定义
在$Rt\triangle ABC$中，如下有六个三角函数的定义：
正弦：
$$\sin A = \frac{a}{c}$$
级数表示：$\sin (x)==\sum_{k=0}^{\infty} \frac{(-1)^{k} x^{1+2k}}{(1+2k)!}$
余弦：
$$\cos A = \frac{b}{c}$$
级数表示：$\cos (x)=\sum_{k=0}^{\infty} \frac{(-1)^{k} x^{2 k}}{(2 k) !}$
正切：
$$\tan A = \frac{a}{b}$$
级数表示：$\tan (x)=i+2 i \sum_{k=1}^{\infty}(-1)^{k} q^{2 k} \color{gray}\textrm{ for } q=e^{i x}$
余切：
$$\cot A = \frac{b}{a}$$
级数表示：$\cot (x)=-i-2 i \sum_{k=1}^{\infty} q^{2 k} \color{gray}\text { for } q=e^{i x}$
正割：
$$\sec A = \frac{c}{b}$$
级数表示：$\sec (x)=-2 \sum_{k=1}^{\infty}(-1)^{k} q^{-1+2 k} \color{gray}\text { for } q=e^{i x}$
余割：
$$\csc A = \frac{c}{a}$$
级数表示：$\csc (x)=-2 i \sum_{k=1}^{\infty} q^{-1+2 k} \color{gray}\text { for } q=e^{i x}$
诱导公式
关系 & 定理 & 公式
倒数关系
$$\cos \alpha \cdot \sec \alpha = 1$$
$$\sin \alpha \cdot \csc \alpha = 1$$
$$\tan \alpha \cdot \cot \alpha = 1$$
平方关系
$$1 + \tan ^ 2 \alpha = \sec ^ 2 \alpha$$
$$1 + \cot ^ 2 \alpha = \csc ^ 2 \alpha$$
$$\sin^2 \alpha + cos ^ 2 \alpha = 1$$
商的关系
$$\frac{\sin \alpha}{\cos \alpha} = \frac{\sec \alpha}{\csc \alpha} = \tan \alpha$$
$$\frac{\cos \alpha}{\sin \alpha} = \frac{\csc \alpha}{\sec \alpha} = \cot \alpha$$
正弦定理
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R = D$$
$R$ 为三角形外切圆半径，$D$ 为三角形外切圆直径。
证明：
如图在 $\triangle ABC$ 中可得 $\sin A = \frac{h}{b}$ 和 $\sin B = \frac{h}{a}$ 。
$$\therefore h = \sin A \times b, h = \sin B \times a \\\ \therefore \sin A \times b = \sin B \times a \\\ \therefore \frac{\sin A}{a} = \frac{\sin B}{b} \\\ \therefore \frac{a}{\sin A} = \frac{b}{\sin B} \\\ \textrm{同理：} \frac{a}{\sin A} = \frac{c}{\sin C} \\\ \therefore \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
如图， $\triangle CDB$ 中线段 $CD$ 经过圆心，所以 $\angle CBD = 90 ^ \circ$ ， $CD = 2R$。
$$\therefore \sin A = \sin D = \frac{CB}{CD} = \frac{a}{2R} \\\ \therefore \frac{a}{\sin A} = 2R \\\ \textrm{同理：} \frac{b}{\sin B} = 2R, \frac{c}{\sin C} = 2R \\\ \therefore \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R = D$$
余弦定理
$$a ^ 2 = b ^ 2 + c ^ 2 - 2bc\cos A, b ^ 2 = a ^ 2 + c ^ 2 - 2ac\cos B, c ^ 2 = a ^ 2 + b ^ 2 - 2ab\cos C \\\ \rm{或} \\\ \cos A = \frac{b ^ 2 + c ^ 2 - a ^ 2}{2bc}, \cos B = \frac{a ^ 2 + c ^ 2 - b ^ 2}{2ac}, \cos C = \frac{a ^ 2 + b ^ 2 - c ^ 2}{2ab}$$
证明：
如图，在 $\triangle ABC$ 中，令$\vec{AB} = \vec{c}, \vec{CB} = \vec{a}, \vec{CA} = \vec{b}$。
$$\therefore \vec{c} = \vec{AB} = \vec{CB} - \vec{CA} = \vec{a} - \vec{b} \\\ \therefore (\vec{c}) ^ 2 = (\vec{a} - \vec{b}) ^ 2 = \vec{a} ^ 2 + \vec{b} ^ 2 - 2 \vec{a} \cdot \vec{b} \\\ \therefore |\vec{c}| ^ 2 = |\vec{a}| ^ 2 + |\vec{b}| ^ 2 - 2 |\vec{a}| \cdot |\vec{b}| \cdot \cos C \\\ \therefore c ^ 2 = a ^ 2 + b ^ 2 - 2ab\cos C \\\ 同理：\cos A = \frac{b ^ 2 + c ^ 2 - a ^ 2}{2bc}, \cos B = \frac{a ^ 2 + c ^ 2 - b ^ 2}{2ac}$$
和角公式
$$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$
$$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$$
$$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$$
