1.证明：(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/2
(sinx+cosx-1)(sinx-cosx+1)/sin2x
=[sinx+(cosx-1)][sinx-(cosx-1)]/sin2x
=[(sinx)^2-(cosx-1)^2]/sin2x
=[(sinx)^2-(cosx)^2-1+2cosx]/sin2x
=[2cosx-2(cosx)^2]/sin2x
=2cosx(1-cosx)/2sinxcosx
=(1-cox)/sinx
=tanx/2 得证
2.化简:
（1+sin4x-cos4x)/(1+sin4x+cos4x) - (1+sin4x+cos4x)/(1+sin4x-cos4x) 通分
=[（1+sin4x-cos4x)^2+(1+sin4x+cos4x)^2]/[(1+sin4x+cos4x)*(1+sin4x-cos4x)]
=[1+(sin4x)^2+(cos4x)^2+2sin4x-2cos4x-2sin4xcos4x+1+(sin4x)^2+(cos4x)^2+2sin4x+2cos4x+2sin4xcos4x]/[(1+sin4x)^2-(cos4x)^2]
=[2+2(sin4x)^2+2(cos4x)^2+4sin4x]/[1+2sin4x+(sin4x)^2-(cos4x)^2]
=[2+2+4sin4x]/[2sin4x+2(sin4x)^2]
=4[1+sin4x]/[2sin4x(1+sin4x)]
=2/sin4x
3.已知sin(π/4 -x)=5/13,0