List of trigonometric identities

The basic relationship between the sine and cosine is given by the Pythagorean identity:

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

where ${\displaystyle \sin ^{2}\theta }$ means ${\displaystyle {(\sin \theta )}^{2}}$ and ${\displaystyle \cos ^{2}\theta }$ means ${\displaystyle {(\cos \theta )}^{2}.}$

This can be viewed as a version of the Pythagorean theorem, and follows from the equation ${\displaystyle x^{2}+y^{2}=1}$ for the unit circle. This equation can be solved for either the sine or the cosine:

$${\displaystyle {\begin{aligned}\sin \theta &=\pm {\sqrt {1-\cos ^{2}\theta }},\\\cos \theta &=\pm {\sqrt {1-\sin ^{2}\theta }}.\end{aligned}}}$$

where the sign depends on the quadrant of ${\displaystyle \theta .}$

Dividing this identity by ${\displaystyle \sin ^{2}\theta }$, ${\displaystyle \cos ^{2}\theta }$, or both yields the following identities: $${\displaystyle {\begin{aligned}&1+\cot ^{2}\theta =\csc ^{2}\theta \\&1+\tan ^{2}\theta =\sec ^{2}\theta \\&\sec ^{2}\theta +\csc ^{2}\theta =\sec ^{2}\theta \csc ^{2}\theta \end{aligned}}}$$

Using these identities, it is possible to express any trigonometric function in terms of any other (up to a plus or minus sign):

Each trigonometric function in terms of each of the other five.^[1]

in terms of	${\displaystyle \sin \theta }$	${\displaystyle \csc \theta }$	${\displaystyle \cos \theta }$	${\displaystyle \sec \theta }$	${\displaystyle \tan \theta }$	${\displaystyle \cot \theta }$
${\displaystyle \sin \theta =}$	${\displaystyle \sin \theta }$	${\displaystyle {\frac {1}{\csc \theta }}}$	${\displaystyle \pm {\sqrt {1-\cos ^{2}\theta }}}$	${\displaystyle \pm {\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}}$	${\displaystyle \pm {\frac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}}}$	${\displaystyle \pm {\frac {1}{\sqrt {1+\cot ^{2}\theta }}}}$
${\displaystyle \csc \theta =}$	${\displaystyle {\frac {1}{\sin \theta }}}$	${\displaystyle \csc \theta }$	${\displaystyle \pm {\frac {1}{\sqrt {1-\cos ^{2}\theta }}}}$	${\displaystyle \pm {\frac {\sec \theta }{\sqrt {\sec ^{2}\theta -1}}}}$	${\displaystyle \pm {\frac {\sqrt {1+\tan ^{2}\theta }}{\tan \theta }}}$	${\displaystyle \pm {\sqrt {1+\cot ^{2}\theta }}}$
${\displaystyle \cos \theta =}$	${\displaystyle \pm {\sqrt {1-\sin ^{2}\theta }}}$	${\displaystyle \pm {\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}}$	${\displaystyle \cos \theta }$	${\displaystyle {\frac {1}{\sec \theta }}}$	${\displaystyle \pm {\frac {1}{\sqrt {1+\tan ^{2}\theta }}}}$	${\displaystyle \pm {\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}}$
${\displaystyle \sec \theta =}$	${\displaystyle \pm {\frac {1}{\sqrt {1-\sin ^{2}\theta }}}}$	${\displaystyle \pm {\frac {\csc \theta }{\sqrt {\csc ^{2}\theta -1}}}}$	${\displaystyle {\frac {1}{\cos \theta }}}$	${\displaystyle \sec \theta }$	${\displaystyle \pm {\sqrt {1+\tan ^{2}\theta }}}$	${\displaystyle \pm {\frac {\sqrt {1+\cot ^{2}\theta }}{\cot \theta }}}$
${\displaystyle \tan \theta =}$	${\displaystyle \pm {\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}}$	${\displaystyle \pm {\frac {1}{\sqrt {\csc ^{2}\theta -1}}}}$	${\displaystyle \pm {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}}$	${\displaystyle \pm {\sqrt {\sec ^{2}\theta -1}}}$	${\displaystyle \tan \theta }$	${\displaystyle {\frac {1}{\cot \theta }}}$
${\displaystyle \cot \theta =}$	${\displaystyle \pm {\frac {\sqrt {1-\sin ^{2}\theta }}{\sin \theta }}}$	${\displaystyle \pm {\sqrt {\csc ^{2}\theta -1}}}$	${\displaystyle \pm {\frac {\cos \theta }{\sqrt {1-\cos ^{2}\theta }}}}$	${\displaystyle \pm {\frac {1}{\sqrt {\sec ^{2}\theta -1}}}}$	${\displaystyle {\frac {1}{\tan \theta }}}$	${\displaystyle \cot \theta }$

By examining the unit circle, one can establish the following properties of the trigonometric functions.

