(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)=
来源:学生作业帮 编辑:大师作文网作业帮 分类:综合作业 时间:2024/09/22 12:29:21
(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 =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,则sin(θ-5π)*sin(3π/2-θ)=
若(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)等于?
sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=_____
已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=6/11,求5cos^2θ/(sin^2θ+2sinθcos
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)
求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ
若sin θ-cos θ 分之sin θ+cos θ=2 则sin θcos θ 是
2sinθ+cosθ/sinθ-3cosθ=-5,求cos2θ+4sinθ
sinθ-cosθ=1/2,则sin^3θ-cos^3θ=?.
为什么sin2θ+sinθ=2sinθcosθ+sinθ=sinθ(2cosθ+1)
已知tanθ=3,求(3cosθ-5sin^2θcosθ)/sin(π-θ)的值
已知θ为第三象限角,1-sinθcosθ-3cos^2=0则5sin^2θ+3sinθcosθ=?