三角恒等式

在数学中，三角恒等式是对出现的所有值都为實变量，涉及到三角函数的等式。这些恒等式在表达式中有些三角函数需要简化的时候是很有用的。一个重要应用是非三角函数的积分：一个常用技巧是首先使用使用三角函数的代换规则，则通过三角恒等式可简化结果的积分。

在几何上依据以O为中心的单位圆可以构造角θ的很多三角函数

三角函數示意圖

幾個三角函數的圖形，分別為正弦、餘弦、正切、餘切、正割、餘割和正矢。配色與上圖相同

单位圆的角度

符号

为了避免由于${\displaystyle \sin ^{-1}x}$ 的不同意思所带来的混淆，我們經常用下列兩個表格來表示三角函数的倒数和反函数。另外在表示余割函数時，'${\displaystyle \csc }$ '有时會寫成比較长的'${\displaystyle \mathrm {cosec} }$ '。

函数			反函數			倒数
中文	全寫	簡寫	中文	全寫	簡寫	中文	全寫	簡寫
正弦	sine	sin	反正弦	arcsine	arcsin	餘割	cosecant	csc
餘弦	cosine	cos	反餘弦	arccosine	arccos	正割	secant	sec
正切	tangent	tan	反正切	arctangent	arctan	餘切	cotangent	cot
餘切	cotangent	cot	反餘切	arccotangent	arccot	正切	tangent	tan
正割	secant	sec	反正割	arcsecant	arcsec	餘弦	cosine	cos
餘割	cosecant	csc	反餘割	arccosecant	arccsc	正弦	sine	sin

不同的角度度量适合于不同的情况。本表展示最常用的系统。弧度是缺省的角度量并用在指数函数中。所有角度度量都是无单位的。另外在計算機中角度的符號為D，弧度的符號為R，梯度的符號為G。

相同角度的轉換表

角度單位	值								計算機中代號
轉	${\displaystyle 0}$	${\displaystyle {\frac {1}{12}}}$	${\displaystyle {\frac {1}{8}}}$	${\displaystyle {\frac {1}{6}}}$	${\displaystyle {\frac {1}{4}}}$	${\displaystyle {\frac {1}{2}}}$	${\displaystyle {\frac {3}{4}}}$	${\displaystyle 1}$	無
角度	${\displaystyle 0^{\circ }}$	${\displaystyle 30^{\circ }}$	${\displaystyle 45^{\circ }}$	${\displaystyle 60^{\circ }}$	${\displaystyle 90^{\circ }}$	${\displaystyle 180^{\circ }}$	${\displaystyle 270^{\circ }}$	${\displaystyle 360^{\circ }}$	D
弧度	${\displaystyle 0}$	${\displaystyle {\frac {\pi }{6}}}$	${\displaystyle {\frac {\pi }{4}}}$	${\displaystyle {\frac {\pi }{3}}}$	${\displaystyle {\frac {\pi }{2}}}$	${\displaystyle \pi }$	${\displaystyle {\frac {3\pi }{2}}}$	${\displaystyle 2\pi }$	R
梯度	${\displaystyle 0^{g}}$	${\displaystyle 33{\frac {1}{3}}^{g}}$	${\displaystyle 50^{g}}$	${\displaystyle 66{\frac {2}{3}}^{g}}$	${\displaystyle 100^{g}}$	${\displaystyle 200^{g}}$	${\displaystyle 300^{g}}$	${\displaystyle 400^{g}}$	G

基本關係

 

