基本关系
\[\begin{array}{} \csc\alpha=\dfrac{1}{\sin\alpha}& \sec\alpha=\dfrac{1}{\cos\alpha}& \cot\alpha=\dfrac{1}{\tan\alpha}& \tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}& \cot\alpha=\dfrac{\cos\alpha}{\sin\alpha} \end{array}\] \[\begin{array}{} \sin^2\alpha+\cos^2\alpha=1& 1+\tan^2\alpha=\sec^2\alpha& 1+\cot^2\alpha=\csc^2\alpha \end{array}\]诱导公式
$\sin(\dfrac{\pi}{2}\pm\alpha)=\cos\alpha$
$\cos(\dfrac{\pi}{2}\pm\alpha)=\mp\sin\alpha$
$\sin(\pi\pm\alpha)=\mp\sin\alpha$
$\cos(\pi\pm\alpha)=-\cos\alpha$
和差公式
$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$
$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$
倍角公式
\[\begin{align*} \sin 2\alpha&=2\sin\alpha\cos\alpha\\\\ &\cos 2\alpha\\ =&\cos^2\alpha-\sin^2\alpha\\ =&(\sin\alpha+\cos\alpha)(\cos\alpha-\sin\alpha)\\ =&1-2\sin^2\alpha\\ =&2\cos^2\alpha-1 \end{align*}\] \[\begin{array}{lrlrl} &1+\sin2\alpha&=(\sin\alpha+\cos\alpha)^2 &1-\sin2\alpha&=(\sin\alpha-\cos\alpha)^2\\\\ &\sin 3\alpha&=-4\sin^3\alpha_3\sin\alpha &\cos 3\alpha&=4\cos^3\alpha-3\cos\alpha\\\\ &\tan 2\alpha&=\dfrac{2\tan\alpha}{1-\tan^2\alpha} &\cot 2\alpha&=\dfrac{\cot^2\alpha-1}{2\cot\alpha}\\ \end{array}\]半角公式
$\sin^2\dfrac{\alpha}{2}=\dfrac{1}{2}(1-\cos\alpha)$
$\cos^2\dfrac{\alpha}{2}=\dfrac{1}{2}(1+\cos\alpha)\text{（降幂公式）}$
$\sin\dfrac{\alpha}{2}=\pm\sqrt{\dfrac{1-\cos\alpha}{2}}$
$\cos\dfrac{\alpha}{2}=\pm\sqrt{\dfrac{1+\cos\alpha}{2}}$
$\tan\dfrac{\alpha}{2}=\dfrac{1-\cos\alpha}{\sin\alpha}=\dfrac{\sin\alpha}{1+\cos\alpha}=\pm\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha}}=\dfrac{1}{\cot\dfrac{\alpha}{2}}$
推导
\[\begin{align*} &\cos(2\alpha)=1-2\sin^2\alpha=2\cos^2\alpha-1\\ &\to\\ &\sin^2\alpha=\dfrac{1}{2}(1-\cos 2\alpha)\\ &\cos^2\alpha=\dfrac{1}{2}(1+\cos 2\alpha)\\ &\to\\ &\sin\alpha=\pm\sqrt{\dfrac{1-\cos 2\alpha}{2}}\\ &\cos\alpha=\pm\sqrt{\dfrac{1+\cos 2\alpha}{2}} \end{align*}\]积化和差公式/和差化积公式
积化和差：和角公式的组合相消 和差化积：积化和差的换元/和角公式中对角度的对称分解
\[\begin{array}{ll|l} &\sin\alpha\cos\beta=\dfrac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)] &\cos\alpha\sin\beta=\dfrac{1}{2}[\sin(\alpha+\beta)-\sin(\alpha-\beta)]\\\\ &\cos\alpha\cos\beta=\dfrac{1}{2}[\cos(\alpha+\beta)+\cos(\alpha-\beta)] &\sin\alpha\sin\beta=\dfrac{1}{2}[\cos(\alpha-\beta)-\cos(\alpha+\beta)]\\\\ &\sin\alpha+\sin\beta=2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2} &\sin\alpha-\sin\beta=2\sin\dfrac{\alpha-\beta}{2}\cos\dfrac{\alpha+\beta}{2}\\\\ &\cos\alpha+\cos\beta=2\cos\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2} &\cos\alpha-\cos\beta=-2\sin\dfrac{\alpha+\beta}{2}\sin\dfrac{\alpha-\beta}{2}\\ \end{array}\]推导
