三角函數示意圖

由于已知的技术原因，图表暂时不可用。带来不便，我们深表歉意。

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

三角学
历史（英语：History of trigonometry） 三角函数 (反三角函数) 广义三角函数（英语：Generalized trigonometry）
参考
恒等式 精确值 三角表（英语：Trigonometric tables） 单位圆
定理
正弦 餘弦 正切 餘切 勾股定理
微积分
三角换元法 积分 (反三角函数) 微分
查 论 编

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。

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'${\displaystyle \mathrm {cosec} }$'。

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

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

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	${\displaystyle \arg x}$	${\displaystyle \operatorname {Im} (\ln x)}$

通過檢視單位圓，可確立三角函數的下列性質，這些性質也被稱為誘導公式：

當三角函数反射自某个特定的${\displaystyle \theta }$值，結果經常是另一个其他三角函數。這導致了下列恆等式：

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

通過旋轉特定角度移位三角函數，經常可以找到更簡單的表達结果的不同的三角函數。例如通過旋轉${\displaystyle {\tfrac {\pi }{2}}}$、${\displaystyle \pi }$和${\displaystyle 2\pi }$弧度移位函數。因爲這些函數的周期要麼是${\displaystyle \pi }$要麼是${\displaystyle 2\pi }$，所以新函數和沒有移位的舊函數完全一樣。

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

又稱做“和差定理”、“和差公式”或“和角公式”。最簡要的檢定方式是使用歐拉公式^{[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	${\displaystyle \csc(\alpha \pm \beta )={\frac {\csc \alpha \csc \beta }{\cot \beta \pm \cot \alpha }}}$
注意正負號的對應。 ${\displaystyle {\begin{aligned}x\pm y=a\pm b&\Rightarrow \ x+y=a+b\\&{\mbox{and}}\ x-y=a-b\end{aligned}}}$ ${\displaystyle {\begin{aligned}x\pm y=a\mp b&\Rightarrow \ x+y=a-b\\&{\mbox{and}}\ x-y=a+b\end{aligned}}}$

根据${\displaystyle sin{\frac {\pi }{4}}=cos{\frac {\pi }{4}}={\frac {1}{\sqrt {2}}}}$，以及和差恒等式，可以得到同角的正弦余弦的和差关系，例如，

${\displaystyle \sin \alpha +\cos \alpha ={\sqrt {2}}\left(\sin \alpha \cos {\frac {\pi }{4}}+\sin {\frac {\pi }{4}}\cos \alpha \right)={\sqrt {2}}\sin \left(\alpha +{\frac {\pi }{4}}\right)={\sqrt {2}}\cos \left(\alpha -{\frac {\pi }{4}}\right)}$

${\displaystyle \sin \alpha -\cos \alpha ={\sqrt {2}}\left(\sin \alpha \cos {\frac {\pi }{4}}-\sin {\frac {\pi }{4}}\cos \alpha \right)={\sqrt {2}}\sin \left(\alpha -{\frac {\pi }{4}}\right)={\sqrt {2}}\cos \left(\alpha +{\frac {\pi }{4}}\right)}$

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

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

這里的"${\displaystyle |A|=k}$"意味著索引${\displaystyle A}$遍歷集合${\displaystyle \left\{1,2,3,\ldots \right\}}$的大小為${\displaystyle k}$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。

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。

${\displaystyle T_{n}}$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。）

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

${\displaystyle n}$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”。

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

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

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	${\displaystyle \cos ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}{\binom {n}{k}}\cos {[(n-2k)\theta ]}}$	${\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{\left({\frac {n}{2}}-k\right)}{\binom {n}{k}}\cos {[(n-2k)\theta ]}}$

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

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

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

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

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