作业帮1.已知8cos(2x+y)+5cosy=0,求tan(x+y)*tanx的值.2.已知关于x的方程x^2+px+q=0的两根是tanx,tany,求sin(x+y)/cos(x-y)的值.3.化简:sin(x-y)/(sinx*siny)+sin(y-z)/(siny*sinz)+sin(z-x)/(sinz*sinx).
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1.已知8cos(2x+y)+5cosy=0,求tan(x+y)*tanx的值.2.已知关于x的方程x^2+px+q=0的两根是tanx,tany,求sin(x+y)/cos(x-y)的值.3.化简:sin(x-y)/(sinx*siny)+sin(y-z)/(siny*sinz)+sin(z-x)/(sinz*sinx).
1.已知8cos(2x+y)+5cosy=0,求tan(x+y)*tanx的值.
2.已知关于x的方程x^2+px+q=0的两根是tanx,tany,求sin(x+y)/cos(x-y)的值.
3.化简:sin(x-y)/(sinx*siny)+sin(y-z)/(siny*sinz)+sin(z-x)/(sinz*sinx).
1.已知8cos(2x+y)+5cosy=0,求tan(x+y)*tanx的值.2.已知关于x的方程x^2+px+q=0的两根是tanx,tany,求sin(x+y)/cos(x-y)的值.3.化简:sin(x-y)/(sinx*siny)+sin(y-z)/(siny*sinz)+sin(z-x)/(sinz*sinx).
3:通分,将3个分式的分母化成相同的,即
sin(x-y)sinz/(sinxsinysinz) +sin(y-z)sinx/(sinxsinysinz) +sin(z-x)siny/(sinxsinysinz)
再将分子展开,即
sinxcosysinz-cosxsinysinz+sinycoszsinx-cosysinzsinx+sinzcosxsiny-coszsinxsiny=0
所以结果是0
本人能力不怎样,先给第3题的答案
(这道题要根据已知来构建角,也就是 x+y 和 x)
1:8cos(2x+y) +5cosy =0
8cos(x+y+x) +5cos(x+y-x) =0
8cos(x+y)cosx -8sin(x+y)sinx + 5cos(x+y)cosx + 5sin(x+y)*sinx =0
整理得:13cos(x+y)cosx = 3sin(x+y)sinx
sin(x+y)sinx/cos(x+y)cosx = 13/3
tan(x+y)tanx = sin(X+y)sinx/cos(x+y)cosx
= 13/3
2:由已知可得:tanx+tany = -p ,tanxtany = q
sin(x+y)/cos(x-y) =(sinxcosy +cosxsiny)/(cosxcosy+sinxsiny)
右边分子,分母同除以cosxcosy得:
tanx+tany/1+tanxtany
所以,sin(x+y)/cos(x-y) = tanx+tany/1+tanxtany
= -p/1+q