\[\begin{align*} \sin\alpha+\sin\beta =&\sin\left(\dfrac{\alpha+\beta}{2}+\dfrac{\alpha-\beta}{2}\right)+\sin\left(\dfrac{\alpha+\beta}{2}-\dfrac{\alpha-\beta}{2}\right)\\ =&\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}+\cos\dfrac{\alpha+\beta}{2}\sin\dfrac{\alpha-\beta}{2}+\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}\\ =&-\cos\dfrac{\alpha+\beta}{2}\sin\dfrac{\alpha-\beta}{2}\\ =&2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2} \end{align*}\]辅助角公式
$a\sin x+b\cos x=\sqrt{a^2+b^2}\sin(x+\phi)$，$\sin\phi=\dfrac{b}{\sqrt{a^2+b^2}}$，$\cos\phi=\dfrac{a}{\sqrt{a^2+b^2}}$。
导数与积分
三角与反三角函数
https://math.fandom.com/zh/wiki/%E5%8F%8D%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0
\[\begin{array}{ll|ll|ll} (\sin x)'&=\cos x & (\cos x)'&=-\sin x & (\tan x)'&=\dfrac{1}{\cos^2 x}=\sec^2 x\\ (\csc x)'&=-\csc x \sin^{-2} x=-\csc x\cot x & (\sec x)'&=\sin x \cos^{-2} x=\sec x\tan x & (\cot x)'&=-csc^2 x\\ (\arcsin x)'&=\dfrac{1}{\sqrt{1-x^2}} & (\arccos x)'&=-\dfrac{1}{\sqrt{1-x^2}} & (\arctan x)'&=\dfrac{1}{1+x^2} \\ (\operatorname{arccsc} x)'&=-\dfrac{1}{x\sqrt{x^2-1}} & (\operatorname{arcsec} x)'&=\dfrac{1}{x\sqrt{x^2-1}} & (\operatorname{arccot} x)'&=-\dfrac{1}{1+x^2} \end{array}\]特殊积分
\[\begin{align*} \int\sec x\,dx =& \int \dfrac{\sec x (\sec x + \tan x)}{ (\sec x + \tan x)}\,dx = \int \dfrac{d(\sec x + \tan x)}{ (\sec x + \tan x)}\\ =& ln|\sec x + \tan x| + C\\ \int\csc x\,dx =& \int \dfrac{\csc x (\csc x + \cot x)}{ (\csc x + \cot x)}\,dx = \int -\dfrac{d(\csc x + \cot x)}{ (\csc x + \cot x)}\\ =& -ln|\csc x + \cot x| + C\\ =& ln|\csc x - \cot x| + C \end{align*}\]$\sec x + \tan x $, $\csc x - \cot x $ 等具有性质$f’(x) = g(x) f(x)$ 的函数。$g(x) dx = \dfrac{f’(x)}{f(x)} dx = d\ln f(x)$
双曲与反双曲函数
\[\begin{array}{lll|lll} \operatorname{sinh}x&=\operatorname{sh}x=\dfrac{e^{x}-e^{-x}}{2}&& \operatorname{arsinh} x &= \ln(x+\sqrt{x^2+1}) \\\\ \operatorname{cosh}x&=\operatorname{ch}x=\dfrac{e^{x}+e^{-x}}{2}&& \operatorname{arcosh} x &= \ln(x+\sqrt{x^2-1})&,x\in[1,\infty) \\\\ \operatorname{tanh}x&=\operatorname{th}x=\dfrac{\operatorname{sinh}x}{\operatorname{cosh}x}=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}&& \operatorname{artanh} x &= \dfrac{1}{2} \ln\left|\dfrac{1+x}{1-x}\right|&,|x| < 1 \\\\ \operatorname{coth}x&=\dfrac{\operatorname{cosh}x}{\operatorname{sinh}\,x}=\dfrac{e^{x}+e^{-x}}{e^{x}-e^{-x}}&,x\in\mathbb{R} - \{0\}& \operatorname{arcoth} x &= \dfrac{1}{2} \ln\left|\dfrac{1+x}{1-x}\right|&,|x| > 1 \\\\ \operatorname{csch}x&=\dfrac{1}{\operatorname{sinh}x}=\dfrac{2}{e^{x}-e^{-x}}&,x\in\mathbb{R}- \{0\}& \operatorname{arcsch} x &= \ln(\dfrac{1}{x}+\dfrac{\sqrt{1+x^2}}{|x|}) = \begin{cases} \ln\dfrac{1+\sqrt{1+x^2}}{x}\\\\ \ln\dfrac{1-\sqrt{1+x^2}}{x} \end{cases}& \begin{array}{} ,x\in(0,\infty)\\\\\\ ,x\in(\infty,0) \end{array} \\\\ \operatorname{sech}x&=\dfrac{1}{\operatorname{cosh}x}=\dfrac{2}{e^{x}+e^{-x}}&& \operatorname{arsech} x &= \ln\dfrac{1+\sqrt{1-x^2}}{x}&,x\in(0,1] \end{array}\]note: 从 $\operatorname{arsinh} x$ 到 $\operatorname{arcsch} x$ 或 $\operatorname{arcosh} x$ 到 $\operatorname{arsech} x$ 可以由
- 转换为 $\dfrac{1}{x}$ 的形式(辅助)
- 将 $\dfrac{1}{x}$ 换元为 $x$ 得到(本质是$x$换元为$\frac{1}{x}$)
导数如 从 $(\operatorname{arsinh} x)’$ 到 $(\operatorname{arcsch} x)’$ 要额外乘以 $-\dfrac{1}{x^2}$ ($\dfrac{1}{x}$的导数，复合函数求导法则)
性质
\[\begin{align*} &\cosh x + \sinh x = e^x\\ &\cosh x - \sinh x = e^{-x}\\ &\cosh^2 x - \sinh^2 x = 1\\ \end{align*}\] \[\begin{array}{ll|ll|ll} (\sinh x)'&=\cosh x & (\cosh x)'&=\sinh x & (\tanh x)'&=1-\tanh^2 x\\ (\operatorname{csch} x)'&=-\coth x \operatorname{csch}& (\operatorname{sech} x)'&=-\tanh x \operatorname{sech}& (\operatorname{coth} x)'&=1-\operatorname{coth}^2 x\\ (\operatorname{arsinh} x)'&=\dfrac{1}{\sqrt{x^2+1}}& (\operatorname{arcosh} x)'&=\dfrac{1}{\sqrt{x^2-1}}& (\operatorname{artanh} x)'&=\dfrac{1}{1-x^2}\\ (\operatorname{arcsch} x)'&=-\dfrac{1}{|x|\sqrt{1+x^2}}& (\operatorname{arsech} x)'&=-\dfrac{1}{x\sqrt{1-x^2}}& (\operatorname{arcoth} x)'&=\dfrac{1}{1-x^2} \end{array}\]需要注意的
\[\begin{array}{ll} \begin{align*} \int\dfrac{1}{\sqrt{x^2+1}}\,dx &= \operatorname{arcsinh}\,x + C \\&= \ln\left(x+\sqrt{x^2+1}\right)+C\\ \end{align*} & \begin{align*} \int\dfrac{1}{\sqrt{x^2-1}}\,dx &= \operatorname{arcosh}\,x + C\,,(x\geq 1)\\ &= \ln\left|x+\sqrt{x^2-1}\right| + C\\ \end{align*}\\ \begin{align*} \int\dfrac{1}{1-x^2}\,dx &= \operatorname{arctanh}\,x + C \,,(-1<x<1)\\ &= \dfrac{1}{2}\ln\left|\dfrac{1+x}{1-x}\right| + C \end{align*} \end{array}\]https://math.fandom.com/zh/wiki/%E5%8F%8D%E5%8F%8C%E6%9B%B2%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0?variant=zh
