高中数学：函数 知识汇总

关系式：\( y = ax^2 + bx + c \quad (a \neq 0) \)
配方全过程推导： $$ \begin{align*} y &= a\left(x^2 + \frac{b}{a}x\right) + c \\ &= a\left[x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 \right] + c \\ &= a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a} \\ &= a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a} \end{align*} $$ 顶点坐标：\( (h, k) = \left( -\frac{b}{2a}, \frac{4ac - b^2}{4a} \right) \)

二次方程根公式： \( x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
韦达定理 1:1 代数证明： $$ \begin{align*} x_1 + x_2 &= \frac{-b + \sqrt{\Delta}}{2a} + \frac{-b - \sqrt{\Delta}}{2a} = \frac{-2b}{2a} = -\frac{b}{a} \\ x_1 x_2 &= \frac{-b + \sqrt{\Delta}}{2a} \cdot \frac{-b - \sqrt{\Delta}}{2a} = \frac{(-b)^2 - (\sqrt{\Delta})^2}{4a^2} = \frac{b^2 - (b^2 - 4ac)}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a} \end{align*} $$

定义：如果 \( a^y = x \ (a > 0, a \neq 1) \)，那么数 \( y \) 叫做以 \( a \) 为底 \( x \) 的对数，记作 \( y = \log_a x \)。

指对数运算法则 1:1 推导： (设 \( a^m = M, a^n = N \))
$$ \begin{align*} 1. &\quad \because a^m \cdot a^n = a^{m+n} \implies \log_a(M \cdot N) = \log_a M + \log_a N \\ 2. &\quad \because a^m \div a^n = a^{m-n} \implies \log_a \frac{M}{N} = \log_a M - \log_a N \\ 3. &\quad \because a^0 = 1 \implies \log_a 1 = 0, \quad a^1 = a \implies \log_a a = 1 \\ 4. &\quad \because (a^m)^{\frac{n}{m}} = a^n \implies \log_M N = \frac{\log_a N}{\log_a M} \quad (\text{换底公式}) \\ 5. &\quad a^{\log_a M} = M \iff \log_a a^m = m \quad (\text{恒等关系}) \end{align*} $$

1. 坐标系定义 (编号对照)：
设角 $\theta$ 终边上一点 $P(x, y)$，距离 $r = \sqrt{x^2 + y^2}$，则： $$ \begin{align*} \sin \theta &= \frac{y}{r} \quad \dots (1) & \cos \theta &= \frac{x}{r} \quad \dots (2) & \tan \theta &= \frac{y}{x} \quad \dots (3) \\ \cot \theta &= \frac{x}{y} \quad \dots (4) & \sec \theta &= \frac{r}{x} \quad \dots (5) & \csc \theta &= \frac{r}{y} \quad \dots (6) \end{align*} $$ 2. 视觉模型与几何意义：
弦切割分类：
正系：$\sin$ (正弦), $\tan$ (正切), $\sec$ (正割)   |   余系：$\cos$ (余弦), $\cot$ (余切), $\csc$ (余割)

