(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)=

李俊yaya 1年前 已收到2个回答 举报

Firment 幼苗

共回答了21个问题采纳率:81% 举报

(sinθ+cosθ)/(sinθ-cosθ)=2
sinθ+cosθ =2sinθ-2cosθ
3cosθ =sinθ
sinθ与cosθ同号
两边平方:
9cos^2θ =sin^2θ
9cos^2θ =1-cos^2θ
cos^2θ =1/10
|cosθ|=1/根号10
|sinθ|=3|cosθ|=3/根号10
sin(θ-5π)*sin(3π/2-θ)
= sin(π+θ-6π)*sin(π+π/2-θ)
= sin(π+θ)*{-sin(π/2-θ)}
={- sinθ)}*{-sin(π/2-θ)}
= sinθ*sin(π/2-θ)
=sinθ*cosθ
=|sinθ|*|cosθ|
=3/根号10 * 1/根号10
=3/10

1年前

8

lovediving 幼苗

共回答了430个问题 举报

(sinθ+cosθ)/(sinθ-cosθ)=2
sinθ+cosθ =2sinθ-2cosθ
3cosθ =sinθ
tanθ=3 cotθ=1/3
sin(θ-5π)*sin(3π/2-θ)
= sin(π+θ)*sin(3π/2-θ)
= (-sinθ)*(-cosθ)=sin2θ/2=1/(tanθ+cotθ)=1/(3+1/3)=3/10

1年前

1
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