三角恒等式

符号

sine sin cosecant csc
cosine cos secant sec
tangent tan cotangent cot

sine sin arcsine arcsin
cosine cos arccosine arccos
tangent tan arctangent arctan
cotangent cot arccotangent arccot
secant sec arcsecant arcsec
cosecant csc arccosecant arccsc

$\mathbf{0}$ $\tfrac{1}{12}$ $\tfrac{1}{8}$ $\tfrac{1}{6}$ $\tfrac{1}{4}$ $\tfrac{1}{2}$ $\tfrac{3}{4}$ $\mathbf{1}$

基本關係

 $\sin^2 \theta + \cos^2 \theta = 1\,$ $\tan^2 \theta + 1\, = \sec^2 \theta$ $1\, + \cot^2 \theta = \csc^2 \theta$

$\sin \theta$ $\sin \theta\$ $\sqrt{1 - \cos^2\theta}$ $\frac{\tan\theta}{\sqrt{1 + \tan^2\theta}}$ $\frac{1}{\sqrt{1+\cot^2\theta}}$ $\frac{\sqrt{\sec^2 \theta - 1}}{\sec \theta}$ $\frac{1}{\csc \theta}$
$\cos \theta$ $\sqrt{1 - \sin^2\theta}$ $\cos \theta\$ $\frac{1}{\sqrt{1 + \tan^2 \theta}}$ $\frac{\cot \theta}{\sqrt{1 + \cot^2 \theta}}$ $\frac{1}{\sec \theta}$ $\frac{\sqrt{\csc^2\theta - 1}}{\csc \theta}$
$\tan \theta$ $\frac{\sin\theta}{\sqrt{1 - \sin^2\theta}}$ $\frac{\sqrt{1 - \cos^2\theta}}{\cos \theta}$ $\tan \theta\$ $\frac{1}{\cot \theta}$ $\sqrt{\sec^2\theta - 1}$ $\frac{1}{\sqrt{\csc^2\theta - 1}}$
$\cot \theta$ ${\sqrt{1 - \sin^2\theta} \over \sin \theta}$ ${\cos \theta \over \sqrt{1 - \cos^2\theta}}$ ${1 \over \tan\theta}$ $\cot\theta\$ ${1 \over \sqrt{\sec^2\theta - 1}}$ $\sqrt{\csc^2\theta - 1}$
$\sec \theta$ ${1 \over \sqrt{1 - \sin^2\theta}}$ ${1 \over \cos \theta}$ $\sqrt{1 + \tan^2\theta}$ ${\sqrt{1 + \cot^2\theta} \over \cot \theta}$ $\sec\theta\$ ${\csc\theta \over \sqrt{\csc^2\theta - 1}}$
$\csc \theta$ ${1 \over \sin \theta}$ ${1 \over \sqrt{1 - \cos^2 \theta}}$ ${\sqrt{1 + \tan^2\theta} \over \tan \theta}$ $\sqrt{1 + \cot^2 \theta}$ ${\sec \theta \over \sqrt{\sec^2\theta - 1}}$ $\csc \theta\$

其他函數的基本關係

$\operatorname{vers}(\theta)$
$\operatorname{ver}(\theta)$
$1 - \cos (\theta)$

$\operatorname{cvs}(\theta)$
$1 - \sin(\theta)$

cohaversine
$\operatorname{hacoversin}(\theta)$ $\frac{1 - \sin (\theta)}{2}$

cohavercosine
$\operatorname{hacovercosin}(\theta)$ $\frac{1 + \sin (\theta)}{2}$

, chord $\operatorname{crd}(\theta)$ $2\sin\left(\frac{\theta}{2}\right)$

cosine and imaginary unit sine
$\operatorname{cis}(\theta)$ $\cos \theta + i\;\sin \theta$

