设a属于(π,2π),若tan(a+π/6)=2,则cos(π/6-a)=
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/04/30 19:25:51
设a属于(π,2π),若tan(a+π/6)=2,则cos(π/6-a)=
问题是不是错了
a∈(π,2π),tan(a+π/6)=2,则cos(π/6-2a)=?
tan(a+π/6)=2
∴
sin(2a+π/3)
=sin2(a+π/6)
=2sin(a+π/6)cos(a+π/6)
=2sin(a+π/6)cos(a+π/6)/[sin²(a+π/6)+cos²(a+π/6)]
=2tan(a+π/6)/[1+tan²(a+π/6)]
=2*2/(1+2²)
=4/5
∴ cos(π/6-2a)
= cos[π/2-(2a+π/3)]
=sin(2a+π/3)
=4/5
a∈(π,2π),tan(a+π/6)=2,则cos(π/6-2a)=?
tan(a+π/6)=2
∴
sin(2a+π/3)
=sin2(a+π/6)
=2sin(a+π/6)cos(a+π/6)
=2sin(a+π/6)cos(a+π/6)/[sin²(a+π/6)+cos²(a+π/6)]
=2tan(a+π/6)/[1+tan²(a+π/6)]
=2*2/(1+2²)
=4/5
∴ cos(π/6-2a)
= cos[π/2-(2a+π/3)]
=sin(2a+π/3)
=4/5
已知f(a)=sin(π+a)tan(π+a)cos(-a)/cos(3π-a)tan(2π-a),若a∈(0,π),c
设0<θ<π/2,向量a=(sin2θ,cos),b=(cosθ,1),若a平行b,则tanθ=
若a=[cot(4π+a)cos(a+π)sin^2(3π+a)]/[tan(π+a)cos^3(-a-π)]则a^2+
求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) =
已知a属于(2分之π,2分之3π),tan(a-7π)=-4分之3,则sin a+cos a 的值为多少
已知f(a)=sin(π-a)cos(2π-a)tan(-a+3π/2)/tan(π/2+a)sin(-π-a)若a是第
tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3π+a)cos^2(2/5π+2)=0.5,求tan
已知f(a)={sin(π-a)cos(2π-a)tan[-a+(3π/2)]}/cos(-π-a)
求证:tan(2π-a) sin(-2π-a) cos(6π-a) / cos(a-π) sin(5π-a)=-tana
求证:tan(2π-a)sin(-2π-a)cos(6π-a)/cos(a-π)sin(5π-a)=-tana
化简f(a)=sin(π\2-a)*cos(2π-a)*tan(-a+3π)
cos(π-a)tan(a-2π)tan(2π-a) /sinπ+a = -tana求过程