三角函数问题4

来源：百度知道 编辑：UC知道 时间：2024/06/10 14:57:58
（1）化简f(x)=(1+cosx-sinx)/(1-sinx-cosx)+(1-cosx-sinx)/(1-sinx+cosx)
具体步骤
（2）是否存在x,使得tan(x/2) * f(x)与（1+tan平方(2/x)）/sinx相等？若存在，求x的值，若不存在，请说明理由

(1+cosx-sinx)/(1-cosx-sinx)+(1-cosx-sinx)/(1+cosx-sinx)
=[(1+cosx-sinx)(1+cosx+sinx)]/[(1-cosx-sinx)(1+cosx+sinx)]
+[(1-cosx-sinx)(1-cosx+sinx)]/[(1+cosx-sinx)(1-cosx+sinx)]

=(1+2cosx+cos2x)/(-sin2x)+(1-2cosx+cos2x)/(sin2x)

=(-4cosx)/sin2x

=-2/sinx

2)若tan(x/2)*(-2/sinx)=[1+tan^2(x/2)]/sinx＝>
tan^2(x/2)+2tan(x/2)+1=0 =>[tan(x/2)+1]^2=0 =>tan(x/2)=-1
=>x/2=kpai-pai/4 =>x=2kpai-pai/2

f(x)=(1+cosx-sinx)/(1-sinx-cosx)+(1-cosx-sinx)/(1-sinx+cosx)
=[2(cos(x/2))^2-2sin(x/2)cos(x/2)]/[2(sin(x/2))^2-2sin(x/2)cos(x/2)]+[2(sin(x/2))^2-2sin(x/2)cos(x/2)]/[2(cos(x/2))^2-2sin(x/2)cos(x/2)]
=2cos(x/2)[cos(x/2)-sin(x/2)]/2sin(x/2)[sin(x/2)-cos(x/2)]+2sin(x/2)[sin(x/2)-cos(x/2)]/2cos(x/2)[cos(x/2)-sin(x/2)]
=-cos(x/2)/sin(x/2)-sin(x/2)/cos(x/2)
=-[(cos(x/2))^2+(sin(x/2))^2]/sin(x/2)cos(x/2)=-2/sinx

解:f(x)=(1+cosx-sinx)/(1-sinx-cosx)+(1-cosx-sinx)/(1-sinx+cosx) =f(x)=(1+cosx-sinx)/