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Question

If cosα+2cosβ+3cosγ=sinα+2sinβ+3sinγ=0, then

A
cos3α+8cos3β+27cos3γ=18cos(α+β+γ)
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B
cos3α+8cos3β+27cos3γ=18sin(α+β+γ)
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C
sin3α+8sin3β+27sin3γ=18sin(α+β+γ)
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D
sin3α+8sin3β+27sin3γ=18cos(α+β+γ)
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Solution

The correct option is C sin3α+8sin3β+27sin3γ=18sin(α+β+γ)
Given,
cosα+2cosβ+3cosγ=0sinα+2sinβ+3sinγ=0

Let, a=cosα+isinα=eiα
b=cosβ+isinβ=eiβ
c=cosγ+isinγ=eiγ

Now,
a+2b+3c=(cosα+2cosβ+3cosγ)+i(sinα+2sinβ+3sinγ)
a+2b+3c=0
{ If A+B+C=0, then A3+B3+C3=3ABC}

a3+8b3+27c3=18abc
e3iα+8e3iβ+27e3iγ=18ei(α+β+γ)
By equating real and imaginary parts, we have
cos3α+8cos3β+27cos3γ=18cos(α+β+γ)
and, sin3α+8sin3β+27sin3γ=18sin(α+β+γ)

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