对称、移位和周期

对称

\begin{align} \sin(0-\theta) &= -\sin \theta \\ \cos(0-\theta) &= +\cos \theta \\ \tan(0-\theta) &= -\tan \theta \\ \cot(0-\theta) &= -\cot \theta \\ \sec(0-\theta) &= +\sec \theta \\ \csc(0-\theta) &= -\csc \theta \end{align} \begin{align} \sin(\tfrac{\pi}{2} - \theta) &= +\cos \theta \\ \cos(\tfrac{\pi}{2} - \theta) &= +\sin \theta \\ \tan(\tfrac{\pi}{2} - \theta) &= +\cot \theta \\ \cot(\tfrac{\pi}{2} - \theta) &= +\tan \theta \\ \sec(\tfrac{\pi}{2} - \theta) &= +\csc \theta \\ \csc(\tfrac{\pi}{2} - \theta) &= +\sec \theta \end{align} \begin{align} \sin(\pi - \theta) &= +\sin \theta \\ \cos(\pi - \theta) &= -\cos \theta \\ \tan(\pi - \theta) &= -\tan \theta \\ \cot(\pi - \theta) &= -\cot \theta \\ \sec(\pi - \theta) &= -\sec \theta \\ \csc(\pi - \theta) &= +\csc \theta \end{align} \begin{align} \sin(\tfrac{3\pi}{2} - \theta) &= -\cos \theta \\ \cos(\tfrac{3\pi}{2} - \theta) &= -\sin \theta \\ \tan(\tfrac{3\pi}{2} - \theta) &= +\cot \theta \\ \cot(\tfrac{3\pi}{2} - \theta) &= +\tan \theta \\ \sec(\tfrac{3\pi}{2} - \theta) &= -\csc \theta \\ \csc(\tfrac{3\pi}{2} - \theta) &= -\sec \theta \end{align}

移位和周期

$\tan$$\cot$的周期

$\sin$, $\cos$, $\csc$$\sec$的周期
\begin{align} \sin(\theta + \tfrac{\pi}{2}) &= +\cos \theta \\ \cos(\theta + \tfrac{\pi}{2}) &= -\sin \theta \\ \tan(\theta + \tfrac{\pi}{2}) &= -\cot \theta \\ \cot(\theta + \tfrac{\pi}{2}) &= -\tan \theta \\ \sec(\theta + \tfrac{\pi}{2}) &= -\csc \theta \\ \csc(\theta + \tfrac{\pi}{2}) &= +\sec \theta \end{align} \begin{align} \sin(\theta + \pi) &= -\sin \theta \\ \cos(\theta + \pi) &= -\cos \theta \\ \tan(\theta + \pi) &= +\tan \theta \\ \cot(\theta + \pi) &= +\cot \theta \\ \sec(\theta + \pi) &= -\sec \theta \\ \csc(\theta + \pi) &= -\csc \theta \end{align} \begin{align} \sin(\theta + \tfrac{3\pi}{2}) &= -\cos \theta \\ \cos(\theta + \tfrac{3\pi}{2}) &= +\sin \theta \\ \tan(\theta + \tfrac{3\pi}{2}) &= -\cot \theta \\ \cot(\theta + \tfrac{3\pi}{2}) &= -\tan \theta \\ \sec(\theta + \tfrac{3\pi}{2}) &= +\csc \theta \\ \csc(\theta + \tfrac{3\pi}{2}) &= -\sec \theta \end{align} \begin{align} \sin(\theta + 2\pi) &= +\sin \theta \\ \cos(\theta + 2\pi) &= +\cos \theta \\ \tan(\theta + 2\pi) &= +\tan \theta \\ \cot(\theta + 2\pi) &= +\cot \theta \\ \sec(\theta + 2\pi) &= +\sec \theta \\ \csc(\theta + 2\pi) &= +\csc \theta \end{align}

