已知0<x<π,cosx-sinx=-(√10)/5,求:(sin2x-cos2x+1)/(1+tanx)的值
熱心網(wǎng)友
已知0 原式=(sin2x-cos2x+1)/(1+tanx)=[sin2x+2(sinx)^2]/[(cosx+sinx)/cosx]=[sin2xcosx-2cosx(sinx)^2]/(cosx+sinx)=sin2x(cosx+sinx)/(cosx+sinx)=sin2x由已知0 00cosx-sinx=-(√10)/5, == 1-sin2x=2/5, == sin2x=3/5所以原式=3/5 (cosx-sinx)^2=1-2sinxcosx=2/5,所以2sinxcosx=3/5(cosx+sinx)^2=1+2sinxcosx=8/5因?yàn)?0,所以sinx0,cosx0,所以cosx+sinx=2(√10)/5(sin2x-cos2x+1)/(1+tanx)=[2sinxcosx-1+2(sinx)^2+1]/(1+sinx/cosx)=2sinxcosx(cosx-sinx)/(sinx+cosx)=(3/5)*[-(√10)/5]/[2(√10)/5]=-3/10熱心網(wǎng)友
熱心網(wǎng)友