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Question

If Cosα+Cosβ+Cosγ=0=Sinα+Sinβ+Sinγ prove that
Cos2α+Cos2β+Cos2γ=32=Sin2α+Sin2β+Sin2γ

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Solution

Given that:
Cosα+Cosβ+Cosγ=0=Sinα+Sinβ+Sinγ
To prove:
Cos2α+Cos2β+Cos2γ=32=Sin2α+Sin2β+Sin2γ
Solution:
Cosα+Cosβ+Cosγ=0=Sinα+Sinβ+Sinγ
Let, Z=e(α+β+γ)=0
Therefore, Cosα+Cosβ+Cosγ=0=Sinα+Sinβ+Sinγ
or, Z2=0 (Z=0)
or, Z2=e2(α+β+γ)=0
or, Z2=e(2α+2β+2γ)=0
or, Cos2α+Cos2β+Cos2γ=0=Sin2α+Sin2β+Sin2γ
or, Cos2α+Cos2β+Cos2γ=0
or, 2Cos2α1+2Cos2β1+2Cos2γ1=0 (Cos2θ=2Cos2θ1)
or, 2(Cos2α+Cos2β+Cos2γ)=3
or, Cos2α+Cos2β+Cos2γ=32
Similarly,
or, Sin2α+Sin2β+Sin2γ=0
or, 12Sin2α+12Sin2β+12Sin2γ=0 (Cos2θ=12Sin2θ)
or, 2(Sin2α+Sin2β+Sin2γ)=3
or, Sin2α+Sin2β+Sin2γ=32
Therefore, Cos2α+Cos2β+Cos2γ=32=Sin2α+Sin2β+Sin2γ



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