tan+2π+x

辛诚炭3758tan(x+5π/4)=2 tanx=? -
阮张奋15771651819 ______[答案] tan(x+5π/4)=2 tan(x+π/4)=2 即[tanx+tan(π/4)]/[1-tanxtan(π/4)]=2 1+tanx=2-2tanx 3tanx=1 tanx=1/3

辛诚炭3758已知tanα tanβ是方程x∧2+3√3x+4=0的两个根,且α,β∈( - π/2,π/2)则(α+β)是 -
阮张奋15771651819 ______ tanα tanβ是方程x²+3√3x+4=0的两个根,tanα+ tanβ=-3√3,tanα ·tanβ=4,tan(α+β)=(tanα+ tanβ)/(1-tanα ·tanβ)=√3,又α,β∈(-π/2,π/2),故α+β=π/3或4π/3

辛诚炭3758高中一年级数学,三角函数恒等式1+sinx/cosx=tan(π/4+x/2) -
阮张奋15771651819 ______[答案] tan(π/4+x/2)=(tanπ/4+tanx/2)/(1-tanπ/4*tanx/2)=(1+tanx/2)/(1-tanx/2)=(cosx/2+sinx/2)/(cosx/2-sinx/2)=(cosx/2+sinx/2)^2/((cosx/2)^2-(sinx/2)^2)=(1+2sinx/2*cosx/2)/cosx=(1+sinx)/cosx

辛诚炭3758三角函数题.已知tanx=2.求(sin(π - x)cos(2π - x)sin( - x+3π/2))/tan( - x - π)sin( - π - x)的值已知tanx=2.求(sin(π - x)cos(2π - x)sin( - x+3π/2))/tan( - x - π)sin( - π - x)的值.过... -
阮张奋15771651819 ______[答案] 因为sin(π-x)=sinx,cos(2π-x)=cosx,sin(-x+3π/2)=sin(3π/2-x)=-cosx,tan(-x-π)=-tan(π+x)=-tanx,sin(-π-x)=-sin(π+x)=sinx所以原式=[sinx*cosx*(-cosx)]/(-tanx*sinx)=cos²x/tanx因为tanx=sinx/cosx...

辛诚炭3758利用函数极限求数列极限例题,求:limtan^n(π/4+2/n)(n趋近于∞)记:f(x)=(tan(π/4+2/x))^x,则f(n)=tan^n(π/4+2/n)limf(x)=e^limxlntan(π/4+2/x),(x趋近于+∞);……... -
阮张奋15771651819 ______[答案] 为什么x趋近于正无穷?这一步是怎么变化来的?因为n为正整数,趋于+∞,通过求函数的极限求数列的极限是利用收敛函数的子列必收敛,且极限相同.lntan(π/4+2/x)= ln(1+tan(π/4+2/x)-1)~ tan(π/4+2/x)-1∵t...

辛诚炭3758(已知tanx/2=2,求tan(x+π/4)的值) -
阮张奋15771651819 ______[答案] tanx=2tanx/(1-tan^2x)=2*2/(1-2^2)=-4/3 tan(x+π/4)=(tanx+tanπ/4)/(1-tanx*tanπ/4)=(-4/3+1)/[1-(-4/3)*1]=-1/7 说明:tan^2x是tanx的平方,2^2就是2的平方.

辛诚炭3758化简tan(x+π/4) - tan(x - π/4) -
阮张奋15771651819 ______ tan(x+π/4)-tan(x-π/4) =(tanx+tan(π/4))/(1-tanx*tan(π/4))-(tanx-tan(π/4))/(1+tanx*tan(π/4)) =(tanx+1)/(1-tanx)-(tanx-1)/(1+tanx) =((tanx+1)²+(tanx-1)²)/(1-tan²x) =2(tan²x+1)/(1-tan²x)

辛诚炭3758lim x→无穷 [tan(πx/2x+1)]^1/x 次幂用洛必达 -
阮张奋15771651819 ______[答案] πx/(2x+1)=(π/2)(2x+1-1)/(2x+1)=(π/2)(1-1/(2x+1))=π/2-π/(4x+2) tan[πx/(2x+1)]=ctan[π/(4x+2)] 设y={ctan[π/(4x+2)]}^(1/x) 两边取对数 lny={lncos[π/(4x+2)]-lnsin[π/(4x+2)]}/x->-lnsin[π/(4x+2)]/x x->∞时,π/(4x+2)->0,lncos[π/(4x+2)]->0,lnsin[π/(4x+2)->-∞,∞/∞型不定...

辛诚炭3758已知x∈(π/2,π)tan(x+π/4)=1/7 求sinx+cosx
阮张奋15771651819 ______ 因为tan(x+π/4)=1/7,0<x<π,π<x+π/4<3π/2,cos(x+π/4)=-7V2/10,sin(x+π/4)=-V2/10sinx+cosx=V2sin(x+π/4)=-1/5

辛诚炭3758已知tan(x+π/4)=2,则cos2x=? -
阮张奋15771651819 ______[答案] tan(x+π/4) = (tanx+tanπ/4)/(1-tanxtanπ/4) = (tanx+1)/(1-tanx) = 2tanx+1 = 2-2tanx3tanx = 1tanx=1/3cosx = ±1/根号(1+tan^2x) = ±1/根号(1+1/9) = ± 3/根号10cos2x = 2cos^2x - 1 = 2*(±3/根号10)^2 - ...