3. 六边形关系全量清单及对照证明 (基于定义编号)： A. 倒数关系（对角线）
$$ \begin{align*} \sin \theta \cdot \csc \theta = 1 &\quad \Longleftarrow \quad (1) \cdot (6) = \frac{y}{r} \cdot \frac{r}{y} = 1 \\ \cos \theta \cdot \sec \theta = 1 &\quad \Longleftarrow \quad (2) \cdot (5) = \frac{x}{r} \cdot \frac{r}{x} = 1 \\ \tan \theta \cdot \cot \theta = 1 &\quad \Longleftarrow \quad (3) \cdot (4) = \frac{y}{x} \cdot \frac{x}{y} = 1 \end{align*} $$ B. 乘积关系（邻角相乘）
$$ \begin{align*} \sin \theta = \tan \theta \cdot \cos \theta &\quad \Longleftarrow \quad (3) \cdot (2) = \frac{y}{x} \cdot \frac{x}{r} = \frac{y}{r} = (1) \\ \cos \theta = \sin \theta \cdot \cot \theta &\quad \Longleftarrow \quad (1) \cdot (4) = \frac{y}{r} \cdot \frac{x}{y} = \frac{x}{r} = (2) \\ \tan \theta = \sin \theta \cdot \sec \theta &\quad \Longleftarrow \quad (1) \cdot (5) = \frac{y}{r} \cdot \frac{r}{x} = \frac{y}{x} = (3) \\ \cot \theta = \cos \theta \cdot \csc \theta &\quad \Longleftarrow \quad (2) \cdot (6) = \frac{x}{r} \cdot \frac{r}{y} = \frac{x}{y} = (4) \\ \csc \theta = \cot \theta \cdot \sec \theta &\quad \Longleftarrow \quad (4) \cdot (5) = \frac{x}{y} \cdot \frac{r}{x} = \frac{r}{y} = (6) \\ \sec \theta = \csc \theta \cdot \tan \theta &\quad \Longleftarrow \quad (6) \cdot (3) = \frac{r}{y} \cdot \frac{y}{x} = \frac{r}{x} = (5) \end{align*} $$ C. 商数关系（顺/逆时针）
$$ \begin{align*} \tan \theta = \frac{\sin \theta}{\cos \theta} &\quad \Longleftarrow \quad \frac{(1)}{(2)} = \frac{y/r}{x/r} = \frac{y}{x} = (3) \\ \sin \theta = \frac{\cos \theta}{\cot \theta} &\quad \Longleftarrow \quad \frac{(2)}{(4)} = \frac{x/r}{x/y} = \frac{y}{r} = (1) \end{align*} $$ \( \dots \) 其余同理，可依此类推

D. 平方关系（阴影倒三角）
$$ \begin{align*} \sin^2 \theta + \cos^2 \theta = 1 &\quad \Longleftarrow \quad \frac{y^2}{r^2} + \frac{x^2}{r^2} = \frac{x^2+y^2}{r^2} = \frac{r^2}{r^2} = 1 \\ \tan^2 \theta + 1 = \sec^2 \theta &\quad \Longleftarrow \quad \frac{y^2}{x^2} + 1 = \frac{y^2+x^2}{x^2} = \frac{r^2}{x^2} = (5)^2 \\ 1 + \cot^2 \theta = \csc^2 \theta &\quad \Longleftarrow \quad 1 + \frac{x^2}{y^2} = \frac{y^2+x^2}{y^2} = \frac{r^2}{y^2} = (6)^2 \end{align*} $$

1. 和差角基础公式 (六组独立展开)： $$ \begin{align*} \sin(\alpha + \beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \quad \dots (1) \\ \sin(\alpha - \beta) &= \sin \alpha \cos \beta - \cos \alpha \sin \beta \quad \dots (2) \\ \cos(\alpha + \beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \quad \dots (3) \\ \cos(\alpha - \beta) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \quad \dots (4) \\ \tan(\alpha + \beta) &= \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \quad \dots (5) \\ \tan(\alpha - \beta) &= \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \quad \dots (6) \end{align*} $$ 2. 积化和差推导 (由 (1)(2)(3)(4) 加减得出)：
• (1) + (2) $ \implies \sin(\alpha + \beta) + \sin(\alpha - \beta) = 2\sin \alpha \cos \beta \implies \sin \alpha \cos \beta = \frac{1}{2}[\sin(\alpha + \beta) + \sin(\alpha - \beta)] $
• (1) - (2) $ \implies \sin(\alpha + \beta) - \sin(\alpha - \beta) = 2\cos \alpha \sin \beta \implies \cos \alpha \sin \beta = \frac{1}{2}[\sin(\alpha + \beta) - \sin(\alpha - \beta)] $
• (3) + (4) $ \implies \cos(\alpha + \beta) + \cos(\alpha - \beta) = 2\cos \alpha \cos \beta \implies \cos \alpha \cos \beta = \frac{1}{2}[\cos(\alpha + \beta) + \cos(\alpha - \beta)] $
• (4) - (3) $ \implies \cos(\alpha - \beta) - \cos(\alpha + \beta) = 2\sin \alpha \sin \beta \implies \sin \alpha \sin \beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)] $

