Q.

If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ , 
then match the following

 LIst I List II
I)cos 3α + cos 3β + cos 3γ = 0
II)sin 2α + sin 2β + sin 2γ = 3
III)cos2 α + cos2 β+ cos2 γ = 3/2
IV)cos(2α-β-γ) + cos(2β-γ-α)+ cos(2γ-α-β) = 3cos(α+β+γ)


                                                                            

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a

d,c,b,a

b

b,c,a,d

c

d,a,c,b

d

d,a,b,c

answer is A.

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Detailed Solution

Let  a=cis α,  b=cis β,  c=cis γ

cos α+cos β+cos γ=0,  sin α+sin β+sin γ=0

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

a+b+c=0

 1a+1b+1c=i(cos α+cos β+cos γ)-1(sin α+sin β+sin γ)

1a+1b+1c=0  ab+bc+ca=0

(a+b+c)2=  a2+b2+c2+2(ab+bc+ca)=0

   a2+b2+c2=0

  (cos α+i sin α)2+(cos β+i sin β)2+(cos γ+i sin γ)2=0

  cos2α+i sin2α+cos2β+i sin2β+cos2γ+i sin2γ=0

  cos2α+cos2β+cos2γ=0,  sin2α+sin2β+sin2γ=0

2cos2α-1+2cos2α-2+2cos2γ-1=0

 cos2α+cos2β+cos2γ=3/2

a+b+c=0    a3+b3+c3=3abc

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

  cos3α+cos3β+cos3γ=3cos(α+β+γ)

   cos3α+i sin3α+cos3β+i sin3β+cos3γ+i sin3γ=3cis(α+β+γ)  (cos3α+cos3β+cos3γ)+i(sin3α+sin3β+sin3γ) =3(cos(α+β+γ)+i sin(α+β+γ))

a+b+c=0    a3+b3+c3=3abc

a2bc+b2ca+c2ab=3

(cis α)2cis β cis γ+(cis β)2cis γ cis α+(cis γ)2cis α cis β=3

cis(2α-β-γ)+cis(2β-γ-α)+cis(2α-α-β)=3

   cos(2-α-β-γ)+cos(2β-γ-α)+cos(2γ-α-β)=3

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