tanX,tanY是方程X²+(4M+1)X+2M=0的俩根,且M≠-1/2求sin(X+Y)/cos(X-Y)

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/30 15:40:27
tanX,tanY是方程X²+(4M+1)X+2M=0的俩根,且M≠-1/2求sin(X+Y)/cos(X-Y)

tanX,tanY是方程X²+(4M+1)X+2M=0的俩根,且M≠-1/2求sin(X+Y)/cos(X-Y)
tanX,tanY是方程X²+(4M+1)X+2M=0的俩根,且M≠-1/2求sin(X+Y)/cos(X-Y)

tanX,tanY是方程X²+(4M+1)X+2M=0的俩根,且M≠-1/2求sin(X+Y)/cos(X-Y)
sin(X+Y)/cos(X-Y)=(tanx+tany)/(1+tanxtany)
由于tanx,tany为方程X²+(4M+1)X+2M=0的俩根
所以,tanx+tany=-(4M+1),tanxtany=2M
所以sin(X+Y)/cos(X-Y)=(tanx+tany)/(1+tanxtany)=-(4M+1)/(2M+1)

tanX,tanY是方程X²+(4M+1)X+2M=0的俩根,根据韦达定理有:
tanX+tanY = -(4M+1)
tanX tanY = 2M
M≠-1/2
sin(X+Y)/cos(X-Y) = (sinXcosY+cosXsinY)/(cosXcosY+sinxsinY)
分子分母同除以cosXcosY:
= (tanX+tanY)/(1+tanXtanY) = -(4M+1)/(1+2M)

sin(X+Y)/cos(X-Y)=(sinxcosy+cosxsiny)/(cosxcosy+sinxsiny)
分式上下同除cosxcosy得 原式=(tanx+tany)/(1+tanxtany)
因为tanx和tany是方程X²+(4M+1)X+2M=0的俩根,根据韦达定理
tanx+tany=-4m-1
tanxtany=2m
M≠-1...

全部展开

sin(X+Y)/cos(X-Y)=(sinxcosy+cosxsiny)/(cosxcosy+sinxsiny)
分式上下同除cosxcosy得 原式=(tanx+tany)/(1+tanxtany)
因为tanx和tany是方程X²+(4M+1)X+2M=0的俩根,根据韦达定理
tanx+tany=-4m-1
tanxtany=2m
M≠-1/2,所以1+2m≠0
所以原式=(-4m-1)/(1+2m)=-(4m+1)/(2m+1)

收起