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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
∵ 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
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