性质1 复合关系
\[\begin{array}{ll|l} &\sinh(\operatorname{arcsinh} x) = x \quad & \cosh(\operatorname{arcsinh} x) = \sqrt{x^2 + 1} \\\\ &\sinh(\operatorname{arcosh} x) = \sqrt{x^2 - 1} \quad & \cosh(\operatorname{arcosh} x) = x \\\\ &\sinh(\operatorname{arctanh} x) = \dfrac{x}{\sqrt{1 - x^2}} \quad & \cosh(\operatorname{arctanh} x) = \dfrac{1}{\sqrt{1 - x^2}} \\\\ &\sinh(\operatorname{arsech} x) = \dfrac{1}{\sqrt{x^2 - 1}} \quad & \cosh(\operatorname{arsech} x) = \dfrac{|x|}{\sqrt{x^2 - 1}} \\\\ &\sinh(\operatorname{arsech} x) = \dfrac{\sqrt{1 - x^2}}{x} \quad & \cosh(\operatorname{arsech} x) = \dfrac{1}{x} \\\\ &\sinh(\operatorname{arcsch} x) = \dfrac{1}{x} \quad & \cosh(\operatorname{arcsch} x) = \dfrac{\sqrt{x^2 + 1}}{|x|} \end{array}\]像三角复合问题一样，有时候我们在积分中做了双曲代换，但最后又不得不回代原变量，尤其是不定积分情形，这时我们就要解决类似$x = \sinh t, \cosh t = ???$的问题，三角代换情形可以使用直角三角形辅助，也可使用三角函数线，对应双曲情形也可以使用单位双曲线的右支——即双曲函数线解决，当然也可以用反双曲函数的对数定义，再用双曲函数的指数定义。 \(\begin{array}{l|l|l|l} 函数& \sinh t = & \cosh t = & \tanh t = & \\\\ x = \sinh t& x& \sqrt{x^2+1}& \dfrac{x}{\sqrt{x^2+1}}& \\\\ x = \cosh t& \sqrt{x^2-1}& x& \dfrac{\sqrt{x^2-1}}{x}& \\\\ x = \tanh t& \dfrac{x}{\sqrt{1-x^2}}& \dfrac{1}{\sqrt{1-x^2}}& x& \end{array}\)
这里仅列出九个主要的，其余都可以用复合函数化为上表的类型，不过要特别注意如下从根式中提取$x$的操作的符号问题： $\cosh(\operatorname{arcoth} x) = \cosh \left( \operatorname{artanh} \dfrac{1}{x} \right) = \dfrac{1}{\sqrt{1-\left(\frac{1}{x}\right)^2}} = \dfrac{|x|}{\sqrt{x^2-1}}$
性质2 平方恒等关系
\[\begin{aligned} &\cosh^2 x - \sinh^2 x = 1 \\ &1 - \tanh^2 x = \operatorname{sech}^2 x \\ &\coth^2 x - 1 = \operatorname{csch}^2 x \end{aligned}\]根据公式(4)中的第一个等式，我们可以发现这个形式与双曲线很相似。实际上，由此式可以说明，参数方程
\[\begin{cases} x = a \cosh t \\ y = b \sinh t \\ \end{cases}, \quad a, b > 0\]恰可表示双曲线
\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\] \[\begin{align*} I=\int_0^1 x^{2}\sqrt{1+4x^{2}}dx\\ \overset{2x=\sinh t}{=} &\int_0\operatorname{arcsinh} 2(\frac{\sinh t}{2})^{2\cosh td(frac{\sinh t}{2})}\\ =&\frac{1}{8}\int_0^{\operatorname{arcsinh} 2}(\sinh t\cosh t)^2dt\\ =&\frac{1}{32}\int_0^{\operatorname{arcsinh} 2}\sinh^22tdt\\ =&\frac{1}{64}\int_0^{\operatorname{arcsinh} 2}(\cosh4t-1)dt\\ =&\frac{1}{64}(\frac{\sinh4t}4-t)|_{0}^{\operatorname{arcsinh} 2}\\ =&\frac{1}{256}\sinh(4\operatorname{arcsinh}2)-\frac{1}{64}\operatorname{arcsinh}2\\ =&\frac{9\sqrt{5}}{32}-\frac{1}{64}\ln(2+\sqrt5) \end{align*}\]