化简（1+sinx）/cosx*sin2x/2cos（π/4-x/2）

2. 因为,2{cos[(π/4)-(x/2)]}^2 =cos2[(π/4)-(x/2)] 1 =sinx 1 xtan[(π/4)-(x/2)] =xsin[(π/4)-(x/2)]/cos[(π/4)-(x/2)] =2xsin[(π/4)-(x/2)]*cos[(π/4)-(x/2)]/2{cos[(π/4)-(x/2)]}^2 =xsin2[(π/4)-(x/2)]/(sinx 1) =xcosx/(sinx 1) 所以, 原式=(sinx 1){3cosx/2{cos[(π/4)-1)]}^2-2tan[(π/4)-1]} =(sinx 1)[3cosx/(sinx 1)-2cosx/(sinx 1)] =3cosx-2cosx =cosx