Question

函数的证明题。。。。

Answer 1

用求导法比较方便
设f(x)=arctanx-1/2arcos(2x/1+x^2)
求得: f'(x) =1/(1+x^2)-1/2*(-1)*1/√1-(2x/(1+x^2))^2*(2x/(1+x^2))'=0
那么f(x)=C
f(1) = π/4 - (1/2) arccos[2/(1+1)]=π/4 - (1/2)arccos1= π/4 - 0= π/4.
那么f(x) = π/4.
即知命题成立.

追问

f'(x)的算法能写详细点么？？
看不懂。。。

追答

f'(x)=(arctanx-1/2arcos(2x/1+x^2))'
=(arctanx)'-1/2(arccos(2x/(1+x^2))'
=1/(1+x^2)-1/2*(-1)*1/√1-(2x/(1+x^2))^2*(2x/(1+x^2))'
=1/(1+x^2)-1/2*(-1)*(x^2-1)/(x^2+1)*2*(1-x^2)/(x^2+1)^2
=1/(1+x^2)-1/(1+x^2)
=0
够详细了吧

Answer 2

令a=arctanx,b=arccos2x/(1+x^2),则左边=a-b/2
0<cosb=2x/(1+x^2)<=1,故0<=b<π/2,sinb=(1-cos^2b)^(1/2)=(x^2-1)/(1+x^2)
代入tan(b/2)=(1-cosb)/sinb得tan(b/2)=(x-1)/(x+1)
tan(a-b/2)=[tana-tan(b/2)]/[1+tanatan(b/2)]
将tana=x,tan(b/2)=(x-1)/(x+1)代入计算tan(a-b/2)=1
估计角度：π/4<=a<π/2,0<=b<π/2故0<a-b/2<π/2为y=tanx单调区间
因此a-b/2=π/4
证毕。