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Question

If cos α+cos β+cos γ=sin α +sin β +sin γ =0 then cos 3α+cos 3 β + cos 3 γ equals to

A
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C
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Solution

The correct option is C
cos α+cos β+cos γ=0 and sinα i sinβ + sin γ=0 Leta=cos α i sin α; b=cos β i i sin β and c=cos γ +i sin γ. Therefore a+b+c=(cosα+cosβ+cosγ)+i(sinα+sinβ+sinγ) =0+i0=0 if a+b+c=0, then a3+b3+c3=3abc or (cosα+i sinα)3+(cosβ+i sinβ)3+(cosγ+i sinγ)3 =3(cosα+i sin α)(cosβ+i sin β)(cos γ+i sin γ) (cos 3γ+i sin 3γ)+(cos 3β+i sin 3β)+(cos 3γ+i sin 3γ) =3[cos(α+β+γ)+i sin(α+β+γ)] or cos 3α+cos 3β+cos 3) γ=3 cos(α+β+γ).

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