求“已知sinα=15/17,α∈(π/2,π),求cos(π/3-α)的值”的答案和“已知cosθ=-5/13,α∈(π,3/2π),求cos(θ-π/6)的值”的答案,谢谢额……可以加上过程么谢谢

sinα=15/17,α∈(π/2,π),求cos(π/3-α)的值
cosα=-8/17
cos(π/3-α)=cosπ/3cosα+sinπ/3sinα
=-1/2*(-8/17)+√3/2*15/17
=（8+15√3）/34
cosθ=-5/13,θ∈(π,3/2π),求cos(θ-π/6)的值
sinθ=-12/13
cos(θ-π/6)=cosθcosπ/6+sinθsinπ/6
=-5/13*√3/2-12/13*1/2
=-（5√3+12）/26

1、(15倍根号3-8)/34
2、-(5倍根号3+12)/26

因为sinα=15/17,α∈(π/2,π),所以cosa所以cosa=-8/17
cos(π/3-α)=cosπ/3cosa+sinπ/3sina=-4/17+15根号3/34=(15根号3-8)/34
因为cosθ=-5/13,α∈(π,3/2π),
sina=-12/13
cos(θ-π/6)=cosacosπ/6+sinasinπ/6=-5根号3/26-12/26=-(5根号3+12）/26