﻿ Given that \cosα+\cosβ+\cosγ=0 and \sinα+\sinβ+\sinγ=0, prove that \sin3α+\sin3β+\sin3γ=3\sin(α+β+γ)

#### Question

Given that \cosα+\cosβ+\cosγ=0 and \sinα+\sinβ+\sinγ=0, prove that \sin3α+\sin3β+\sin3γ=3\sin(α+β+γ)

Collected in the board: Imaginary number

Steven Zheng posted 1 week ago

Given conditions

\cosα+\cosβ+\cosγ=0
(1)
\sinα+\sinβ+\sinγ=0
(2)

Let

a=\cosα+i\sinα

b=\cosβ+i\sinβ

c=\cosγ+i\sinγ

Then

a+b+c=0

Then

According to the identity

a^3+b^3+c^3−3abc = (a+b+c)(a^2+b^2+c^2−ab−bc−ca)

We get

a^3+b^3+c^3 = 3abc

that is,

(\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)

Apply power and multiplication rules for complex numbers

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

Compare the real and imaginary parts of both sides, we get

\cos 3α +\cos 3β+\cos 3γ = 3\cos (α+β+γ)

and

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

Steven Zheng posted 1 week ago

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