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lilihurry

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[求助] 请大师帮忙解决一中学不等式问题,谢谢!

已知0<x<180,求证:cot(x/8)-cotx>3
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feixiaolin

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引用回帖:
3楼: Originally posted by lilihurry at 2013-11-18 22:11:36
可以详细点吗?...

cot(x/8)=cos(x/8)/sin(x/8);
分子分母同乘2*cos(x/8)/*2*cos(x/4)/*2*cos(x/2),
于是,cot(x/8)=(略)/sinx,
cot(x/8)-cotx-3
=(略)/sinx-cosx/sinx-3*sinx/sinx
整理,即得。
4楼2013-11-18 22:18:03
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feixiaolin

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cot(x/8)=cos(x/8)/sin(x/8);
分子分母同乘2*cos(x/8)/*2*cos(x/4)/*2*cos(x/2),整理,即得。
2楼2013-11-18 17:59:54
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lilihurry

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引用回帖:
2楼: Originally posted by feixiaolin at 2013-11-18 17:59:54
cot(x/8)=cos(x/8)/sin(x/8);
分子分母同乘2*cos(x/8)/*2*cos(x/4)/*2*cos(x/2),整理,即得。

可以详细点吗?
3楼2013-11-18 22:11:36
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zywang1999

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【答案】应助回帖

★ ★ ★ ★ ★
感谢参与,应助指数 +1
lilihurry: 金币+5, ★有帮助 2013-11-19 08:55:01
细一看, 命题是错误的. 反例: x=18.85时, cot(x/8) - cot(x) = - 2252.8543. 看来, 就是院士也帮不了你.
正确的命题是:
0度<x<180度, cot(x/8) - cot(x) >3.
证明:
cot(x/8) -cotx
= [8cos(x/8)*cos(x/8)*cos(x/4)*cos(x/2) - cos(x)] / sin(x)
= [4(cos(x/4)+1) * cos(x/4) *cos(x/2) - 2 cos(x/2) * cos(x/2) +1] / sinx
= [2cos(x/2)* (2 cos(x/4)^2 + 2cos(x/4) - cos(x/2)) + 1]/sinx
= [2cos(x/2)* (1 + 2cos(x/4) ) + 1]/sinx
= [sinx/sin(x/2) + sinx/sin(x/4) + 1]/sinx
= 1/sin(x/2) + 1/sin(x/4) + 1/sinx > 3 (0度<x<180度)
我是这么深爱你啊,我的中国
5楼2013-11-18 22:54:35
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