角的和差恒等式

正弦 $\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\,$ $\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta\,$ $\tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}$ $\cot(\alpha\pm\beta)=\frac{\cot\alpha\cot\beta\mp1}{\cot\beta\pm\cot\alpha}$ $\sec(\alpha\pm\beta)=\frac{\sec\alpha\sec\beta}{1\mp\tan\alpha\tan\beta}$ $\csc(\alpha\pm\beta)=\frac{\csc\alpha\csc\beta}{\cot\beta\pm\cot\alpha}$ 注意正负号的对应。 \begin{align}x \pm y = a \pm b &\Rightarrow \ x + y = a + b \\ &\mbox{and} \ x -y = a -b \end{align} \begin{align} x \pm y = a \mp b &\Rightarrow \ x + y = a - b \\ &\mbox{and}\ x - y = a + b\end{align}

正弦与余弦的无限多项和

$\sin\left(\sum_{i=1}^\infty \theta_i\right) =\sum_{\mathrm{odd}\ k \ge 1} (-1)^{\frac{k-1}{2}} \sum_{ |A| = k } \left(\prod_{i \in A} \sin\theta_i \prod_{i \not \in A} \cos\theta_i\right)$
$\cos\left(\sum_{i=1}^\infty \theta_i\right) =\sum_{\mathrm{even}\ k \ge 0} ~ (-1)^{\frac{k}{2}} ~~ \sum_{ |A| = k } \left(\prod_{i \in A} \sin\theta_i \prod_{i \not \in A} \cos\theta_i\right)$

正切的有限多项和

xi = tan(θi )，对于i = 1, ..., n。设ek是变量xi, i = 1, ..., n, k = 0, ..., nk基本对称多项式。则

$\tan(\theta_1+\cdots+\theta_n) = \frac{e_1 - e_3 + e_5 -\cdots}{e_0 - e_2 + e_4 - \cdots},$

\begin{align} \tan(\theta_1 + \theta_2 + \theta_3) &{}= \frac{e_1 - e_3}{e_0 - e_2} = \frac{(x_1 + x_2 + x_3) \ - \ (x_1 x_2 x_3)}{ 1 \ - \ (x_1 x_2 + x_1 x_3 + x_2 x_3)}, \\ \\ \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) &{}= \frac{e_1 - e_3}{e_0 - e_2 + e_4} \\ \\ &{}= \frac{(x_1 + x_2 + x_3 + x_4) \ - \ (x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4)}{ 1 \ - \ (x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4) \ + \ (x_1 x_2 x_3 x_4)},\end{align}

多倍角公式

Tn是n次切比雪夫多项式 $\cos n\theta =T_n \cos \theta \,$ $\sin^2 n\theta = S_n \sin^2\theta\,$ $\cos n\theta +i\sin n\theta=[\cos(\theta)+i\sin(\theta)]^n \,$
$1+2\cos(x) + 2\cos(2x) + 2\cos(3x) + \cdots + 2\cos(nx) = \frac{\sin\left[\left(n +\frac{1}{2}\right)x\right]}{\sin \frac{x}{2}}$

（這個x的函數是狄利克雷核。）

二倍角、三倍角和半角公式

\begin{align} \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ &= 2 \cos^2 \theta - 1 \\ &= 1 - 2 \sin^2 \theta \\ &= \frac{1 - \tan^2 \theta} {1 + \tan^2 \theta} \end{align} $\cot 2\theta = \frac{\cot \theta - \tan \theta}{2}\,$ $\csc 2\theta = \frac{\csc^2\theta}{2\cot\theta}$
$=\frac{\sec\theta\csc\theta}{2}$

$\cos 3\theta = 4 \cos^3\theta - 3 \cos \theta \,$ $\cot 3\theta =\frac{\cot^3\theta - 3\cot\theta}{3 \cot^2\theta-1}$ $\csc 3\theta =\frac{\csc^3\theta}{3\csc^2\theta-4}$

