题抄错了,要么是tan(π/4-a)=(1-sin2a)/cos2a 要么是tan(π/4-a/2)=(1-sina)/cosa
tan(π/4-a)=[tan(π/4)-tan(a)]/[1+tan(π/4)tan(a)]
=[1-sina/cosa][1+sina/cosa]=[cosa-sina]/[cosa+sina]=[cosa-sina]²/[cos²a-sin²a]
=(1-2sinacosa)/cos2a=(1-sin2a)/cos2a
也可如此证明（1-sina)/cosa==[1-cos(π/2-a)]/sin(π/2-a)
=2sin²(π/4-a/2)/[2sin(π/4-a/2)cos(π/4-a/2)]=tan(π/4-a/2)