∵C-A=π/2,A+B+C=π,
∴2A=π/2-B
则cos(2A)=cos(π/2-B)=sinB=1/3
即1-2(sinA)^2=1/3
(sinA)^2=1/3
∵C=A+π/2>π/2,0<A<π/2,∴sinA>0
∴sinA=√3/3
∵C-A=π/2,A+B+C=π,
∴2A=π/2-B
则cos(2A)=cos(π/2-B)=sinB=1/3
即1-2(sinA)^2=1/3
(sinA)^2=1/3
∵C=A+π/2>π/2,0<A<π/2,∴sinA>0
∴sinA=√3/3