tan2a=[2tana]/[1－tan²a]=－4/3sin(π/4＋a)/cos(π/4＋a)=[sin(π/4)cosa＋cos(π/4)sina]/[cos(π/4)cosa－sin(π/4)sina]=[cosa＋sina]/[cosa－sina]=[1＋tana]/[1－tana]=－3则：原式=[－3]/[－4/3]=9/4...化简:sin2α/sin(π+α)+cos2α+1/cosα已知tanα=2,求sin(π/4+α)/cos(π/4+α)·tan2α的值求值:sin^2α+cos^2(π/6+α)+1/2sin(2α+π/6)1、=[2sinacosa/sina]＋(2cos²a－1＋1)/cosa=(2cosa)＋(2cosa)=4cosa 2、解决了。 3、=－(1/2)(1－cos2a)＋(1/2)[cos(π/3＋2a)]＋(1/2)sin(2a＋π/6) =－(1/2)＋(1/2)cos2a＋(1/2)[(1/2)cos2a－(√3/2)sin2a]＋(1/2)[(√3/2)sin2a＋(1/2)cos2a] =cos2a－(1/2)