【2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3】/【2+2sin²(π/2
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/10/02 12:27:27
【2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3】/【2+2sin²(π/2+θ)-sin(3π/2-θ)】
求f(π/3)的值.
求f(π/3)的值.
f(θ)=【2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3】/【2+2sin²(π/2+θ)-sin(3π/2-θ)】
=(2cos²θ+sin²θ+cosθ-3)/(2+2cos²θ+cosθ)
=(cos²θ+cosθ-2)/(2cos²θ+cosθ+2)
f(π/3)=[cos²(π/3)+cos(π/3)-2]/[2cos²(π/3)+cos(π/3)+2]
=(1/4+1/2-2)/(2*1/4+1/2+2)
=(-5/4)/3
=-5/12
=(2cos²θ+sin²θ+cosθ-3)/(2+2cos²θ+cosθ)
=(cos²θ+cosθ-2)/(2cos²θ+cosθ+2)
f(π/3)=[cos²(π/3)+cos(π/3)-2]/[2cos²(π/3)+cos(π/3)+2]
=(1/4+1/2-2)/(2*1/4+1/2+2)
=(-5/4)/3
=-5/12
θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小
若(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)等于?
(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)=
α属于(0,π/2)且2sin²α-sinαcosα-3cos²α=0求sin(α+π/4)/sin
已知tanθ=根号2,求(1)(cosθ+sinθ)/(cosθ-sinθ);(2)sin²θ-sinθcos
2sinα=sinθ+cosθ,sin²β==sinθcosθ.求证cos2β=2cos2α=2cos
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
1.已知2sin(3π+θ)=cos(π+θ),求2sin
sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=_____
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)
已知α、β≠kπ+π2(k∈Z),且sinθ+cosθ=2sinα , sinθcosθ=sin
设f(α)=2sinαcosα+cosα/1+sin²α+cos(3π/2+α)-sin²(π/2+