已知sin(π/4-α)=-2/3,π/4<α<π/2,求sinα

由题意： -π/4 <π/4-α <0
因为 sin(π/4-α)=-2/3
所以cos(π/4-α)=√ 5/3
sinα=sin(α-π/4+π/4)=sin(α-π/4)cosπ/4+cos(α-π/4)sinπ/4
=-sin(π/4-α)cosπ/4+cos(π/4-α)sinπ/4
=(√10+2√2)/6

π/4<α<π/2
-π/4<π/4-α<0
cos(π/4-α)=√5/3
sinα=-sin(π/4-α-π/4)=sinπ/4cos(π/4-α)-sin(π/4-α)cosπ/4=(√10+2√3)/6

解：因为 sin(π/4-α)=-2/3,π/4<α<π/2所以 -π/4 <π/4-α <0
又 sin(π/4-α)=-2/3 所以cos(π/4-α)=√ 5/3
而sinα=sin(α-π/4+π/4)=sin(α-π/4)cosπ/4+cos(α-π/4)sinπ/4
=-sin(π/4-α)cosπ/4+cos(π/4-α)sinπ/4
=(√10+2√3)/6