已知cosθ=5/13,θ∈(π,2π),求sin(θ-π/6),cos(θ-π/6)及tan(θ-π/6)的值.

已知cosθ=5/13,θ∈(π,2π)
所以θ一定是第4象限角
所以sinθ=-12/13 tanθ=-12/5
然后sin(θ-π/6),cos(θ-π/6)及tan(θ-π/6) 这些都用公式把他们拆开来
到最后再把cosθ=5/13,sinθ=-12/13代入,就可求得答案

θ∈(π,2π)
cosθ=5/13,sinθ=-12/13
sin(θ-π/6)
=sinnθ*cosπ/6)-cosθ*sinπ/6
=(-12/13)*√3/2-(5/13)*1/2
=-(5+12√3)/26

cos(θ-π/6)
=cosθ*cos(π/6)+sinθ*sin(π/6)
=(5/13)*√3/2+(-12/13)*1/2
=(5√3-12)/26

tan(θ-π/6)
=sin(θ-π/6)/cos(θ-π/6)
=[-(5+12√3)/26]/[(5√3-12)/26]
=(240+169√3)/69