$\cos \frac{\theta}{2} = \pm\, \sqrt{\frac{1 + \cos\theta}{2}}$ \begin{align} \cot \frac{\theta}{2} &= \csc \theta + \cot \theta \\ &= \pm\, \sqrt{1 + \cos \theta \over 1 - \cos \theta} \\ &= \frac{\sin \theta}{1 - \cos \theta} \\ &= \frac{1 + \cos \theta}{\sin \theta}\\ &= \frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1} \end{align} $\csc \frac{\theta}{2} = \pm\, \sqrt{\frac{2\sec\theta}{\sec\theta - 1}}$

n倍角公式

n倍角公式
$\sin n\theta = \sum_{k=0}^n \binom{n}{k} \cos^k \theta\,\sin^{n-k} \theta\,\sin\left[\frac{1}{2}(n-k)\pi\right] =\sin \theta \sum_{k=0}^{\lfloor \frac{n-1}{2}\rfloor}(-1)^k \binom{n-1-k}{k}~(2\cos \theta)^{n-1-2k}$

（第二类切比雪夫多项式

$\cos n\theta = \sum_{k=0}^n \binom{n}{k} \cos^k \theta\,\sin^{n-k} \theta\,\cos\left[\frac{1}{2}(n-k)\pi\right] =\frac{1}{2}\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}(-1)^k \frac{n}{n-k} \binom{n-k}{k}~(2\cos \theta)^{n-2k}$

（第一类切比雪夫多项式

$\tan n\theta = \frac{\displaystyle \sum_{k=1}^{\left[\frac{n}{2}\right]} (-1)^{k+1} \binom{n}{2k-1} \tan^{2k-1}\theta}{\displaystyle \sum_{k=1}^{\left[\frac{n+1}{2}\right]} (-1)^{k+1} \binom{n}{2(k-1)} \tan^{2(k-1)}\theta}$
n倍遞迴公式
$\tan\,n\theta = \frac{\tan (n{-}1)\theta + \tan \theta}{1 - \tan (n{-}1)\theta\,\tan \theta}$$\cot\,n\theta = \frac{\cot (n{-}1)\theta\,\cot \theta - 1}{\cot (n{-}1)\theta + \cot \theta}$。(遞迴關係)

其他函數的倍半角公式

• $\operatorname{versin} 2\theta = 2\sin^2\theta$

• $\operatorname{cvs} 2\theta = (\sin\theta-\cos\theta)^2$

幂简约公式

$\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ $\sin^2\theta \cos^2\theta = \frac{1 - \cos 4\theta}{8}$
$\sin^3\theta = \frac{3 \sin\theta - \sin 3\theta}{4}$ $\cos^3\theta = \frac{3 \cos\theta + \cos 3\theta}{4}$ $\sin^3\theta \cos^3\theta = \frac{3\sin 2\theta - \sin 6\theta}{32}$
$\sin^4\theta = \frac{3 - 4 \cos 2\theta + \cos 4\theta}{8}$ $\cos^4\theta = \frac{3 + 4 \cos 2\theta + \cos 4\theta}{8}$ $\sin^4\theta \cos^4\theta = \frac{3-4\cos 4\theta + \cos 8\theta}{128}$
$\sin^5\theta = \frac{10 \sin\theta - 5 \sin 3\theta + \sin 5\theta}{16}$ $\cos^5\theta = \frac{10 \cos\theta + 5 \cos 3\theta + \cos 5\theta}{16}$ $\sin^5\theta \cos^5\theta = \frac{10\sin 2\theta - 5\sin 6\theta + \sin 10\theta}{512}$

数值连乘

$\prod_{k=0}^{n-1}\cos 2^k\theta = \frac{\sin2^n \theta}{2^n\sin\theta}$[2]

$\prod_{k=0}^{n-1}\sin\left(x+\frac{k\pi}{n}\right) = \frac{\sin nx}{2^{n-1}}$[2]

$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$,$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{2n}\right) = \frac{\sqrt{n}}{2^{n-1}}$,$\prod_{k=1}^{n}\sin\left(\frac{k\pi}{2n+1}\right) = \frac{\sqrt{2n+1}}{2^n}$
$\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right) = \frac{\sin \frac{n\pi}{2}}{2^{n-1}}$,$\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{2n}\right) = \frac{\sqrt{n}}{2^{n-1}}$,$\prod_{k=1}^n \cos\left(\frac{k\pi}{2n+1}\right) = \frac{1}{2^n}$
$\prod_{k=1}^{n-1}\tan\left(\frac{k\pi}{n}\right) = \frac{n}{\sin \frac{n\pi}{2}}$,$\prod_{k=1}^{n-1}\tan\left(\frac{k\pi}{2n}\right) = 1$,$\prod_{k=1}^n \tan\frac{k\pi}{2n+1} = \sqrt{2n+1}$

