\documentclass[UTF,noindent,fancyhdr,titlepage]{ctexart}
\usepackage{delarray,booktabs}
\usepackage[usenames,dvipsnames]{xcolor}
\usepackage{tabu}
\usepackage{colortbl}
\usepackage[margin=1cm]{geometry}
\usepackage{amsmath}
\definecolor{mypink}{rgb}{.99,.91,.95}
\newcommand{\envert}[1]{\left\lvert#1\right\rvert}
\let\abs=\envert
\begin{document}
\begin{table}[htb]
\centering
\tabulinesep=1ex
\taburulecolor |mypink|{VioletRed}
\arrayrulewidth = 0.1mm \doublerulesep=5mm
\everyrow{\hline \hline }
\begin{tabu} to \linewidth{>{\columncolor{VioletRed}}X[-1.5,c,]|X[9,l,m]|}
\large\heiti\textcolor{white}{1.集合}&
含$n$个元素的所有子集有$2^n$个
\\
\large\heiti\textcolor{white}{2.对数}&
\begin{description}
\item[(2) 对数运算性质] 如果$a>0,\mbox{且}\;a \neq 1,\;M>0,\;N>0$,那么：
\begin{itemize}
\item[] 积的对数：$\log_a(MN) = \log_aM +\log_aN$
\item[] 商的对数：$\log_a(\dfrac{M}{N})=\log_aM -\log_aN$
\item[] 幂的对数：$\log_aM^n = n\log_aM\;(n \in \mathbf{R})$
\end{itemize}
\item[(2) 对数恒等式:] \qquad{\large $a^{\log_aN}=N$}$\;(a>0,\mbox{且}a\neq1,\;N>0)$
\item[(3) 对数换底公式:]\quad $\log_ab = \dfrac{\log_cb}{\log_ca}(a>0,\;\mbox{且}\;a\neq1,c>0\;\mbox{且}\;c\neq1,b>0)$
\end{description}
\\
\large\heiti\textcolor{white}{3.等差数列}&
\begin{description}
\item[(1)通项公式:] {\large $a_n = a_1 + (n-1)d$}\;\;(其中首项是$a_1$,公差是$d$)
\item[(2)前$n$项和公式:]{\large $S_n = \dfrac{n(a_1+a_n)}{2} = na_1 + \dfrac{n(n-1)}{2}d$}
\item[(3)等差中项:] $\mathit{A}$是$a$与$b$的等差中项：$\mathit{A}=\dfrac{a+b}{2}$
\end{description}
\\
\large\heiti\textcolor{white}{4.等比数列}&
\begin{description}
\item[(1)通项公式:] {\large $a_n = a_1q^{n-1}d$}\;\;(其中首项是$a_1$,公比是$q$)
\item[(2)前$n$项和公式:]{
\large $S_n = \begin{cases}
na_1 & (q=1)\\
\dfrac{a_1(1-q^n)}{1-q}& (q \neq 1)
\end{cases}
$}
\item[(3)等比中项:] $\mathbf{G}$是$a$与$b$的等比中项：$\dfrac{\mathbf{G}}{a} = \dfrac{b}{\mathbf{G}}$ 即 $\mathbf{G}^2 = ab$ (或$\mathbf{G} = \pm\sqrt{ab}$,等比中项有两个)
\end{description}
\\
\large\heiti\textcolor{white}{5.同角三角函数基本关系式}&
{\large
$\sin^2\alpha +\cos^2\alpha = 1 \qquad \tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$
}
\\
\large\heiti\textcolor{white}{6.诱导公式}&
{\tabureset \tabulinesep=1ex
\begin{tabu}{>{\heiti}X[-1]>{\large}X[2]}
公式一：&
$\sin(\alpha + k \cdot 2\pi)=\sin\alpha,\;(k\in \mathbf{Z})$ \qquad
$\cos(\alpha + k \cdot 2\pi)=\cos\alpha,\;(k\in \mathbf{Z})$ \qquad\newline
$\tan(\alpha + k \cdot 2\pi)=\tan\alpha,\;(k\in \mathbf{Z})$ \qquad
\\
公式二：& $\sin(\pi + \alpha) = -\sin\alpha$ \quad
$\cos(\pi + \alpha) = -\cos\alpha$ \quad
$\tan(\pi + \alpha) = -\tan\alpha$ \quad
\\
公式三：& $\sin(-\alpha) = -\sin\alpha$ \quad\quad
$\cos(-\alpha) = \cos\alpha$ \quad\quad\quad
$\tan(-\alpha) = -\tan\alpha$ \quad
\\
公式四：& $\sin(\pi - \alpha) = \sin\alpha$ \qquad
$\cos(\pi - \alpha) = -\cos\alpha$ \quad
$\tan(\pi - \alpha) = -\tan\alpha$ \quad
\\
公式五：& $\sin(\frac{\pi}{2} - \alpha) = \cos\alpha$ \qquad
