tanx/tany等于
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tanx-(tanx)平方
sin(x+y)=sinxcosy+cosxsiny=1/2,sin(x-y)=sinxcosy-cosxsiny=1/3,两式作商得(sinxcosy+cosxsiny)/(sinxcosy-cos
利用tanx=sinx/cosx的定义,左边的一定可以化为sinx和cosx的式子同理右边的可以分解为sinx和cosx的式子二者一定相等
题目有问题再问:已知sinx+cosy=3/4,cosx+cosy=1/2,求tanx*tany,题目打错了,不用和差化积公式怎么做?再答:你只要求出cosy为多少即可以,只要把俩个等式的COSY都移
cotx+coty=301/tanx+1/tany=30(tanx+tany)/tanxtany=3025/tanxtany=30tanxtany=25/30=5/6,tan(x+y)=(tanx+t
sin(x+y)=sinxcosy+cosxsiny=1/2sin(x-y)=sinxconsy-cosxsiny=1/3sinxcosy=5/12,cosxsiny=1/12tanx/tany=si
这道题是这样的:原式=tan(2x+2y)=(tan2x+tan2y)/(1-tan2x.tan2y);其中tan2x=tan(x+x)=(tanx+tanx)/(1-tanx.tanx)=-4/3t
tan(x+y)=(tanx+tany)/(1-tanxtany)=-1x+y=3π/4
sinx+siny=2sin(x+y)2sin(x/2+y/2)cos(x/2-y/2)=4sin(x/2+y/2)cos(x/2+y/2)cos(x/2-y/2)=2cos(x/2+y/2)cos(
cotx+coty=1/tanx+1/tany=(tanx+tany)/(tnaxtany)即30=25/(tanxtany),所以tanxtnay=5/6tan(x+y)=(tanx+tany)/(
要求tanx/tany则x,y≠π/2+kπ,且y≠kπ(k为整数),则cosxcosy≠0sin(x+y)/cos(x-y)=m/n(sinxcosy+cosxsiny)/(cosxcosy+sin
当x≠kπ±π/2时,tanx=sinx/cosx当x=kπ±π/2时,tanx无意义(k∈Z)
再答:你也可以简单计算!如果是填空或者选择题的话再问:请问填空题怎么简单计算啊
cos(x+y)+cos(x-y)=2cos[(x+y+x-y)/2]cos[(x+y-x+y)/2]=2cosx*cosy=4/5cos(x+y)-cos(x-y)=-2sin[(x+y+x-y)/
大师教学有过程,先彩后教详解,彩吶我第一时间教谢谢!
对边除以邻边
tan(x+y)=[tanx+tany]/[1-tanx*tany]=(1/4-3)/(1+3/4)=-11/7
(tanx+tany(/(tanx-tany)分子分母同乘以cosxcosy:=(sinxcosy+cosxsiny)/(sinxcosy-cosxsiny)=sin(x+y)/sin(x-y)
tan(x+y)=(tanx+tany)/(1-tanxtany)0.4=(tany+0.5)/(1-0.5*tany)就可以解出来了.希望对你有所帮助!
可采用凑角变换:这是标准的答题模式,对你今后学习三角函数有很大的帮助.tany=[tan(x+y)-x]tan(x+y)-tanx=-----------------1+tan(x+y)tanx1+3