(1)已知cosθ=﹣3/5,θ∈(π/2,π),求(θ+π/3)(2)已知tanα=2,求tan(α-π/4)

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/19 08:58:38
(1)已知cosθ=﹣3/5,θ∈(π/2,π),求(θ+π/3)(2)已知tanα=2,求tan(α-π/4)(1)已知cosθ=﹣3/5,θ∈(π/2,π),求(θ+π/3)(2)已知tanα=2

(1)已知cosθ=﹣3/5,θ∈(π/2,π),求(θ+π/3)(2)已知tanα=2,求tan(α-π/4)
(1)已知cosθ=﹣3/5,θ∈(π/2,π),求(θ+π/3)(2)已知tanα=2,求tan(α-π/4)

(1)已知cosθ=﹣3/5,θ∈(π/2,π),求(θ+π/3)(2)已知tanα=2,求tan(α-π/4)
cos(απ/4)=3/5 因为sin^x cos^x=1 00 sin(απ/4)=4/5 sinα=sin(απ/4-π/4)=sin(απ/4)cos(π/4)-cos(απ/4)csin(π/4)=(4/5)*(根号2/2)-(3/5)*(根号2/2)=根号2/10 cosα=7*根号2/10 tan α=sinα/cosα=1/7 很高兴为你解答

求值:已知sinθ+cosθ=1/5,已知θ∈(0,π).求(1)sinθ*cosθ(2)sinθ-cosθ(3)tanθ 已知cosθ=1/3,θ∈(0,π),则cos(π+2θ)等于 已知sinθ+cosθ=1/5,θ∈(o,π),求cos(θ-π/3)+cotθ 已知sin(3π+θ)=lg1/(10开3次方0{cos(π+θ)/cosθ[cos(π-θ)-1]}+{ cos(θ-π)/cos(π-θ)+cos(θ已知sin(3π+θ)=lg1/(10开3次方)求{cos(π+θ)/cosθ[cos(π-θ)-1]}+{ cos(θ-2π)/cos(π-θ)+cos(θ-2π)}(最后结果 已知sin(3π+θ)=lg1/(10开3次方),求值cos(π+θ)/cosθ{cos(π-θ)-1} + cos(θ-π)/cos(π-θ)+cos(更正:已知sin(3π+θ)=lg1/(10开3次方),求值cos(π+θ)/cosθ{cos(π-θ)-1} + cos(θ-π)/cos(π-θ)+cos(θ-2π) 高一数学 已知sinθsinβ=-4/7,则cosθcosβ∈已知tan(π-θ)=3,求下式的值 (5xin^3θ+cosθ)/(2cos^3θ+sin^2θcosθ) 已知sinθ+cosθ=1/5,θ∈(0,π),求下列各式的值1)tanθ(2)sinθ-cosθ(3)sin^3θ+cos^3θ 已知sin(θ+kπ)=2cos[θ+(k+1)π],k∈Ζ,求4sinθ-2cosθ/5cosθ+3sinθ的值 已知sinθ+cosθ=1/5,θ∈(0,π)求值sin³θ+cos³θ 已知cos(π/2+θ)=1/2,求cos(3π+θ)/cosθ(cos(π+θ)-1) + cos(θ-4π)/cos(θ+2π)cos(3π+θ)+cos(-θ)求值~ 已知 cosθ=-(3/5).且θ∈(π,3π/2)求cosθ/2的值. 已知cos θ=1/3,θ∈(0,π),则cos(3π/2+2θ) 已知cos(π/6+θ)=根号3/3,求cos(5π/6-θ)的值 已知tanΘ=2,计算①sinΘ+2cosΘ/5cosΘ-sinΘ②1/sinΘcosΘ-cos∧Θ 已知cos(-θ)=-1/2,则cos(-3π+θ)= 已知sinθ-cosθ=1/5,且θ∈(0,π)(1)求sinθ+cosθ;(2)求tanθ 已知cos(π/4 - α)=- 3/5,α∈(π/2,π),求cosα 已知cosθ= - 3/5.π