三角函数基本关系和公式

July 09, 2018

《数学手册》

作者	《数学手册》编写组编
ISBN	978-7-04-003401-1
CIP	（2011）第240483号
出版社	高等教育出版社
版次	1979年5月第1版 2016年10月第24次印刷

三角函数

常见的三角函数

三角函数基本关系和公式

三角函数基本关系和公式

诱导公式

三角函数的诱导公式表
函数角	$\sin$	$\cos$	$\tan$	$\cot$	$\sec$	$\csc$
$-\alpha$	$-\sin\alpha$	$\cos\alpha$	$-\tan\alpha$	$-\cot\alpha$	$\sec\alpha$	$-\csc\alpha$
$\frac \pi 2 \pm \alpha$	$\cos\alpha$	$\mp\sin\alpha$	$\mp\cot\alpha$	$\mp\tan\alpha$	$\mp\csc\alpha$	$\sec\alpha$
$\pi \pm \alpha$	$\mp\sin\alpha$	$-\cos\alpha$	$\pm\tan\alpha$	$\pm\cot\alpha$	$-\sec\alpha$	$\mp\csc\alpha$
$\frac {3\pi} 2 \pm \alpha$	$-\cos\alpha$	$\pm\sin\alpha$	$\mp\cot\alpha$	$\mp\tan\alpha$	$\pm\csc\alpha$	$-\sec\alpha$
$2\pi \pm \alpha$	$\pm\sin\alpha$	$\cos\alpha$	$\pm\tan\alpha$	$\pm\cot\alpha$	$\sec\alpha$	$\pm\csc\alpha$
$n\pi \pm \alpha$	$\pm(-1)^n\sin\alpha$	$(-1)^n\cos\alpha$	$\pm\tan\alpha$	$\pm\cot\alpha$	$(-1)^n\sec\alpha$	$\pm(-1)^n\csc\alpha$

表中 $n$ 为整数。

基本关系

$\sin ^2 \alpha + \cos ^2 \alpha = 1$ $\tan \alpha = \frac {\sin \alpha}{\cos \alpha}$ $\cot \alpha = \frac {\cos \alpha}{\sin \alpha}$ $\tan \alpha \cdot \cot \alpha = 1$ $\sin \alpha \cdot \csc \alpha = 1$ $\cos \alpha \cdot \sec \alpha = 1$ $\sec ^2 \alpha - \tan ^2 \alpha = 1$ $\csc ^2 \alpha - \cot ^2 \alpha = 1$

三角函数的相互关系表

	$\sin \alpha = x$	$\cos \alpha = x$	$\tan \alpha = x$	$\cot \alpha = x$	$\sec \alpha = x$	$\csc \alpha = x$
$\sin \alpha = $	$x$	$\pm \sqrt{1 - x^2}$	$\pm \frac{x}{\sqrt{1 + x^2}}$	$\pm \frac{1}{\sqrt{1 + x^2}}$	$\pm \frac{\sqrt{x^2 - 1}}{x}$	$\frac{1}{x}$
$\cos \alpha = $	$\pm \sqrt{1 - x^2}$	$x$	$\pm \frac{1}{\sqrt{1 + x^2}}$	$\pm \frac{x}{\sqrt{1 + x^2}}$	$\frac{1}{x}$	$\pm \frac{\sqrt{x^2 - 1}}{x}$
$\tan \alpha = $	$\pm \frac{x}{\sqrt{1 - x^2}}$	$\pm \frac{\sqrt{1 - x^2}}{x}$	$x$	$\frac{1}{x}$	$\pm \sqrt{x^2 - 1}$	$\pm \frac{1}{\sqrt{x^2 - 1}}$
$\cot \alpha = $	$\pm \frac{\sqrt{1 - x^2}}{x}$	$\pm \frac{x}{\sqrt{1 - x^2}}$	$\frac{1}{x}$	$x$	$\pm \frac{1}{\sqrt{x^2 - 1}}$	$\pm \sqrt{x^2 - 1}$
$\sec \alpha = $	$\pm \frac{1}{\sqrt{1 - x^2}}$	$\frac{1}{x}$	$\pm \sqrt{1 + x^2}$	$\pm \frac{\sqrt{1 + x^2}}{x}$	$x$	$\pm \frac{x}{\sqrt{x^2 - 1}}$
$\csc \alpha = $	$\frac{1}{x}$	$\pm \frac{1}{\sqrt{1 - x^2}}$	$\pm \frac{\sqrt{1 + x^2}}{x}$	$\pm \sqrt{1 + x^2}$	$\pm \frac{x}{\sqrt{x^2 - 1}}$	$x$

