抗战电视剧太行山突围:三角函数问题 高手快进来!!!

来源:百度文库 编辑:杭州交通信息网 时间:2024/05/06 05:21:38
4(sinX)^2+6sinX-(cosX)^2-3cosX=0,求(cos2X-sin2X)/[(1-cos2X)(1-tan2X)]

4(sinX)^2+6sinX-(cosX)^2-3cosX
=(2sinx+cosx)(2sinx-cosx)+3*(2sinx-cosx)
=(2sinx-cosx)(2sinx+cosx+3)
=0
2sinx-cosx=0
2sinx=cosx
4sinx^2=cosx^2
5sinx^2=cosx^2+sinx^2=1
sinx^2=1/5
cos2x=1-2sinx^2=1-2*1/5=3/5

(cos2X-sin2X)/[(1-cos2X)(1-tan2X)]
=(cos2X-sin2X)/[(1-cos2X)(cos2x-sin2x)/cos2x]
=cos2x/(1-cos2x)
=3/5/(1-3/5)
=3/2