常見的恆等式

积化和差与和差化积恆等式

$\sin\theta\sin\phi={\cos(\theta-\phi)-\cos(\theta+\phi)\over2}$
$\cos\theta\cos\phi={\cos(\theta-\phi)+\cos(\theta+\phi)\over2}$
$\sin\theta\cos\phi={\sin(\theta+\phi)+\sin(\theta-\phi)\over2}$
$\cos\theta\sin\phi={\sin(\theta+\phi)-\sin(\theta-\phi)\over2}$

$\sin\theta+\sin\phi=2\sin\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}$
$\cos\theta+\cos\phi=2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}$
$\cos\theta-\cos\phi=-2\sin{\theta+\phi\over2}\sin{\theta-\phi\over2}$
$\sin\theta-\sin\phi=2\cos{\theta+\phi\over2}\sin{\theta-\phi\over2}$

其他恆等式

$\cot x\cot y + \cot y\cot z + \cot z\cot x = 1\,$

$\cot x + \cot y + \cot z = \cot x\cot y\cot z\,$

$\sin(x+y)\sin(x-y) = \sin^2 {x} - \sin^2 {y} = \cos^2 {y} - \cos^2 {x}\,$

$\cos(x+y)\cos(x-y) = \cos^2 {x} - \sin^2 {y} = \cos^2 {y} - \sin^2 {x}\,$

托勒密定理

\begin{align} & \sin(w + x)\sin(x + y) \\ &{} = \sin(x + y)\sin(y + z) \\ &{} = \sin(y + z)\sin(z + w) \\ &{} = \sin(z + w)\sin(w + x) \\ &{} = \sin w \sin y + \sin x \sin z \end{align}

（前三个等式是一般情况；第四个是本质。）

三角函數與雙曲函數的恆等式

$e^{i x} = \cos x + i \;\sin x , \; e^{-i x} = \cos x - i \;\sin x$

$e^x = \cosh x + \sinh x\! , \; e^{-x} = \cosh x - \sinh x \!$

$\cosh ix = \tfrac12(e^{i x} + e^{-i x}) = \cos x$

$\sinh ix = \tfrac12(e^{i x} - e^{-i x}) = i \sin x$

$\sin \theta =-i\sinh{i\theta}\,$ $\sinh{\theta}=i\sin{(-i\theta)}\,$
$\cos{\theta}=\cosh{i\theta}\,$ $\cosh{\theta}=\cos{(-i\theta)}\,$
$\tan \theta =-i\tanh{i\theta}\,$ $\tanh{\theta}=i\tan{(-i\theta)}\,$
$\cot{\theta}=i\coth{i\theta}\,$ $\coth \theta =-i\cot{(-i\theta)}\,$
$\sec{\theta}=\operatorname{sech}{\,i\theta}\,$ $\operatorname{sech}{\theta}=\sec{(-i\theta)}\,$
$\csc{\theta}=i\;\operatorname{csch}{\, i\theta}\,$ $\operatorname{csch} \theta =-i\csc{(-i\theta)}\,$
• 其他恆等式：
$\cosh ix = \tfrac12(e^{i x} + e^{-i x}) = \cos x$
$\sinh ix = \tfrac12(e^{i x} - e^{-i x}) = i \sin x$
$\cosh(x+iy) = \cosh(x) \cos(y) + i \sinh(x) \sin(y) \,$
$\sinh(x+iy) = \sinh(x) \cos(y) + i \cosh(x) \sin(y) \,$
$\tanh ix = i \tan x \,$
$\cosh x = \cos ix \,$
$\sinh x = -i \sin ix \,$
$\tanh x = -i \tan ix \,$

