函数的证明题.

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函数的证明题.
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函数的证明题.
函数的证明题.

函数的证明题.
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.

令a=arctanx,b=arccos2x/(1+x^2),则左边=a-b/2
0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+tan...

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令a=arctanx,b=arccos2x/(1+x^2),则左边=a-b/2
0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因此a-b/2=π/4
证毕。

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