Question
If cosα+cosβ+cosγ=0=sinα+sinβ+sinγ then
cos2α+cos2β+cos2γ equals
cos2α+cos2β+cos2γ equals
Answer: Option C
:
C
cosα+cosβ+cosγ=0 ......(i)
sinα+sinβ+sinγ=0 .......(ii)
Let a=cosα+isinα, b=cosβ+isinβ
c=cosγ+isinγ
⇒ a+b+c=0 .......(iii)
1a+1b+1c
=[cosα+isinα]−1+[cosβ+isinβ]−1+[cosγ+isinγ]−1
=cosα-isinα+cosβ-isinβ+cosγ-isinγ
⇒ ab+bc+ca=0 .....(iv) [by (i) and (ii)]
Squaring both sides of equations (iii),
We get a2+b2+c2+2ab+2bc+2ca=0
or a2+b2+c2 [by (iv)]
[cosα+isinα]2+[cosβ+isinβ]2+[cosγ+isinγ]2=0
⇒ (cos2α+cos2β+cos2γ) + i(sin2α+sin2β+sin2γ)
Equating the real part to zero, we have
⇒cos2α+cos2β+cos2γ=0
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:
C
cosα+cosβ+cosγ=0 ......(i)
sinα+sinβ+sinγ=0 .......(ii)
Let a=cosα+isinα, b=cosβ+isinβ
c=cosγ+isinγ
⇒ a+b+c=0 .......(iii)
1a+1b+1c
=[cosα+isinα]−1+[cosβ+isinβ]−1+[cosγ+isinγ]−1
=cosα-isinα+cosβ-isinβ+cosγ-isinγ
⇒ ab+bc+ca=0 .....(iv) [by (i) and (ii)]
Squaring both sides of equations (iii),
We get a2+b2+c2+2ab+2bc+2ca=0
or a2+b2+c2 [by (iv)]
[cosα+isinα]2+[cosβ+isinβ]2+[cosγ+isinγ]2=0
⇒ (cos2α+cos2β+cos2γ) + i(sin2α+sin2β+sin2γ)
Equating the real part to zero, we have
⇒cos2α+cos2β+cos2γ=0
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