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我们的思想是这样的：先把 1-\cos x 提负号并整理成 \cos x-1 的形式，再把这个形式重新安在 \ln (1+x) 上，接着化成各种 \ln\cos x 系列作和的形式，最后再次利用等价无穷小和极限的四则运算算出最后结果。 比如这两个题有如下过程。

To solve a trigonometric simplify the equation using trigonometric identities. Then, write the equation in a standard form, and isolate the variable using algebraic manipulation to solve for the variable. Use inverse trigonometric functions to find the …

Call f (x) = \cos\sqrt {x} f' (0) f' (x) = -\frac {\sin\sqrt {x}} {2\sqrt {x}} So that \lim_ {x\to 0 }f' (x) = -\frac {1} {2} where I used the well known limit \lim_ {t\to 0}\sin t/t = 1 ...

\sec x = \frac {1} {\cos {x}} \csc {x} = \frac {1} {\sin {x}} \cot {x} = \frac {1} {\tan {x}} \sin^ {2}x + \cos^ {2}x = 1. 1 + \tan^ {2} x = \sec^ {2}x. 1 + \cot^ {2}x = \csc^ {2}x. \tan {x} = \frac {\sin {x}} {\cos {x}} = \frac {\sec {x}} {\csc {x}} \cot {x} = \frac {\cos {x}} {\sin {x}} = \frac {\csc {x}} {\sec {x}}

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6 天之前 · 一、引言 高考的比大小题目中，除了 e^x 和 \ln {x} ，出现较多的就是 \sin {x},\cos {x},\tan {x} 这几个三角函数。导数大题中，也有可能出现这些函数。因此，我们需要研究这几个函数的性质，其中， x\in (0,\frac {\pi} {2}) 是研究这几个函数性质的重要区间。根据三角函数的周期性和对称性，其他区间上的 ...

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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music…