tanx/tany等于

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 07:16:03
tanx/tany等于
(1-tanx)×tanx等于什么

tanx-(tanx)平方

若sin(x+y)=1/2,sin(x-y)=1/3,则tanx/tany等于?

sin(x+y)=sinxcosy+cosxsiny=1/2,sin(x-y)=sinxcosy-cosxsiny=1/3,两式作商得(sinxcosy+cosxsiny)/(sinxcosy-cos

证明 (tanX+tanY)/(tanX-tanY)=(sin(X+Y))/(sin(X-Y))

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

已知sinx+cosy=3/4,cosx+cosy+1/2,求tanx*tany

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

tanx+tany=25 cotx+coty=30,那么tan(x+y)=?

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)=1\2,sin(x—y)=1\3,求[tan(x+y)-tanx-tany]\[tany的平方tan(

sin(x+y)=sinxcosy+cosxsiny=1/2sin(x-y)=sinxconsy-cosxsiny=1/3sinxcosy=5/12,cosxsiny=1/12tanx/tany=si

已知tanx=2,tany=1/3,则tan2(x+y)=

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

已知tanx=2,tany=3,x,y∈(0,π/2),求x+y.

tan(x+y)=(tanx+tany)/(1-tanxtany)=-1x+y=3π/4

已知sinx+siny=2sin(x+y)≠0,则tanx/2tany/2等于 是三角函数和差化积部分的

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(

tanx+tany=25,cotx+coty=30.tan(x+y)=?

cotx+coty=1/tanx+1/tany=(tanx+tany)/(tnaxtany)即30=25/(tanxtany),所以tanxtnay=5/6tan(x+y)=(tanx+tany)/(

已知sin(x+y)/cos(x-y)=m/n,则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

tanx等于多少

当x≠kπ±π/2时,tanx=sinx/cosx当x=kπ±π/2时,tanx无意义(k∈Z)

若x、y是锐角且tanx=2/3tany=3/4则sin(x+y)等于

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)=1/5,cos(x-y)=3/5则tanx.tany=?

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)/

己知tanx,tany,是方程7㎡-8m+1=0的两根,则tan[(x+y)÷2]等于多少.求详解.

大师教学有过程,先彩后教详解,彩吶我第一时间教谢谢!

tanx等于

对边除以邻边

已知tanx=1/4,tany=-3,求tan(x+y)的值

tan(x+y)=[tanx+tany]/[1-tanx*tany]=(1/4-3)/(1+3/4)=-11/7

求证tanx+tany/tanx-tany=sin(x+y)/sin(x-y)(详细步骤)

(tanx+tany(/(tanx-tany)分子分母同乘以cosxcosy:=(sinxcosy+cosxsiny)/(sinxcosy-cosxsiny)=sin(x+y)/sin(x-y)

tanX=2分之1,tan(x+y)=5分之2,那么tanY=?

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!

已知tanx=-3/4,且tan(x+y)=1,求tany的值

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