If cosα+cosβ=0=sinα+sinβ, then which of the following statement(s) is/are true?
A
cos(α−β)=−1
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B
cos2α+cos2β+2cos(α+β)=0
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C
sin2α+sin2β+2sin(α+β)=0
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D
sin2(α+π2)+sin2(β+π2)=1−cos(α+β)
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Solution
The correct options are Acos(α−β)=−1 Bcos2α+cos2β+2cos(α+β)=0 Csin2α+sin2β+2sin(α+β)=0 Dsin2(α+π2)+sin2(β+π2)=1−cos(α+β)
cosα+cosβ=0……(1) sinα+sinβ=0……(2) (1)2+(2)2⇒2+2cos(α−β)=0⇒cos(α−β)=−1 (1)2−(2)2⇒cos2α+cos2β+2cos(α+β)=0 ((1)−(2))2⇒(cosα+cosβ+sinα+sinβ)2=0 ⇒2+2cos(α−β)+2cosαsinα+2cosβsinβ+2cosβsinα=0 Using cos(α−β)=−1 we get 2−2+sin2α+sin2β+2sin(α+β)=0 sin2α+sin2β+2sin(α+β)=0 Lastly, sin2(α+π2)+sin2(β+π2)=cos2α+cos2β =2cos2α=2cos2β[using|cosα|=|cosβ|] =1+cos2α=1+cos2β =1+cos2α+cos2β2 =1−cos(α+β) [using result in (B)]