数学题三角函数已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/10/07 07:27:32
数学题三角函数
已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
2sin(3π+θ)=cos(π+θ)
2sinθ=cosθ
tanθ=1/2
2sin^2 θ+3sinθcosθ-cos^2 θ 2sin^2 θ+3sinθcosθ-cos^2 θ
___________________________= ____________________________=
1 sin^2 θ +cos^2 θ
分子分母同除以cos^2 θ
2tan^2θ+3tanθ-1
________________ 将tanθ=1/2带入
tan^2θ+1
=4/5
2sinθ=cosθ
tanθ=1/2
2sin^2 θ+3sinθcosθ-cos^2 θ 2sin^2 θ+3sinθcosθ-cos^2 θ
___________________________= ____________________________=
1 sin^2 θ +cos^2 θ
分子分母同除以cos^2 θ
2tan^2θ+3tanθ-1
________________ 将tanθ=1/2带入
tan^2θ+1
=4/5
已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=6/11,求5cos^2θ/(sin^2θ+2sinθcos
已知2sinθ+3cosθ=2,求sinθ+cosθ的值
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)
已知tanθ=3,求(3cosθ-5sin^2θcosθ)/sin(π-θ)的值
1.已知2sin(3π+θ)=cos(π+θ),求2sin
已知sinθ+2cosθsinθ−cosθ=3,求值:
已知向量a=(sinθ,cosθ-2sinθ),向量b=(1,2) 求tanθ 求sinθ*cosθ-3cos^2θ
①化简[sin(2π+α)*cos(7π-α)]/cos(-α)②已知cosθ=-3/5,θ∈(π/2,π),求sin(
怎样用matlab解三角函数cos(π/2*cosθ)/sin(θ)=0.707?
θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小
2sinθ+cosθ/sinθ-3cosθ=-5,求cos2θ+4sinθ
三角函数化简[2cos^3θ+sin^2(π-θ)+sin(π/2+θ)-3]/[2+2cos^2(π+θ)+cos(-