Question

数学 三角函数，这道题怎么做

Answer 1

解析：

∵ π/2<α<π
∴ cosα<0

∵ π/2<α<π
∴ π/4<α/2<π/2
∴ sin(α/2)>0，cos(α/2)>0
∴
cosα
=-√(1-sin²α)
=-√[1-(5/13)²]
=-12/13

cosα=2cos²(α/2)-1
⇒
cos(α/2)
=√[(1+cosα)/2]
=√(1/26)
=(√26)/26

cosα=1-2sin²(α/2)
⇒
sin(α/2)
=√[(1-cosα)/2]
=(5√26)/26

sin(α-π/6)
=sinαcos(π/6)-cosαsin(π/6)
=(5/13)*(√3/2)-(-12/13)*(1/2)
=(5√3+12)/26