例如，若 $\sin \alpha = x$，则 $\cos \alpha = \pm \sqrt{1 - x^2}$。

加法公式

$\sin(\alpha \pm \beta) = \sin\alpha \cos\beta \pm \cos\alpha \sin\beta$ $\cos(\alpha \pm \beta) = \cos\alpha \cos\beta \mp \sin\alpha \sin\beta$ $\tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha \tan\beta}$ $\cot(\alpha \pm \beta) = \frac{\cot\alpha \cot\beta \mp 1}{\cot\beta \pm \cot\alpha}$ $\begin{eqnarray} \sin(\alpha + \beta + \gamma) &=& \sin\alpha \cos\beta \cos\gamma + \cos\alpha \sin\beta \cos\gamma + \cos\alpha \cos\beta \sin\gamma - \sin\alpha \sin\beta\sin\gamma \\ &=& \sin\alpha \sin\beta \sin\gamma(\cot\beta \cot \gamma + \cot\gamma \cot\alpha + \cot\alpha \cot\beta - 1) \\ &=& \cos\alpha \cos\beta \cos\gamma(\tan\alpha + \tan\beta + \tan\gamma - \tan\alpha \tan\beta \tan \gamma) \end{eqnarray}$ $\begin{eqnarray} \cos(\alpha + \beta + \gamma) &=& \cos\alpha \cos\beta \cos\gamma - \cos\alpha \sin\beta \sin\gamma - \sin\alpha \cos\beta \sin\gamma - \sin\alpha \sin\beta \cos\gamma \\ &=& \cos\alpha \cos\beta \cos\gamma (1 - \tan\beta \tan\gamma - \tan\gamma \tan\alpha - \tan\alpha \tan\beta) \\ &=& - \sin\alpha \sin\beta \sin\gamma (\cot\alpha + \cot\beta + \cot\gamma - \cot\alpha \cot\beta \cot\gamma) \end{eqnarray}$ $\tan(\alpha + \beta + \gamma) = \frac{\tan\alpha + \tan\beta + \tan\gamma - \tan\alpha \tan\beta \tan\gamma} {1 - \tan\beta \tan\gamma - \tan\gamma \tan\alpha - \tan\alpha \tan\beta}$ $\cot(\alpha + \beta + \gamma) = \frac{\cot\alpha \cot\beta \cot\gamma - \cot\alpha - \cot\beta - \cot\gamma} {\cot\beta \cot\gamma + \cot\gamma \cot\alpha + \cot\alpha \cot\beta - 1}$

和差化积公式

$\sin\alpha + \sin\beta = 2 \sin\frac{\alpha + \beta}{2} \cos\frac{\alpha - \beta}{2}$ $\sin\alpha - \sin\beta = 2 \cos\frac{\alpha + \beta}{2} \sin\frac{\alpha - \beta}{2}$ $\cos\alpha + \cos\beta = 2 \cos\frac{\alpha + \beta}{2} \cos\frac{\alpha - \beta}{2}$ $\cos\alpha - \cos\beta = -2 \sin\frac{\alpha + \beta}{2} \sin\frac{\alpha - \beta}{2}$ $\tan\alpha \pm \tan\beta = \frac{\sin(\alpha \pm \beta)}{\cos\alpha \cos\beta}$ $\cot\alpha \pm \cot\beta = \pm \frac{\sin(\alpha \pm \beta)}{sin\alpha \sin\beta}$ $\tan\alpha \pm \cot\beta = \pm \frac{\cos(\alpha \mp \beta)}{\cos\alpha \sin\beta}$

积化和差公式

$\sin\alpha \sin\beta = - \frac 1 2 [\cos(\alpha + \beta) - \cos(\alpha - \beta)]$ $\cos\alpha \cos\beta = \frac 1 2 [\cos(\alpha + \beta) + \cos(\alpha - \beta)]$ $\sin\alpha \cos\beta = \frac 1 2 [\sin(\alpha + \beta) + \sin(\alpha - \beta)]$ $\cos\alpha \sin\beta = \frac 1 2 [\sin(\alpha + \beta) - \sin(\alpha - \beta)]$

倍角公式

$\sin 2\alpha = 2 \sin\alpha \cos\alpha = \frac{2 \tan\alpha}{1 + \tan^2\alpha}$ $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = 2 \cos^2\alpha - 1 = 1 - 2 \sin^2\alpha = \frac{1 - \tan^2\alpha}{1 + \tan^2\alpha}$ $\tan 2\alpha = \frac{2 \tan\alpha}{1 - \tan^2\alpha}$ $\cot 2\alpha = \frac{\cot^2\alpha - 1}{2 \cot\alpha}$ $\sec 2\alpha = \frac{\sec^2\alpha}{1 - \tan^2\alpha} = \frac{\cot\alpha + \tan\alpha}{\cot\alpha - \tan\alpha}$ $\csc 2\alpha = \frac 1 2 \sec\alpha \csc\alpha = \frac 1 2 (\tan\alpha + \cot\alpha)$ $\sin 3\alpha = -4\sin^3\alpha + 3\sin\alpha$ $\cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha$ $\tan 3\alpha = \frac{3\tan\alpha - \tan^3\alpha}{1 - 3\tan^2\alpha}$ $\cot 3\alpha = \frac{\cot^3\alpha - 3\cot\alpha}{3\cot^2\alpha - 1}$ $\sin n\alpha = n\cos^{n - 1}\alpha \sin\alpha - C_n^3\cos^{n - 3}\alpha \sin^3\alpha + C_n^5\cos^{n - 5}\alpha \sin^5\alpha - \cdots$ $\cos n\alpha = \cos^n\alpha - C_n^2\cos^{n - 2}\alpha \sin^2\alpha + C_n^4\cos^{n - 4}\alpha \sin^4\alpha - C_n^6\cos^{n - 6}\alpha \sin^6\alpha + \cdots$

