已知sin(π/3+α)=12/13,α∈(π/6,2π/3),求cosα的值

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已知sin(π/3+α)=12/13,α∈(π/6,2π/3),求cosα的值
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已知sin(π/3+α)=12/13,α∈(π/6,2π/3),求cosα的值
已知sin(π/3+α)=12/13,α∈(π/6,2π/3),求cosα的值

已知sin(π/3+α)=12/13,α∈(π/6,2π/3),求cosα的值
sin(π/3+α)=12/13,α∈(π/6,2π/3),
π/3+α∈(π/2,π)
所以
cos(π/3+α)=-5/13
cosα=cos(π/3+α-π/3)
=cos(π/3+α)cosπ/3+sin(π/3+α)sinπ/3
=-5/13×1/2+12/13×√3/2
=(-5+12√3)/26