已知tan(π/4+θ)+tan(π/4-θ)=4,且-π<θ<-π/2,求sin²θ-2sinθcosθ-cos²θ的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/09 03:20:22
已知tan(π/4+θ)+tan(π/4-θ)=4,且-π<θ<-π/2,求sin²θ-2sinθcosθ-cos²θ的值
已知tan(π/4+θ)+tan(π/4-θ)=4,且-π<θ<-π/2,求sin²θ-2sinθcosθ-cos²θ的值
已知tan(π/4+θ)+tan(π/4-θ)=4,且-π<θ<-π/2,求sin²θ-2sinθcosθ-cos²θ的值
tan( π/4 + θ ) + tan( π/4 - θ ) = 4
[ sin( π/4 + θ )cos( π/4 - θ ) + cos( π/4 + θ )sin( π/4 - θ )] / [cos( π/4 + θ )cos( π/4 - θ )] = 4
sin( π/2 ) / [cos( π/4 + θ )cos( π/4 - θ )] = 4
cos( π/4 + θ )cos( π/4 - θ ) = 1 / 4
( cos π/2 + cos 2θ ) / 2= 1 / 4
cos 2θ = 1 / 2
sin 2θ = √( 1 - cos² 2θ ) = √3 / 2
sin² θ - 2sin θ cos θ - cos² θ = - ( sin 2θ + cos 2θ ) = - (√3 + 1) / 2
已知tanα/2=2,求tanα与tan(α+π/4)
已知(1-tanθ)/2+tanθ=1,求证tan2θ=-4tan(θ+π/4)
已知(1-tanθ)/(2+tanθ)=1,求证tan2θ=-4tan(θ+π/4)
求证(1)1+tanθ/1-tanθ=tan(π/4+θ) (2)1-tanθ/1+tanθ=tan(π/4-θ)
已知tanθ=2,则tan(θ+π/4)= ,cos2θ=已知tanθ=2,则tan(θ+π/4)=cos2θ=
已知tan(α-π/4)=1/3,tan(β+π/4),那么tan(α+β)=
tan( x/2+π/4)+tan(x/2-π/4 )=2tanxtan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x
已知(1-tanθ)/(2+tanθ)=1,求证tan2θ=-4tan(θ+π/4) 想得到就不会来问了!乱死了!
已知tan(π/4-θ)+tan(π/4+θ)=4,且-π
已知tan(θ+π/4)=1/2,则tanθ=
已知tan(α+π/4)=2,则tanα/tan2α=
已知(1+tanα)(1+tanβ)=4cos(π/3)0
已知Tan(π/4+尔发)=1/2.求Tan尔发
已知tanα=3,求tan(α-π/4)
已知tanα=-1/2,求tan(α+π/4)
已知tanθ=2.(1)求tan(π/4+θ)的值.(2)求cos2θ的值
求证:(1)1+tanθ/1-tanθ=tan(π/4+θ)(2)1-tanθ/1+tanθ=tan(π/4-θ)
化简:[tan(π/4+a)-tan(π/4-a)]/tan2a-tan(π/4+a)tan(π/4-a)