式中 $n$ 为正整数。

半角公式

下列公式中根号所取符号与等号左边的符号一致。

$\sin \frac \alpha 2 = \pm \sqrt{\frac{1 - \cos\alpha} 2}$ $\cos \frac \alpha 2 = \pm \sqrt{\frac{1 + \cos\alpha} 2}$ $\tan \frac \alpha 2 = \pm \sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}} = \frac{1 - \cos\alpha}{\sin\alpha} = \frac{\sin\alpha}{1 + \cos\alpha}$ $\cot \frac \alpha 2 = \pm \sqrt{\frac{1 + \cos\alpha}{1 - \cos\alpha}} = \frac{1 + \cos\alpha}{\sin\alpha} = \frac{\sin\alpha}{1 - \cos\alpha}$ $\sec \frac \alpha 2 = \pm \sqrt{\frac{2 \sec\alpha}{\sec\alpha + 1}}$ $\csc \frac \alpha 2 = \pm \sqrt{\frac{2 \sec\alpha}{\sec\alpha - 1}}$

降幂公式

$\sin^2\alpha = \frac 1 2 (1 - \cos 2\alpha)$ $\sin^3\alpha = \frac 1 4 (3\sin\alpha - \sin 3\alpha)$ $\sin^4\alpha = \frac 1 8 (3 - 4\cos 2\alpha + \cos 4\alpha)$ $\sin^{2n}\alpha = \frac 1 {2^{2n - 1}} [\sum_{k = 0}^{n - 1}(-1)^{n + k}C_{2n}^k \cos(2n - 2k)\alpha + \frac 1 2 C_{2n}^n]$ $\sin^{2n + 1}\alpha = \frac 1 {2^{2n}} \sum_{k = 0}^n (-1)^{n + k}C_{2n + 1}^k \sin(2n - 2k + 1)\alpha$ $\cos^2\alpha = \frac 1 2(1 + \cos 2\alpha)$ $\cos^3\alpha = \frac 1 4(3\cos\alpha + \cos 3\alpha)$ $\cos^4\alpha = \frac 1 8(3 + 4\cos 2\alpha + \cos 4\alpha)$ $\cos^{2n}\alpha = \frac 1 {2^{2n - 1}}[\sum_{k = 0}^{n - 1} C_{2n}^k \cos(2n - 2k)\alpha + \frac 1 2 C_{2n}^n]$ $\cos^{2n + 1}\alpha = \frac 1 {2^{2n}}\sum_{k = 0}^n C_{2n + 1}^k \cos(2n - 2k + 1)\alpha$

以上式中的 $n$ 为正整数。

三角函数有限和公式

$\sum_{j = 1}^{n - 1} \sin \frac{j\pi} n = \cot \frac \pi {2n}$ $\sum_{j = 1}^n \cos \frac{(2j - 1)\pi} n = 0$ $\sum_{j = 1}^n \cos \frac{(2j - 1)\pi} {2n + 1} = \frac 1 2$ $\sum_{j = 1}^{n - 1} \sin \frac{2j^2\pi} n = \frac{\sqrt n} 2 (1 + \cos\frac {n\pi} 2 - \sin\frac{n\pi} 2)$ $\sum_{j = 1}^{n - 1} \cos \frac{2j^2\pi} n = \frac{\sqrt n} 2 (1 + \cos\frac {n\pi} 2 + \sin\frac{n\pi} 2) - 1$ $\sum_{j = 1}^n \cot^2\frac{j\pi}{2n + 1} = \frac 1 3 n(2n - 1)$ $\sum_{j = 1}^n \cot^2\frac{(2j - 1)\pi}{4n} = n(2n - 1)$ $\sum_{j = 1}^n \sec^2\frac{(4j - 3)\pi}{4n} = 2n^2$ $\sum_{j = 1}^{[\frac{n + 1} 2] - 1} \csc^2\frac{j\pi}{n} = \begin{cases} \frac 1 6 (n^2 - 1), &n为奇数\\ \frac 1 6 (n^2 - 4), &n为偶数 \end{cases}$ $\sum_{j = 1}^{[\frac n 2]} \csc^2\frac{(2j - 1)\pi}{2n} = \begin{cases} \frac 1 2 (n^2 - 1), &n为奇数\\ \frac 1 2 n^2, &n为偶数 \end{cases}$ $\sum_{j = 1}^{2n} \csc^2\frac{j\pi}{2n + 1} = \frac 4 3 n(n + 1)$ $\sum_{j = 1}^n \tan^4\frac{j\pi}{2n + 1} = \frac 1 3 n(2n + 1)(4n^2 + 6n - 1)$ $\sum_{j = 1}^n \cot^4\frac{j\pi}{2n + 1} = \frac 1 {45} n(2n - 1)(4n^2 + 10n - 9)$

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