解：∵α∈(3π/2，2π)，sinα=-3/5∴cosα﹥0cosα=√(1-sin²α)=√[1-(-3/5)²]=√(16/25)=4/5 cos(π/4-α)=cosπ/4cosα+sinπ/4sinα=(√2)/2×4/5+(√2)/2×(-3/5)=(4√2-3√2)/10=(√2)/10
