高一数学已知sin(π/2-b)*cos(a+b)-sin(π+b)*sin(a+b)=3/5其中a∈(3π/2,2π)求tan(π/4-a/2)

解
sin(π/2-b)*cos(a+b)-sin(π+b)*sin(a+b)=3/5
即
cosbcos(a+b)+sinbsin(a+b)
=cos[b-(a+b)]
=cos(-a)
=cosa
∴cosa=3/5
∵a∈(3π/2,2π)
∴sina