Question

高一数学已知函数f(x)=cosx/cos2x^2+sinx/sin2

Answer 1

解：(1)首先化简函数f(x)f(x),得到
f(x)=cos⁡xcos⁡2x2+sin⁡xsin⁡2=cos⁡xcos⁡2x2+1tan⁡2x=cos⁡xcos⁡2x2+cos⁡2xsin⁡2xf(x)=cos2x2cosx+sin2sinx=cos2x2cosx+tan2x1=cos2x2cosx+sin2xcos2x
=cos⁡xsin⁡2x+cos⁡2xcos⁡2xsin⁡2xcos⁡2x=sin⁡(4x+x)sin⁡4x=sin⁡(3x+x)sin⁡4x=sin⁡4xcos⁡x+cos⁡4xsin⁡xsin⁡4x=cos⁡x+cos⁡4xsin⁡xsin⁡4x=sin2xcos2xcosxsin2x+cos2xcos2x=sin4xsin(4x+x)=sin4xsin(3x+x)=sin4xsin4xcosx+cos4xsinx=cosx+sin4xcos4xsinx
=cos⁡x+cos⁡(4x−3x)sin⁡xsin⁡4x=cos⁡x+cos⁡3xsin⁡xsin⁡4x=cos⁡x+2sin⁡xcos⁡2x2sin⁡x⋅cos⁡2x=cos⁡x+12cos⁡2x=32cos⁡x=cosx+sin4xcos(4x−3x)sinx=cosx+sin4xcos3xsinx=cosx+2sinx⋅cos2x2sinxcos2x=cosx+21cos2x=23cosx

Answer 2

设y=(cosx/cos2x)^(1/x^2)
lny
=1/x^2*ln(cosx/cos2x)
=[ln(cos
x)-ln(cos
2x)]/x^2
当x->0时，ln(cosx)=ln(cos
2x)->ln(cos
0)=ln1=0,
x^2->0^2=0
0除0是不定型，必须借助洛必达法则
对于分子分母分别求导
[ln(cos
x)]'=1/cosx
*(-sinx),
-sinx是cosx的导数，运用链式法则
=-tan
x
[ln(cos
2x)]'=1/cos2x
*(-sin2x)*2=-2tan
2x
分母
(x^2)'=2x
一看
[-tanx
+2
tan
2x]/2x,当->0时依旧是-tanx
+2
tan
2x->-tan0+2tan0=tan0=0,2x->2*0=0
0除0，洛必达第二次
(-tan
x)'=-sec^2
x,
(2tan2x)'=2*sec^2
2x*2=4sec^2
2x
(2x)'=2
变成(-sec^2
x+4
sec^2
2x)/2
将x=0带入,sec
0=1所以极限等于(-1^2+4*1^2)/2=3/2