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

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 19:40:50
(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)=

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

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

(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