三角函數間的關係，可分成正函數和餘函數

畢達哥拉斯三角恒等式如下：

${\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1\,}$

${\displaystyle \tan ^{2}\theta +1\,=\sec ^{2}\theta }$

${\displaystyle 1\,+\cot ^{2}\theta =\csc ^{2}\theta }$

由上面的平方關係加上三角函數的基本定義，可以導出下面的表格，即每個三角函數都可以用其他五個表達。（严谨地说，所有根号前都应根据实际情况添加正负号）

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

其他函數的基本關係

正矢、餘矢、半正矢、半餘矢、外正割用於航行。例如半正矢可以計算球體上的兩個點之間的距離，但它們不常用。

名稱	函數	值^[1]
正矢, versine	${\displaystyle \operatorname {versin} \theta }$  ${\displaystyle \operatorname {vers} \theta }$  ${\displaystyle \operatorname {ver} \theta }$	${\displaystyle 1-\cos \theta }$
餘的正矢, vercosine	${\displaystyle \operatorname {vercosin} \theta }$	${\displaystyle 1+\cos \theta }$
餘矢, coversine	${\displaystyle \operatorname {coversin} \theta }$  ${\displaystyle \operatorname {cvs} \theta }$	${\displaystyle 1-\sin \theta }$
餘的餘矢, covercosine	${\displaystyle \operatorname {covercosin} \theta }$	${\displaystyle 1+\sin \theta }$
半正矢, haversine	${\displaystyle \operatorname {haversin} \theta }$	${\displaystyle {\frac {1-\cos \theta }{2}}}$
餘的半正矢, havercosine	${\displaystyle \operatorname {havercosin} \theta }$	${\displaystyle {\frac {1+\cos \theta }{2}}}$
半餘矢, hacoversine cohaversine	${\displaystyle \operatorname {hacoversin} \theta }$	${\displaystyle {\frac {1-\sin \theta }{2}}}$
餘的半餘矢, hacovercosine cohavercosine	${\displaystyle \operatorname {hacovercosin} \theta }$	${\displaystyle {\frac {1+\sin \theta }{2}}}$
外正割，exsecant	${\displaystyle \operatorname {exsec} \theta }$	${\displaystyle \sec \theta -1}$
外餘割，excosecant	${\displaystyle \operatorname {excsc} \theta }$	${\displaystyle \csc \theta -1}$
弦函數, chord	${\displaystyle \operatorname {crd} \theta }$	${\displaystyle 2\sin \left({\frac {\theta }{2}}\right)}$
純虛數指數函數, cosine and imaginary unit sine	${\displaystyle \operatorname {cis} \theta }$	${\displaystyle \cos \theta +i\;\sin \theta }$
輻角,Argument	${\displaystyle \arg x}$	${\displaystyle \operatorname {Im} (\ln x)}$

对称、移位和周期

更多信息：诱导公式

通过检视单位圆，可确立三角函数的下列性质：

对称

当三角函数反射自某个特定的${\displaystyle \theta }$ 值，结果经常是另一个其他三角函数。这导致了下列恒等式：

反射于${\displaystyle \theta =0}$	反射于${\displaystyle \theta ={\tfrac {\pi }{4}}}$	反射于${\displaystyle \theta ={\tfrac {\pi }{2}}}$	反射于${\displaystyle \theta ={\tfrac {3\pi }{4}}}$
${\displaystyle {\begin{aligned}\sin(0-\theta )&=-\sin \theta \\\cos(0-\theta )&=+\cos \theta \\\tan(0-\theta )&=-\tan \theta \\\cot(0-\theta )&=-\cot \theta \\\sec(0-\theta )&=+\sec \theta \\\csc(0-\theta )&=-\csc \theta \end{aligned}}}$	${\displaystyle {\begin{aligned}\sin({\tfrac {\pi }{2}}-\theta )&=+\cos \theta \\\cos({\tfrac {\pi }{2}}-\theta )&=+\sin \theta \\\tan({\tfrac {\pi }{2}}-\theta )&=+\cot \theta \\\cot({\tfrac {\pi }{2}}-\theta )&=+\tan \theta \\\sec({\tfrac {\pi }{2}}-\theta )&=+\csc \theta \\\csc({\tfrac {\pi }{2}}-\theta )&=+\sec \theta \end{aligned}}}$	${\displaystyle {\begin{aligned}\sin(\pi -\theta )&=+\sin \theta \\\cos(\pi -\theta )&=-\cos \theta \\\tan(\pi -\theta )&=-\tan \theta \\\cot(\pi -\theta )&=-\cot \theta \\\sec(\pi -\theta )&=-\sec \theta \\\csc(\pi -\theta )&=+\csc \theta \end{aligned}}}$	${\displaystyle {\begin{aligned}\sin({\tfrac {3\pi }{2}}-\theta )&=-\cos \theta \\\cos({\tfrac {3\pi }{2}}-\theta )&=-\sin \theta \\\tan({\tfrac {3\pi }{2}}-\theta )&=+\cot \theta \\\cot({\tfrac {3\pi }{2}}-\theta )&=+\tan \theta \\\sec({\tfrac {3\pi }{2}}-\theta )&=-\csc \theta \\\csc({\tfrac {3\pi }{2}}-\theta )&=-\sec \theta \end{aligned}}}$

