Discussion Forum : Complex Numbers
Question -


If  sinα+sinβ+sinγ=0= cosα+cosβ+cosγ,


value of  sin2α+sin2β+sin2γ


 

Options:
A .   23
B .   32
C .   12
D .   1
Answer: Option B
:
B

cosα+cosβ+cosγ = 0............(i)


 sinα+sinβ+sinγ  =0...........(ii)


Let a=cosα+isinα, b=cosβ+isinβ


     c=cosγ+isinγ


a+b+c=0 [By (i) and (ii)]....(iii)


1a+1b+1c


=cosα-isinα+cosβ-isinβ+cosγ-isinγ


ab+bc+ca=0 ......(iv) [by (i) and (ii)]


Squaring both sides of equation (iii),


we get a2+b2+c2+2ab+2bc+2ca=0


or a2+b2+c2=0     [by (iv)]


 [cosα+isinα]2 + [cosβ+isinβ]2 + [cosγ+isinγ]2 = 0


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


Separation of real and imaginary part,


cos2α+cos2β+cos2γ=0 .........(v)


And sin2α+sin2β+sin2γ =0 .........(iv)


1-2sin2α+1-2sin2β+1-2sin2γ=0 [By eq.(v)]


Therefore, sin2α+sin2β+sin2γ= 32



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