线性组合

$a\sin x+b\cos x=\sqrt{a^2+b^2}\cdot\sin(x+\varphi)\,$

$\varphi = \arctan \left(\frac{b}{a}\right)$

$a\sin x+b\sin(x+\alpha)= c \sin(x+\beta)\,$

$c = \sqrt{a^2 + b^2 +2ab\cos \alpha},$

$\beta = {\rm arctan} \left(\frac{b\sin \alpha}{a + b\cos \alpha}\right)$

反三角函数

$\arcsin x+\arccos x=\frac{\pi}{2}\;$
$\arctan x+\arccot x=\frac{\pi}{2}.\;$
$\arctan x+\arctan \frac{1}{x}=\left\{\begin{matrix} \frac{\pi}{2}, & \mbox{if }x > 0 \\ -\frac{\pi}{2}, & \mbox{if }x < 0 \end{matrix}\right.$
$\arctan x+\arctan y=\arctan \frac{x+y}{1-xy}+\left\{\begin{matrix} \pi, & \mbox{if }x,y>0 \\ -\pi, & \mbox{if }x,y<0 \\ 0, & \mbox{otherwise } \end{matrix}\right.$
 $\sin ( \arccos x)=\sqrt{1-x^2} \,$ $\sin ( \arctan x)=\frac{x}{\sqrt{1+x^2}}$ $\cos ( \arctan x)=\frac{1}{\sqrt{1+x^2}}$ $\cos ( \arcsin x)=\sqrt{1-x^2} \,$ $\tan ( \arcsin x)=\frac{x}{\sqrt{1 - x^2}}$ $\tan ( \arccos x)=\frac{\sqrt{1 - x^2}}{x}$

无限乘积公式

 $\sin x = x \prod_{n = 1}^\infty\left(1 - \frac{x^2}{\pi^2 n^2}\right)$ $\sinh x = x \prod_{n = 1}^\infty\left(1 + \frac{x^2}{\pi^2 n^2}\right)$ $\frac{\sin x}{x} = \prod_{n = 1}^\infty\cos\left(\frac{x}{2^n}\right)$ $\cos x = \prod_{n = 1}^\infty\left(1 - \frac{x^2}{\pi^2(n - \frac{1}{2})^2}\right)$ $\cosh x = \prod_{n = 1}^\infty\left(1 + \frac{x^2}{\pi^2(n - \frac{1}{2})^2}\right)$ $|\sin x| = \frac1{2}\prod_{n = 0}^\infty \sqrt[2^{n+1}]{\left|\tan\left(2^n x\right)\right|}$

微積分

$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$

$\lim_{x\rightarrow 0}\frac{1-\cos x}{x}=0$

${d \over dx}\sin(x) = \cos(x)$

\begin{align} {d \over dx} \sin x &= \cos x ,& {d \over dx} \arcsin x &={1 \over \sqrt{1 - x^2} } \\ \\ {d \over dx} \cos x &= -\sin x ,& {d \over dx} \arccos x &=-{1 \over \sqrt{1 - x^2}} \\ \\ {d \over dx} \tan x &= \sec^2 x ,& {d \over dx} \arctan x &={ 1 \over 1 + x^2} \\ \\ {d \over dx} \cot x &= -\csc^2 x ,& {d \over dx} \arccot x &=-{1 \over 1 + x^2} \\ \\ {d \over dx} \sec x &= \tan x \sec x ,& {d \over dx} \arcsec x &={ 1 \over |x|\sqrt{x^2 - 1}} \\ \\ {d \over dx} \csc x &= -\csc x \cot x ,& {d \over dx} \arccsc x &=-{1 \over |x|\sqrt{x^2 - 1}} \end{align}

參考文獻

1. ^ Abramowitz and Stegun, p. 78, 4.3.147
2. ^ 2.0 2.1
3. ^ Abramowitz and Stegun, p. 75, 4.3.89–90
4. ^ Abramowitz and Stegun, p. 85, 4.5.68–69