移位和周期

通过旋转特定角度移位三角函数，经常可以找到更简单的表达结果的不同的三角函数。例如通过旋转${\displaystyle {\tfrac {\pi }{2}}}$ 、${\displaystyle \pi }$ 和${\displaystyle 2\pi }$ 弧度移位函数。因为这些函数的周期要么是${\displaystyle \pi }$ 要么是${\displaystyle 2\pi }$ ，新函数和没有移位的旧函数完全一样。

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

角的和差恒等式

 

正弦與餘弦的角和公式的圖形證明法。使用了相似三角形的性質與三角函數的定義，強調的線段是單位長度

 

正切的角和公式的圖形證明法。使用了相似三角形的性質與三角函數的定義，強調的線段是單位長度。

它们也叫做“和差定理”、“和差公式”或“和角公式”。最快速簡要的檢定方式是使用欧拉公式^{[註 1]}。

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

正弦与余弦的无限多项和

${\displaystyle \sin \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{\mathrm {odd} \ k\geq 1}(-1)^{\frac {k-1}{2}}\sum _{|A|=k}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}$ 

${\displaystyle \cos \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{\mathrm {even} \ k\geq 0}~(-1)^{\frac {k}{2}}~~\sum _{|A|=k}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}$ 

这里的"${\displaystyle |A|=k}$ "意味着索引${\displaystyle A}$ 遍历集合${\displaystyle \left\{1,2,3,\ldots \right\}}$ 的大小为${\displaystyle k}$ 的所有子集的集合。

在这两个恒等式中出现了在有限多项中不出现的不对称：在每个乘积中，只有有限多个正弦因子和餘有限多个余弦因子。

如果只有有限多项${\displaystyle \theta _{i}}$ 是非零，则在右边只有有限多项是非零，因为正弦因子将变为零，而在每个项中，所有却有限多的余弦因子将是单位一。

正切的有限多项和

设${\displaystyle x_{i}=\tan \theta _{i}}$ ，对于${\displaystyle i=1,\ldots ,n}$ 。设${\displaystyle e_{k}}$ 是变量${\displaystyle x_{i}}$ ，${\displaystyle i=1,\ldots ,n}$ ，${\displaystyle k=0,\ldots ,n}$ 的${\displaystyle k}$ 次基本对称多项式。则

${\displaystyle \tan(\theta _{1}+\cdots +\theta _{n})={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }},}$ 

项的数目依赖于${\displaystyle n}$ 。例如，

${\displaystyle {\begin{aligned}\tan(\theta _{1}+\theta _{2}+\theta _{3})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\\\\tan(\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\\\&{}={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}$ 

并以此类推。一般情况可通过数学归纳法证明。

多倍角公式

${\displaystyle T_{n}}$ 是${\displaystyle n}$ 次切比雪夫多项式	${\displaystyle \cos n\theta =T_{n}\cos \theta \,}$
${\displaystyle S_{n}}$ 是${\displaystyle n}$ 次伸展多项式	${\displaystyle \sin ^{2}n\theta =S_{n}\sin ^{2}\theta \,}$
棣莫弗定理，${\displaystyle i}$ 是虚单位	${\displaystyle \cos n\theta +i\sin n\theta =(\cos \theta +i\sin \theta )^{n}\,}$

${\displaystyle 1+2\cos x+2\cos 2x+2\cos 3x+\cdots +2\cos(nx)={\frac {\sin \left[\left(n+{\frac {1}{2}}\right)x\right]}{\sin {\frac {x}{2}}}}}$ 。

