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Question

If sinα+sinβ+sinγ=0= cosα+cosβ+cosγ,

value of sin2α+sin2β+sin2γ


A

2/3

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B

3/2

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C

1/2

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D

1

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Solution

The correct option is B

3/2


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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