差角公式
$$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$
$$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$$
$$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$$
和差化积
$$\sin \alpha+\sin \beta=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$$
$$\sin \alpha-\sin \beta=2 \sin \left(\frac{\alpha-\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right)$$
$$\cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$$
$$\cos \alpha-\cos \beta=-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)$$
积化和差
$$\cos \alpha \sin \beta=\frac{1}{2}[\sin (\alpha+\beta)-\sin (\alpha-\beta)]$$
$$\sin \alpha \cos \beta=\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)]$$
$$\cos \alpha \cos \beta=\frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)]$$
$$\sin \alpha \sin \beta=-\frac{1}{2}[\cos (\alpha+\beta)-\cos (\alpha-\beta)]$$
倍角公式
$$\sin 2 \alpha = 2 \sin \alpha \cos \alpha$$
$$\cos 2 \alpha = \cos ^ 2 \alpha - \sin ^ 2 \alpha$$
$$\tan 2 \alpha = \frac{2 \tan \alpha}{1 - \tan ^ 2 \alpha}$$
$$\cot 2 \alpha=\frac{\cot ^{2} \alpha-1}{2 \cot \alpha}$$
$$\sec 2 \alpha=\frac{\sec ^{2} \alpha}{1-\tan ^{2} \alpha}$$
$$\csc 2 \alpha=\frac{1}{2} \sec \alpha \csc \alpha$$
半角公式
$$\sin \left(\frac{\alpha}{2}\right) = \sqrt{\frac{1-\cos \alpha}{2}}$$
$$\cos \left(\frac{\alpha}{2}\right) = \sqrt{\frac{1+\cos \alpha}{2}}$$
$$\tan \left(\frac{\alpha}{2}\right) = \csc \alpha-\cot \alpha$$
$$\cot \left(\frac{\alpha}{2}\right) = \csc \alpha+\cot \alpha$$
$$\sec \left(\frac{\alpha}{2}\right) = \sqrt{\frac{2 \sec \alpha}{\sec \alpha+1}}$$
$$\csc \left(\frac{\alpha}{2}\right) = \sqrt{\frac{2 \sec \alpha}{\sec \alpha-1}}$$
Attachment
常用三角函数值对照表：
| 角$\alpha$ | 弧度 | $\sin$值 | $\cos$值 | $\tan$值 |
|---|---|---|---|---|
| $0^\circ$ | $0$ | $0$ | $1$ | $0$ |
| $15^\circ$ | $\frac{\pi}{12}$ | $\frac{\sqrt{6} - \sqrt{2}}{4}$ | $\frac{\sqrt{6} + \sqrt{2}}{4}$ | $2 - \sqrt{3}$ |
| $22.5^\circ$ | $\frac{\pi}{8}$ | $\frac{\sqrt{2 - \sqrt{2}}}{2}$ | $\frac{\sqrt{2 + \sqrt{2}}}{2}$ | $-1 + \sqrt{2}$ |
| $30^\circ$ | $\frac{\pi}{6}$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ |
| $45^\circ$ | $\frac{\pi}{4}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ |
| $60^\circ$ | $\frac{\pi}{3}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |
| $75^\circ$ | $\frac{5\pi}{12}$ | $\frac{\sqrt{6} + \sqrt{2}}{4}$ | $\frac{\sqrt{6} - \sqrt{2}}{4}$ | $2 + \sqrt{3}$ |
| $90^\circ$ | $\frac{\pi}{2}$ | $1$ | $0$ | $\rm{无}$ |
| $120^\circ$ | $\frac{2\pi}{3}$ | $\frac{\sqrt{3}}{2}$ | $-\frac{1}{2}$ | $-\sqrt{3}$ |
| $135^\circ$ | $\frac{3\pi}{4}$ | $\frac{\sqrt{2}}{2}$ | $-\frac{\sqrt{2}}{2}$ | $-1$ |
| $150^\circ$ | $\frac{5\pi}{6}$ | $\frac{1}{2}$ | $-\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ |
| $180^\circ$ | $\pi$ | $0$ | $-1$ | $0$ |
| $270^\circ$ | $\frac{3\pi}{2}$ | $-1$ | $0$ | $\rm{无}$ |
| $360^\circ$ | $2\pi$ | $0$ | $1$ | $0$ |
Article Author: XiaoHuang