3. 和差化积推导 (令 $ \alpha+\beta=A, \alpha-\beta=B $ 逆推)： $$ \begin{align*} \sin A + \sin B &= 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \\ \sin A - \sin B &= 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} \\ \cos A + \cos B &= 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} \\ \cos A - \cos B &= -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} \end{align*} $$

4. 倍角公式 (令 (1)(3)(5) 中 $ \beta = \alpha $ 推出)： $$ \begin{align*} \sin 2\alpha &= \sin(\alpha + \alpha) = \sin \alpha \cos \alpha + \cos \alpha \sin \alpha = 2\sin \alpha \cos \alpha \\ \cos 2\alpha &= \cos(\alpha + \alpha) = \cos \alpha \cos \alpha - \sin \alpha \sin \alpha = \cos^2 \alpha - \sin^2 \alpha \\ \text{由 } \sin^2 \alpha + \cos^2 \alpha &= 1 \implies \cos 2\alpha = 2\cos^2 \alpha - 1 = 1 - 2\sin^2 \alpha \\ \tan 2\alpha &= \tan(\alpha + \alpha) = \frac{\tan \alpha + \tan \alpha}{1 - \tan \alpha \tan \alpha} = \frac{2\tan \alpha}{1 - \tan^2 \alpha} \end{align*} $$ 5. 半角公式 (由倍角 $ \cos 2\alpha $ 逆推)：
由 $ \cos 2\alpha = 1 - 2\sin^2 \alpha \implies \sin^2 \alpha = \frac{1 - \cos 2\alpha}{2} $。令 $ 2\alpha = \theta \implies \sin^2 \frac{\theta}{2} = \frac{1 - \cos \theta}{2} $
由 $ \cos 2\alpha = 2\cos^2 \alpha - 1 \implies \cos^2 \alpha = \frac{1 + \cos 2\alpha}{2} $。令 $ 2\alpha = \theta \implies \cos^2 \frac{\theta}{2} = \frac{1 + \cos \theta}{2} $
$ \tan^2 \frac{\theta}{2} = \frac{\sin^2(\theta/2)}{\cos^2(\theta/2)} = \frac{1 - \cos \theta}{1 + \cos \theta} $，$ \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta} $

6. 万能公式 (通过“除以 1”技巧推出)：
以 $ \sin \alpha $ 为例，令 $ \alpha = 2 \cdot \frac{\alpha}{2} $，利用倍角公式： $$ \begin{align*} \sin \alpha &= \frac{2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} \quad (\text{同除以 } \cos^2 \frac{\alpha}{2}) = \frac{2\tan \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} \\ \cos \alpha &= \frac{\cos^2 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2}}{\cos^2 \frac{\alpha}{2} + \sin^2 \frac{\alpha}{2}} \quad (\text{同除以 } \cos^2 \frac{\alpha}{2}) = \frac{1 - \tan^2 \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} \end{align*} $$ 设 $ t = \tan \frac{\alpha}{2} $，则 $ \sin \alpha = \frac{2t}{1+t^2}, \cos \alpha = \frac{1-t^2}{1+t^2}, \tan \alpha = \frac{2t}{1-t^2} $

7. 辅助角公式 (Harmonic Addition)：
\( a \sin x + b \cos x = \sqrt{a^2+b^2} \sin(x+\phi) \)
其中 \( \tan \phi = b/a \) (符号看 \( a,b \) 所在象限)。

定义：\( \sinh x = \frac{e^x - e^{-x}}{2}, \quad \cosh x = \frac{e^x + e^{-x}}{2} \)
推导：\( \cosh^2 x - \sinh^2 x = 1 \)
$$ \begin{align*} \cosh^2 x - \sinh^2 x &= \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2 \\ &= \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{4}{4} = 1 \end{align*} $$ 与 \( e \) 的关系：\( e^x = \cosh x + \sinh x, \quad e^{-x} = \cosh x - \sinh x \)

