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Question Number 61799 by alphaprime last updated on 08/Jun/19
If a = Cosα −iSinα and b = Cosβ −iSinβ Prove that (((a+b)(1−ab))/((a−b)(1+ab))) = ((Sinα+Sinβ)/(Sinα−Sinβ))
Ifa=Cosα−iSinαandb=Cosβ−iSinβProvethat(a+b)(1−ab)(a−b)(1+ab)=Sinα+SinβSinα−Sinβ
Answered by tanmay last updated on 09/Jun/19
a=e^(−iα) b=e^(−iβ) (a/b)=e^(−i(∝−β)) ab=e^(−i(α+β)) (a/b)=(e^(−i(α−β)) /1) ((a+b)/(a−b))=((e^(−i(α−β)) +1)/(e^(−i(α−β)) −1)) (1/(ab))=(1/e^(−i(α+β)) )=(e^(i(α+β)) /1) ((1−ab)/(1+ab))=((e^(i(α+β)) −1)/(1+e^(i(α+β)) )) now ((a+b)/(a−b))×((1−ab)/(1+ab)) =((e^(−i(α−β)) +1)/(e^(−i(α−β)) −1))×((e^(i(α+β) )−1)/(e^(i(α+β)) +1)) =((e^(i(α+β−α+β)) +e^(i(α+β)) −e^(−i(α−β)) −1)/(e^(i(α+β−α+β)) −e^(i(α+β)) +e^(−i(α−β)) −1)) =((cos2β+isin2β+cos(α+β)+isin(α+β)−cos(α−β)+isin(α−β)−1)/(cos2β+isi2β−cos(α+β)−isin(α+β)+cos(α−β)−isin(α−β)−1)) =((cos2β−1−2sinαsinβ+i(sin2β+sin(α+β)+sin(α−β)))/(cos2β−1+2siαsinβ+i(sin2β−sin(α+β)−sin(α−β)))) =((−2sin^2 β−2sinαsinβ+i(sin2β+2sinαcosβ))/(−2sin^2 β+2sinαsinβ+i(sin2β−2sinαcosβ))) =((−2sin^2 β−2sinαsinβ+i(2sinβcosβ+2sinαcosβ))/(−2sin^2 β+2sinαsinβ+i(2sinβcosβ−2sinαcosβ))) =((−2sinβ(sinβ+sinα)+i2cosβ(sinα+sinβ))/(−2sinβ(sinβ−sinα)+i2cosβ(sinβ−sinα))) =(((sinα+sinβ)(−2sinβ+i2cosβ))/((sinβ−sinα)(−2sinβ+i2cosβ))) =((sinα+sinβ)/(sinβ−sinα)) pls check mistake if any
a=e−iαb=e−iβab=e−i(∝−β)ab=e−i(α+β)ab=e−i(α−β)1a+ba−b=e−i(α−β)+1e−i(α−β)−11ab=1e−i(α+β)=ei(α+β)11−ab1+ab=ei(α+β)−11+ei(α+β)nowa+ba−b×1−ab1+ab=e−i(α−β)+1e−i(α−β)−1×ei(α+β)−1ei(α+β)+1=ei(α+β−α+β)+ei(α+β)−e−i(α−β)−1ei(α+β−α+β)−ei(α+β)+e−i(α−β)−1=cos2β+isin2β+cos(α+β)+isin(α+β)−cos(α−β)+isin(α−β)−1cos2β+isi2β−cos(α+β)−isin(α+β)+cos(α−β)−isin(α−β)−1=cos2β−1−2sinαsinβ+i(sin2β+sin(α+β)+sin(α−β))cos2β−1+2siαsinβ+i(sin2β−sin(α+β)−sin(α−β))=−2sin2β−2sinαsinβ+i(sin2β+2sinαcosβ)−2sin2β+2sinαsinβ+i(sin2β−2sinαcosβ)=−2sin2β−2sinαsinβ+i(2sinβcosβ+2sinαcosβ)−2sin2β+2sinαsinβ+i(2sinβcosβ−2sinαcosβ)=−2sinβ(sinβ+sinα)+i2cosβ(sinα+sinβ)−2sinβ(sinβ−sinα)+i2cosβ(sinβ−sinα)=(sinα+sinβ)(−2sinβ+i2cosβ)(sinβ−sinα)(−2sinβ+i2cosβ)=sinα+sinβsinβ−sinαplscheckmistakeifany

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