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polinagarres
@polinagarres
August 2022
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помогите с тригонометрией
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kirichekov
Verified answer
2cos²α/(ctgα/2-tgα/2)=
(1/2)*sin2α
1. ctgα/2-tgα/2=(cosα/2)/(sinα/2)-(sinα/2)/(cosα/2)=(cos²(α/2)-sin²(α/2))/(cosα/2*sinα/2)=cos(2*(α/2))/((1/2)*2sinα/2*cosα/2)=2cosα/sinα
2. 2cos²α:(2cosα/sinα)=cosα*sinα=(1/2)*2sinα*cosα=(1/2)sin2α
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Answers & Comments
Verified answer
2cos²α/(ctgα/2-tgα/2)=(1/2)*sin2α1. ctgα/2-tgα/2=(cosα/2)/(sinα/2)-(sinα/2)/(cosα/2)=(cos²(α/2)-sin²(α/2))/(cosα/2*sinα/2)=cos(2*(α/2))/((1/2)*2sinα/2*cosα/2)=2cosα/sinα
2. 2cos²α:(2cosα/sinα)=cosα*sinα=(1/2)*2sinα*cosα=(1/2)sin2α