解sina=3/5∵a∈(0,π/2)∴cosa>0∵sin²a+cos²a=1∴cosa=4/5∴tana=sina/cosa=3/4∵cosa=2coa²a/2-1∴cos²a/2=(cosa+1)/2∵a∈(0,π/2)∴a/2∈(0,π/4)∴cosa/2>0∴cosa/2=√(cosa+1)/2=√(4/5+...