$\cos(\frac{\pi}{2} - \alpha) = \sin\alpha$ \quad
\\
公式五：& $\sin(\frac{\pi}{2}+ \alpha) = \cos\alpha$ \qquad
$\cos(\frac{\pi}{2} + \alpha) = -\sin\alpha$ \quad
\\
\end{tabu}}
\\
\end{tabu}
\end{table}
\begin{table}[t]
\centering
\tabulinesep=1ex
\taburulecolor |mypink|{VioletRed}
\arrayrulewidth = 0.1mm \doublerulesep=5mm
\everyrow{\hline \hline }
\begin{tabu} to \linewidth{>{\columncolor{VioletRed}}X[-1,c,m]|X[9,l,m]|}
\large\heiti\textcolor{white}{7.特殊角的三角函数}&
{\tabureset \tabulinesep=1ex
\taburulecolor |mypink|{VioletRed}
\arrayrulewidth = 0.1mm
\begin{tabu}{>{\columncolor{mypink}}X[2c]|X[c]|X[c]|X[c]|X[c]|X[c]|X[c]|X[c]|X[c]|X[c]|X[c]|X[c]|}
\rowcolor{mypink}
$\alpha$的角度 & 0 & 30 & 45 & 60 & 90 & 120 & 135 & 150 & 180 & 270 & 360 \\\hline
$\alpha$的弧度& 0 & $\dfrac{\pi}{6}$ &$\dfrac{\pi}{4}$&$\dfrac{\pi}{3}$&$\dfrac{\pi}{2}$ &$\dfrac{2\pi}{3}$&$\dfrac{3\pi}{4}$&$\dfrac{5\pi}{6}$ & $\pi$ & $\dfrac{3\pi}{2}$&$2\pi$ \\\hline
$\sin\alpha$ & 0 & $\dfrac{1}{2}$ & $\dfrac{\sqrt{2}}{2}$ &
$\dfrac{\sqrt{3}}{2}$ & $1$ & $\dfrac{\sqrt{3}}{2}$ & $\dfrac{\sqrt{2}}{2}$ & $\dfrac{1}{2}$ & 0 & $-1$ & $0$ \\ \hline
$\cos\alpha$ & $1$ & $\dfrac{\sqrt{3}}{2}$ & $\dfrac{\sqrt{2}}{2}$ & $\dfrac{1}{2}$
& $0$ & $-\dfrac{1}{2}$ & $-\dfrac{\sqrt{2}}{2}$ & $-\dfrac{\sqrt{3}}{2}$ & $-1$ & $0$ &1\\ \hline
$\tan\alpha$ &$0$ & $\dfrac{\sqrt{3}}{3}$ & $1$ & $\sqrt{3}$ & $--$ & $-\sqrt{3}$
& $-1$ & $-\dfrac{\sqrt{3}}{3}$ & $0$ & $--$ & $0$
\\ \hline
\end{tabu}
}
\\
\large\heiti\textcolor{white}{8.两角和与差的正弦、余弦和正切}&
{\tabureset \tabulinesep=1ex
\begin{tabu}{>{\large}X>{\large}X}
$\mathbf{S}_{(\alpha+\beta)}:\sin(\alpha+\beta)=\sin\alpha\cos\beta +\cos\alpha\sin\beta$ & $\mathbf{S}_{(\alpha-\beta)}:\sin(\alpha+\beta)=\sin\alpha\cos\beta -\cos\alpha\sin\beta$ \\
$\mathbf{C}_{(\alpha+\beta)}:\cos(\alpha+\beta)=\cos\alpha\cos\beta -\sin\alpha\sin\beta$ & $\mathbf{C}_{(\alpha-\beta)}:\cos(\alpha+\beta)=\cos\alpha\cos\beta +\sin\alpha\sin\beta$ \\
$\mathbf{T}_{(\alpha+\beta)}:\tan(\alpha+\beta) = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ &
$\mathbf{T}_{(\alpha-\beta)}:\tan(\alpha+\beta) = \dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$\\
\end{tabu}
}
\\
\large\heiti\textcolor{white}{9.辅助角公式}&
\begin{minipage}{6cm}
\begin{eqnarray*}
a\sin x + b\cos x &=&\sqrt{a^2+b^2}\left( \dfrac{a}{\sqrt{a^2+b^2}}+\dfrac{b}{\sqrt{a^2+b^2}}\right)\\
&=&\sqrt{a^2+b^2}(\sin x \cdot \cos \varphi +\cos x \cdot \sin \varphi ) \\
&=&\sqrt{a^2+b^2}\cdot \sin(x+\varphi)\left( \tan\varphi = \dfrac{b}{a}\right)\\
\end{eqnarray*}
\end{minipage}
\\
\large\heiti\textcolor{white}{10.二倍角公式}&
\begin{description}
\item[(1) ] 此公式正用是关键
\begin{description}
\item[$S_{2a}$:] $\sin 2a = 2 \sin a \cos a$
\item[$C_{2a}$:] $\cos 2a = \cos^2 a - \sin^2 a = 2\cos^2 a -1 = 1-2\sin^2 a $