（這個${\displaystyle x}$ 的函數是狄利克雷核。）

雙倍角、三倍角和半角公式

這些公式可以使用和差恒等式或多倍角公式来证明。

		弦	切	割
雙倍角公式	正	${\displaystyle {\begin{aligned}\sin 2\theta &=2\sin \theta \cos \theta \ \\&={\frac {2\tan \theta }{1+\tan ^{2}\theta }}\end{aligned}}}$	${\displaystyle {\begin{aligned}\tan 2\theta &={\frac {2\tan \theta }{1-\tan ^{2}\theta }}\ \\&={\frac {1}{1-\tan \theta }}-{\frac {1}{1+\tan \theta }}\end{aligned}}}$	${\displaystyle {\begin{aligned}\sec 2\theta &={\frac {\sec ^{2}\theta }{1-\tan ^{2}\theta }}\\&={\frac {\sec ^{2}\theta }{2-\sec ^{2}\theta }}\end{aligned}}}$
	餘	${\displaystyle {\begin{aligned}\cos 2\theta &=\cos ^{2}\theta -\sin ^{2}\theta \\&=2\cos ^{2}\theta -1\\&=1-2\sin ^{2}\theta \\&={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}\end{aligned}}}$	${\displaystyle {\begin{aligned}\cot 2\theta &={\frac {\cot ^{2}\theta -1}{2\cot \theta }}\\&={\frac {\cot \theta -\tan \theta }{2}}\end{aligned}}}$	${\displaystyle {\begin{aligned}\csc 2\theta &={\frac {\csc ^{2}\theta }{2\cot \theta }}\\&={\frac {\sec \theta \csc \theta }{2}}\end{aligned}}}$
降次公式	正	${\displaystyle \sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}}$	${\displaystyle \tan ^{2}\theta ={\frac {1-\cos 2\theta }{1+\cos 2\theta }}}$
	餘	${\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}}$	${\displaystyle \cot ^{2}\theta ={\frac {1+\cos 2\theta }{1-\cos 2\theta }}}$
三倍角公式	正	${\displaystyle {\begin{aligned}\sin 3\theta &=3\sin \theta -4\sin ^{3}\theta \\&=4\sin \theta \sin \left({\frac {\pi }{3}}-\theta \right)\sin \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}}$	${\displaystyle {\begin{aligned}\tan 3\theta &={\frac {3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }}\\&=\tan \theta \tan \left({\frac {\pi }{3}}-\theta \right)\tan \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}}$	${\displaystyle {\begin{aligned}\sec 3\theta &={\frac {\sec ^{3}\theta }{4-3\sec ^{2}\theta }}\\&={\dfrac {1}{4\cos \theta \cos \left({\dfrac {\pi }{3}}-\theta \right)\cos \left({\dfrac {\pi }{3}}+\theta \right)}}\end{aligned}}}$
	餘	${\displaystyle {\begin{aligned}\cos 3\theta &=4\cos ^{3}\theta -3\cos \theta \\&=4\cos \theta \cos \left({\frac {\pi }{3}}-\theta \right)\cos \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}}$	${\displaystyle {\begin{aligned}\cot 3\theta &={\frac {\cot ^{3}\theta -3\cot \theta }{3\cot ^{2}\theta -1}}\\&=\cot \theta \cot \left({\frac {\pi }{3}}-\theta \right)\cot \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}}$	${\displaystyle {\begin{aligned}\csc 3\theta &={\frac {\csc ^{3}\theta }{3\csc ^{2}\theta -4}}\\&={\dfrac {1}{4\sin \theta \sin \left({\dfrac {\pi }{3}}-\theta \right)\sin \left({\dfrac {\pi }{3}}+\theta \right)}}\end{aligned}}}$
半角公式	正	${\displaystyle \sin {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1-\cos \theta }{2}}}}$	${\displaystyle {\begin{aligned}\tan {\frac {\theta }{2}}&=\csc \theta -\cot \theta \\&=\pm \,{\sqrt {1-\cos \theta \over 1+\cos \theta }}\\&={\frac {\sin \theta }{1+\cos \theta }}\\&={\frac {1-\cos \theta }{\sin \theta }}\\&={\frac {\cos \theta +\sin \theta -1}{\cos \theta -\sin \theta +1}}\\&={\sqrt {\cot ^{2}\theta +1}}-\cot \theta \end{aligned}}}$	${\displaystyle \sec {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {2\sec \theta }{\sec \theta +1}}}}$
	餘	${\displaystyle \cos {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{2}}}}$	${\displaystyle {\begin{aligned}\cot {\frac {\theta }{2}}&=\csc \theta +\cot \theta \\&=\pm \,{\sqrt {1+\cos \theta \over 1-\cos \theta }}\\&={\frac {\sin \theta }{1-\cos \theta }}\\&={\frac {1+\cos \theta }{\sin \theta }}\\&={\frac {\cos \theta -\sin \theta +1}{\cos \theta +\sin \theta -1}}\\&={\sqrt {\cot ^{2}\theta +1}}+\cot \theta \end{aligned}}}$	${\displaystyle \csc {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {2\sec \theta }{\sec \theta -1}}}}$