1. 图像平移规律 (Translation Rule)
坐标反解证明：
设 \( P(x, y) \) 在 \( y = f(x) \) 上，平移向量 \( \vec{v} = (h, k) \) 得新点 \( P'(X, Y) \)： $$ \begin{cases} X = x + h \\ Y = y + k \end{cases} \xrightarrow{\text{反解}} \begin{cases} x = X - h \\ y = Y - k \end{cases} \xrightarrow{\text{代入}} Y - k = f(X - h) $$ 结论： \( Y = f(X - h) + k \)   (左加右减，上加下减)

2. 图像伸缩规律 (Scaling Rule)
坐标反解证明：
设原点 \( (x, y) \) 满足 \( y = f(x) \)，进行横向 \( \omega \) 倍、纵向 \( A \) 倍伸缩得新点 \( (X, Y) \)： $$ \begin{cases} X = \omega x \\ Y = A y \end{cases} \xrightarrow{\text{反解}} \begin{cases} x = X/\omega \\ y = Y/A \end{cases} \xrightarrow{\text{代入}} \frac{Y}{A} = f\left(\frac{X}{\omega}\right) $$ 结论： \( Y = A \cdot f\left(\frac{X}{\omega}\right) \)

A. 导数定义：
\( [f(x)]' = \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} = f'(x) \)

B. 基本初等函数导数公式及推导：

1. 常数函数 \( (C)' = 0 \)：
$$ (C)' = \lim_{\Delta x \to 0} \frac{C - C}{\Delta x} = 0 $$
2. 幂函数 \( (x^n)' = nx^{n-1} \)：
$$ \begin{align*} (x^n)' &= \lim_{\Delta x \to 0} \frac{(x+\Delta x)^n - x^n}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{x^n + nx^{n-1}\Delta x + \cdots + (\Delta x)^n - x^n}{\Delta x} \\ &= \lim_{\Delta x \to 0} [nx^{n-1} + \cdots + (\Delta x)^{n-1}] = nx^{n-1} \end{align*} $$ 3. 指数函数 \( (a^x)' = a^x \ln a \)，特殊地 \( (e^x)' = e^x \)：
令 \( y = a^x \)，则 \( x = \log_a y \)，由反函数求导法则：
\( y' = \frac{1}{x'} = \frac{1}{\frac{1}{y \ln a}} = y \ln a = a^x \ln a \)

4. 对数函数 \( (\log_a x)' = \frac{1}{x \ln a} \)，特殊地 \( (\ln x)' = \frac{1}{x} \)：
$$ \begin{align*} (\log_a x)' &= \lim_{\Delta x \to 0} \frac{\log_a(x+\Delta x) - \log_a x}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \log_a\left(1+\frac{\Delta x}{x}\right) \\ &= \lim_{\Delta x \to 0} \frac{1}{x} \log_a\left(1+\frac{\Delta x}{x}\right)^{\frac{x}{\Delta x}} \\ &= \frac{1}{x} \log_a e = \frac{1}{x \ln a} \end{align*} $$ 5. 三角函数导数：
\( (\sin x)' = \cos x \)：
$$ \begin{align*} (\sin x)' &= \lim_{\Delta x \to 0} \frac{\sin(x+\Delta x) - \sin x}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{2\cos\left(x+\frac{\Delta x}{2}\right)\sin\left(\frac{\Delta x}{2}\right)}{\Delta x} \\ &= \lim_{\Delta x \to 0} \cos\left(x+\frac{\Delta x}{2}\right) \cdot \frac{\sin\left(\frac{\Delta x}{2}\right)}{\frac{\Delta x}{2}} = \cos x \end{align*} $$ \( (\cos x)' = -\sin x \)（同理可证）
\( (\tan x)' = \left(\frac{\sin x}{\cos x}\right)' = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x \)