3sinθtanθ=8,(0<θ<π),则cos(θ-π/4)=
来源:学生作业帮助网 编辑:六六作业网 时间:2024/08/17 15:13:19
3sinθtanθ=8,(0<θ<π),则cos(θ-π/4)=3sinθtanθ=8,(0<θ<π),则cos(θ-π/4)=3sinθtanθ=8,(0<θ<π),则cos(θ-π/4)=∵0<θ
3sinθtanθ=8,(0<θ<π),则cos(θ-π/4)=
3sinθtanθ=8,(0<θ<π),则cos(θ-π/4)=
3sinθtanθ=8,(0<θ<π),则cos(θ-π/4)=
∵0<θ<π
∴sinθ>0
∵3sinθtanθ=8
==>3(sinθ)^2/cosθ=8
==>(sinθ)^2=8cosθ/3
==>(cosθ)^2+8cosθ/3=1 (∵(sinθ)^2+(cosθ)^2=1)
==>3(cosθ)^2+8cosθ-3=0
==>(3cosθ-1)(cosθ+3)=0
==>3cosθ-1=0 (∵cosθ+3>0)
∴cosθ=1/3,sinθ=√(8cosθ/3)=2√2/3
故cos(θ-π/4)=cosθcos(π/4)+sinθsin(π/4) (应用差角公式)
=(1/3)(√2/2)+(2√2/3)(√2/2)
=(√2+4)/6.
3sinθtanθ=8,(0<θ<π),则cos(θ-π/4)=
①利用公式sin(π-θ)=sinθ和sin(∏+θ)=-sinθ证明:sin(-θ)=-sinθ②证明tanθsinθ∕tanθ-sinθ=1+cosθ∕sinθ③已知sinα-2cosα+1=0,α≠kπ+π∕2,k∈z求:tan(3π-α)和1∕sin2α-sinαcosα+1的值
若sinθ=5分之3且sin2θ<0 ,则tanθ=
问几道高中三角函数题1、求证:tanα*sinα/(tanα-sinα)=(tanα+sinα)/tanα*sinα2、已知sinθ+cosθ=1/5,θ∈(0,π),求(1)sinθ-cosθ;(2)tanθ3、若α角的终边落在第三或第四象限,则α/2的终边落在第______
若0<θ<π/2,化简(sinθ/1-cosθ)*根号tanθ-sinθ/tanθ+sinθ (大根号,后面都包括)
求证 tanθ(1+sinθ )+sinθ /tanθ (1+sinθ )-sinθ =tanθ+sinθ/tanθsinθ
化简(√3cosθ+sinθ)/(√3sinθ-cosθ)不要跳步,内容:两角和差的正切原式=[(√3cosθ+sinθ)/cosθ]/[(√3sinθ-cosθ)/cosθ]=(√3+tanθ)/(√3tanθ-1)= -(tanπ/3+tanθ)/(1-tanπ/3tanθ)= -tan(π/3+θ)
若sinθ+cosθ=√2,则tan(θ+π/3)
已知tan(π/4+θ)+tan(π/4-θ)=4,且-π<θ<-π/2,求sin²θ-2sinθcosθ-cos²θ的值
已知方程2x²-(根号3+1)x+m=0的两个根分别为sinθ,cosθ,求[sin(π-θ)×tan(π+θ)/tanθ-1]+[cos(2π-θ)/1-tanθ]
已知sinθ+cosθ=1/2,且0<θ<π,求tanθ
tan(θ/2)=-1/3,则sin[θ+(π/4)]=?
求值:已知sinθ+cosθ=1/5,已知θ∈(0,π).求(1)sinθ*cosθ(2)sinθ-cosθ(3)tanθ
若tan^2θ-tanθ-2=0,(θ∈(0,π))求sin^2θ+3sinθcosθ-2cos^2θ的值
已知tanθ=(sin α-cos α)/(sin α+cos α) a,θ(0,π/2) 求证cos(3/2兀+ α) -sin(5π/2-α)=根号2sin(θ-4π)
已知sinθcosθ>0 sinθtanθ
证明tan^θ-sin^2θ=tan^2θsin^2θ证明(1)tan^θ-sin^2θ=tan^2θsin^2θ(2)sin^4x+cos^4x=1-2sin^2cos^2x(31-tan^2x/1+tan^2x=cos^2x-sin^2x
sinθ/(1+sinθ)-sinθ/(1-sinθ) 若tanθ=根号2 则原式=