n倍角公式

${\displaystyle n}$ 倍角公式
${\displaystyle \sin n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\sin \left[{\frac {1}{2}}(n-k)\pi \right]=\sin \theta \sum _{k=0}^{\lfloor {\frac {n-1}{2}}\rfloor }(-1)^{k}{\binom {n-1-k}{k}}~(2\cos \theta )^{n-1-2k}}$  （第二类切比雪夫多项式）
${\displaystyle \cos n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\cos \left[{\frac {1}{2}}(n-k)\pi \right]={\frac {1}{2}}\sum _{k=0}^{\lfloor {\frac {n}{2}}\rfloor }(-1)^{k}{\frac {n}{n-k}}{\binom {n-k}{k}}~(2\cos \theta )^{n-2k}}$  （第一类切比雪夫多项式）
${\displaystyle \tan n\theta ={\frac {\displaystyle \sum _{k=1}^{\left[{\frac {n}{2}}\right]}(-1)^{k+1}{\binom {n}{2k-1}}\tan ^{2k-1}\theta }{\displaystyle \sum _{k=1}^{\left[{\frac {n+1}{2}}\right]}(-1)^{k+1}{\binom {n}{2(k-1)}}\tan ^{2(k-1)}\theta }}}$
${\displaystyle n}$ 倍遞迴公式
${\displaystyle \tan \,n\theta ={\frac {\tan(n{-}1)\theta +\tan \theta }{1-\tan(n{-}1)\theta \,\tan \theta }}}$ 、${\displaystyle \cot \,n\theta ={\frac {\cot(n{-}1)\theta \,\cot \theta -1}{\cot(n{-}1)\theta +\cot \theta }}}$ 。(遞迴關係)

参见正切半角公式，它也叫做“万能公式”。

其他函數的倍半角公式

正矢

• ${\displaystyle \operatorname {versin} 2\theta =2\sin ^{2}\theta ={\frac {(\sin 2\theta )(\sin \theta )}{\cos \theta }}=1-\cos 2\theta }$ 

餘矢

• ${\displaystyle \operatorname {cvs} 2\theta =(\sin \theta -\cos \theta )^{2}=1-\sin 2\theta }$ 

幂简约公式

从解余弦二倍角公式的第二和第三版本得到。

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

	餘弦	正弦
如果${\displaystyle n}$ 是奇數	${\displaystyle \cos ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}{\binom {n}{k}}\cos {[(n-2k)\theta ]}}$	${\displaystyle \sin ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}(-1)^{\left({\frac {n-1}{2}}-k\right)}{\binom {n}{k}}\sin {[(n-2k)\theta ]}}$
如果${\displaystyle n}$ 是偶數	${\displaystyle \cos ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}{\binom {n}{k}}\cos {[(n-2k)\theta ]}}$	${\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{\left({\frac {n}{2}}-k\right)}{\binom {n}{k}}\cos {[(n-2k)\theta ]}}$

数值连乘

${\displaystyle \prod _{k=0}^{n-1}\cos 2^{k}\theta ={\frac {\sin 2^{n}\theta }{2^{n}\sin \theta }}}$ ^[2]

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

${\displaystyle \prod _{k=1}^{n-1}\sin \left({\frac {k\pi }{n}}\right)={\frac {n}{2^{n-1}}}}$ ,${\displaystyle \prod _{k=1}^{n-1}\sin \left({\frac {k\pi }{2n}}\right)={\frac {\sqrt {n}}{2^{n-1}}}}$ ,${\displaystyle \prod _{k=1}^{n}\sin \left({\frac {k\pi }{2n+1}}\right)={\frac {\sqrt {2n+1}}{2^{n}}}}$ 

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

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

常見的恆等式

积化和差与和差化积恆等式

數學家韋達在其三角學著作《應用於三角形的數學定律》給出积化和差与和差化积恒等式。积化和差恒等式可以通过展开角的和差恒等式的右手端来证明。

 

积化和差	和差化积
${\displaystyle \sin \alpha \cos \beta ={\sin(\alpha +\beta )+\sin(\alpha -\beta ) \over 2}}$	${\displaystyle \sin \alpha +\sin \beta =2\sin {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}$
${\displaystyle \cos \alpha \sin \beta ={\sin(\alpha +\beta )-\sin(\alpha -\beta ) \over 2}}$	${\displaystyle \sin \alpha -\sin \beta =2\cos {\alpha +\beta \over 2}\sin {\alpha -\beta \over 2}}$
${\displaystyle \cos \alpha \cos \beta ={\cos(\alpha +\beta )+\cos(\alpha -\beta ) \over 2}}$	${\displaystyle \cos \alpha +\cos \beta =2\cos {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}$
${\displaystyle \sin \alpha \sin \beta =-{\cos(\alpha +\beta )-\cos(\alpha -\beta ) \over 2}}$	${\displaystyle \cos \alpha -\cos \beta =-2\sin {\alpha +\beta \over 2}\sin {\alpha -\beta \over 2}}$

平方差公式

一月 01, 1970