\item[$T_{2a}$:] $\tan 2a = \dfrac{2\tan a}{1-\tan^2 a}$
\end{description}
\item[(2) 降次公式] 此公式逆用是关键
\begin{description}
\item[] $\sin^2 a = \dfrac{1-\cos 2a}{2}$
\item[] $\cos^2 a = \dfrac{1+\cos 2a}{2}$
\end{description}
\end{description}
\\
\large\heiti\textcolor{white}{11.解三角形}&
\begin{description}
\item[(1)三角形面积公式：] $\textbf{S}_\bigtriangleup =\dfrac{1}{2}ab\sin C = \dfrac{1}{2}ac\sin B = \dfrac{1}{2}bc\sin A $
\item[(2)正弦定理：] $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2\textbf{R}$ \textbf{用角表示边：}$a=2\textbf{R}\sin A$\quad$b=2\textbf{R}\sin B$\quad $c=2\textbf{R}\sin C$
\item[(3)余弦定理：]
$a^2=b^2+c^2-2bc\cos A$ \quad
$b^2=a^2+c^2-2ac\cos B$ \quad
$c^2=a^2+b^2-2ab\cos C$ \quad
\item[\qquad 求角：\qquad]
$\cos A = \dfrac{b^2+c^2-a^2}{2bc}$\quad
$\cos B = \dfrac{a^2+c^2-b^2}{2ac}$\quad
$\cos C = \dfrac{a^2+b^2-c^2}{2ab}$\quad
\end{description}
\end{tabu}
\end{table}
\begin{table}[t]
\centering
\tabulinesep=1ex
\taburulecolor |mypink|{VioletRed}
\arrayrulewidth = 0.1mm \doublerulesep=5mm
\everyrow{\hline \hline }
\begin{tabu} to \linewidth{>{\columncolor{VioletRed}}X[-1,c,m]|X[9,l,m]|}
\large\heiti\textcolor{white}{12.几种常见函数的导数}&
\begin{tabular}{p{6cm}p{8cm}}
(1)\quad$\mathbf{C}'=0$ (C为常数)&
(5)\quad$(\;\ln x \;)'=\dfrac{1}{x}$ \\
(2)\quad$(\,x^\alpha\,)' = \alpha x^{\alpha-1}\ (\; \alpha \in \mathbf{Q}^\ast\;)$ &
(6)\quad$(\,\log_a x\,)' = \dfrac{1}{x\ln a}(\,a>0\;\mbox{且}\;a\neq 1 \;) $ \\
(3)\quad$(\,\sin x \,)' = \cos x$ &
(7)\quad$(\,e^x \,)' = e^x$\\
(4)\quad$(\,\cos x \,)' = -\sin x$ &
(8)\quad$(\,a^x \,)' = a^x \ln a \;(\, a>0 \,)$\\
\end{tabular}
\\
\large\heiti\textcolor{white}{13.三角函数}&
{\tabureset \tabulinesep=2ex
\taburulecolor |mypink|{VioletRed}
\arrayrulewidth = 0.1mm
\begin{tabu}{X[1.2,c,m]|X[c]|X[0.8,c]|X[c]|X[c]|X[3c]|X[3c]|}
\rowcolor{mypink}\rowfont{\heiti}
函数 & 定义域 & 值域 & 周期性 & 奇偶性 & 递增区间 & 递减区间 \\ \hline
$y=\sin x $ & $x \in \mathbf{R}$ & $[-1,1]$ & $T = 2\pi$ & 奇函数 &
$[-\dfrac{\pi}{2}+2k\pi ,\; \dfrac{\pi}{2}+2k\pi ]$ $\;(k \in \mathbf{Z})$ &
$[\dfrac{\pi}{2}+2k\pi ,\; \dfrac{3\pi}{2}+2k\pi ]$ $\;(k \in \mathbf{Z})$ \\ \hline
$y=\cos x $ & $x \in \mathbf{R}$ & $[-1,1]$ & $T = 2\pi$ & 偶函数 &
$[-\pi+2k\pi ,\; 2k\pi ]$ $\;(k \in \mathbf{Z})$ &
$[2k\pi ,\; \pi+2k\pi ]$ $\qquad(k \in \mathbf{Z})$ \\ \hline
\end{tabu}
}
{\tabureset \tabulinesep=1ex
\taburulecolor |mypink|{VioletRed}
\arrayrulewidth = 0.1mm
\begin{tabu}{X[3.5c,m,p]|X[1.1c]|X[1.5c]|X[0.5c]|X[1.5c]|X[2.5c]|X[1.2c]|X[0.8c]|X[1.2c]|}
\rowcolor{mypink}\rowfont{\heiti}
函数 & 定义域 & 值域 & 振幅 & 周期 & 频率 & 相位 & 初相 &图像 \\ \hline
$y=A\sin (\;\omega x + \varphi \;)$ & $x \in \mathbf{R}$ & $[\ -A,\ A\ ]$ &
A & $T=\dfrac{2\pi}{\abs{\omega}}$ & $\mathbf{f} = \dfrac{1}{T} = \dfrac{\abs{\omega}}{2\pi}$ & $\omega x + \varphi$ & $\varphi$ & 图像法
\\ \hline
\end{tabu}
}
\end{tabu}
\end{table}
\end{document}
