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[函数] 三角函数零点个数问题

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nash posted 2015-7-23 00:06 |Read mode
$f(x)=tan2013x-tan2014x+tan2015x$ 在[0,pai]的零点个数

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shidilin posted 2015-7-23 00:20
2014年全国高中数学联赛安徽省初赛试卷第2题
360截图20150723001922687.jpg

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original poster nash posted 2015-7-23 00:26

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kuing posted 2015-7-23 00:43
直接用 tan 来做不是更简单咩
\begin{align*}
\tan (p-q)+\tan (p+q)-\tan p
&=\frac{\tan p-\tan q}{1+\tan p\tan q}+\frac{\tan p+\tan q}{1-\tan p\tan q}-\tan p \\
&=\frac{2\tan p(1+\tan^2q)}{1-\tan^2p\tan^2q}-\tan p \\
&=\frac{\tan p(1+2\tan^2q+\tan^2p\tan^2q)}{1-\tan^2p\tan^2q}
\end{align*}
下略。

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其妙 posted 2015-7-23 19:52
等差数列?1994高考
2blog图片.png
妙不可言,不明其妙,不着一字,各释其妙!

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kuing posted 2015-7-23 19:57
回复 5# 其妙

0到pi/2以内还用你说么……

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其妙 posted 2015-7-23 19:58
回复 6# kuing
,我不是来解这道题的,是找相关资料

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