e^iγ

a + b + c = (cos α + cos β + cos γ) + i (sin α + sin β + sin γ)

⇒ a + b + c = 0 + i 0

⇒ a + b + c = 0

If a + b + c = 0 then a³ + b³ + c³ = 3abc

`("e"^("i"alpha))^3 + ("e"^("i"beta))^3 + ("e"^("i"γ))^3 = 3"e"^("i"alpha) "e"^("i"beta) "e"^("i"γ)` 

`"e"^("i"3alpha) + "e"^("i"3beta) + "e"^("i"3γ) = 3"e"^("i"(alpha + beta + γ)`

(cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ) = 3[cos (α + β + γ) + i sin (α + β + γ)]

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

Equating real and Imaginary parts

sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)