作业帮已知tan(π÷4+θ)=3,则sin2θ-2cos²θ=
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/30 23:43:36
已知tan(π÷4+θ)=3,则sin2θ-2cos²θ=
已知tan(π÷4+θ)=3,则sin2θ-2cos²θ=
已知tan(π÷4+θ)=3,则sin2θ-2cos²θ=
tan(π/4+θ)=(tanπ/4+tanθ)/(1-tanπ/4tanθ)=(1+tanθ)/(1-tanθ)=3
1+tanθ=3(1-tanθ)
tanθ=1/2
sin2θ-2cos²θ
=sin2θ-(2cos²θ-1)-1
=sin2θ-cos2θ-1
=2tanθ/(1+tan²θ)-(1-tan²θ)/(1+tan²θ)-1
=(2tanθ-1+tan²θ)/(1+tan²θ)-1
=(1-1+1/4)/(1+1/4)-1
=-4/5
tan(π/4+θ)=3
(tanπ/4+tanθ)/(1-tanπ/4tanθ)=3
(1+tanθ)/(1-tanθ)=3
1+tanθ=3-3tanθ
4tanθ=2
tanθ=1/2
sin2θ-2cos²θ
=sin2θ-(1+cos2θ)
=sin2θ-cos2θ-1
=2tanθ/(1+tan²...
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tan(π/4+θ)=3
(tanπ/4+tanθ)/(1-tanπ/4tanθ)=3
(1+tanθ)/(1-tanθ)=3
1+tanθ=3-3tanθ
4tanθ=2
tanθ=1/2
sin2θ-2cos²θ
=sin2θ-(1+cos2θ)
=sin2θ-cos2θ-1
=2tanθ/(1+tan²θ)-(1-tan²θ)/(1+tan²θ)-1
=(2tanθ-1+tan²θ)/(1+tan²θ)-1
=(2*1/2-1+1/2²)/(1+1/2²)-1
=(1-1+1/4)/(1+1/4)-1
=(1/4)/(5/4)-1
=1/5-1
=-4/5
收起
tan(π/4+θ)= 3
sin2θ-2cos²θ= sin2θ-2×（1- cos2θ）/2
=sin2θ-cos2θ-1=- cos(2θ+π/2)- sin(2θ+π/2)-1
= -cos[2(θ+π/4)]- sin[2(θ+π/4)]-1
Cos[2(θ+π/4)]=[1- tan2(π/4+θ)] / [1+ tan2(π/4+θ)]=(1...
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tan(π/4+θ)= 3
sin2θ-2cos²θ= sin2θ-2×（1- cos2θ）/2
=sin2θ-cos2θ-1=- cos(2θ+π/2)- sin(2θ+π/2)-1
= -cos[2(θ+π/4)]- sin[2(θ+π/4)]-1
Cos[2(θ+π/4)]=[1- tan2(π/4+θ)] / [1+ tan2(π/4+θ)]=(1-9)/(1+9)=-8/10
sin[2(θ+π/4)]= 2tan(π/4+θ)] / [1+ tan2(π/4+θ)]=6/10
所以
-cos[2(θ+π/4)]- sin[2(θ+π/4)]-1=8/10-6/10-1= -4/5
即sin2θ-2cos²θ== -4/5
收起
嗨。