已知sin(π/6-α)=-1/3,α∈(π/6,π/2)则sinα=
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已知sin(π/6-α)=-1/3,α∈(π/6,π/2)则sinα=
已知sin(π/6-α)=-1/3,α∈(π/6,π/2)则sinα=
已知sin(π/6-α)=-1/3,α∈(π/6,π/2)则sinα=
解
∵sin(π/6-a)=-sin(a-π/6)=-1/3
∴sin(a-π/6)=1/3
∵a∈(π/6,π/2)
∴a-π/6∈(0,π/3)
∴cos(a-π/6)>0
∴cos(a-π/6)=√1-1/9=2√2/3
∴sina
=sin[(a-π/6)+π/6]
=sin(a-π/6)cosπ/6+cos(a-π/6)sinπ/6
=1/3×√3/2+2√2/3×1/2
=√3/6+√2/3
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