When the direction of a Euclidean vector is represented by an angle ${\displaystyle \theta ,}$ this is the angle determined by the free vector (starting at the origin) and the positive ${\displaystyle x}$-unit vector. The same concept may also be applied to lines in an Euclidean space, where the angle is that determined by a parallel to the given line through the origin and the positive ${\displaystyle x}$-axis. If a line (vector) with direction ${\displaystyle \theta }$ is reflected about a line with direction ${\displaystyle \alpha ,}$ then the direction angle ${\displaystyle \theta ^{\prime }}$ of this reflected line (vector) has the value $${\displaystyle \theta ^{\prime }=2\alpha -\theta .}$$

The values of the trigonometric functions of these angles ${\displaystyle \theta ,\;\theta ^{\prime }}$ for specific angles ${\displaystyle \alpha }$ satisfy simple identities: either they are equal, or have opposite signs, or employ the complementary trigonometric function. These are also known as reduction formulae.^[2]

${\displaystyle \theta }$ reflected in ${\displaystyle \alpha =0}$[3] odd/even identities	${\displaystyle \theta }$ reflected in ${\displaystyle \alpha ={\frac {\pi }{4}}}$	${\displaystyle \theta }$ reflected in ${\displaystyle \alpha ={\frac {\pi }{2}}}$	${\displaystyle \theta }$ reflected in ${\displaystyle \alpha ={\frac {3\pi }{4}}}$	${\displaystyle \theta }$ reflected in ${\displaystyle \alpha =\pi }$ compare to ${\displaystyle \alpha =0}$
${\displaystyle \sin(-\theta )=-\sin \theta }$	${\displaystyle \sin \left({\tfrac {\pi }{2}}-\theta \right)=\cos \theta }$	${\displaystyle \sin(\pi -\theta )=+\sin \theta }$	${\displaystyle \sin \left({\tfrac {3\pi }{2}}-\theta \right)=-\cos \theta }$	${\displaystyle \sin(2\pi -\theta )=-\sin(\theta )=\sin(-\theta )}$
${\displaystyle \cos(-\theta )=+\cos \theta }$	${\displaystyle \cos \left({\tfrac {\pi }{2}}-\theta \right)=\sin \theta }$	${\displaystyle \cos(\pi -\theta )=-\cos \theta }$	${\displaystyle \cos \left({\tfrac {3\pi }{2}}-\theta \right)=-\sin \theta }$	${\displaystyle \cos(2\pi -\theta )=+\cos(\theta )=\cos(-\theta )}$
${\displaystyle \tan(-\theta )=-\tan \theta }$	${\displaystyle \tan \left({\tfrac {\pi }{2}}-\theta \right)=\cot \theta }$	${\displaystyle \tan(\pi -\theta )=-\tan \theta }$	${\displaystyle \tan \left({\tfrac {3\pi }{2}}-\theta \right)=+\cot \theta }$	${\displaystyle \tan(2\pi -\theta )=-\tan(\theta )=\tan(-\theta )}$
${\displaystyle \csc(-\theta )=-\csc \theta }$	${\displaystyle \csc \left({\tfrac {\pi }{2}}-\theta \right)=\sec \theta }$	${\displaystyle \csc(\pi -\theta )=+\csc \theta }$	${\displaystyle \csc \left({\tfrac {3\pi }{2}}-\theta \right)=-\sec \theta }$	${\displaystyle \csc(2\pi -\theta )=-\csc(\theta )=\csc(-\theta )}$
${\displaystyle \sec(-\theta )=+\sec \theta }$	${\displaystyle \sec \left({\tfrac {\pi }{2}}-\theta \right)=\csc \theta }$	${\displaystyle \sec(\pi -\theta )=-\sec \theta }$	${\displaystyle \sec \left({\tfrac {3\pi }{2}}-\theta \right)=-\csc \theta }$	${\displaystyle \sec(2\pi -\theta )=+\sec(\theta )=\sec(-\theta )}$
${\displaystyle \cot(-\theta )=-\cot \theta }$	${\displaystyle \cot \left({\tfrac {\pi }{2}}-\theta \right)=\tan \theta }$	${\displaystyle \cot(\pi -\theta )=-\cot \theta }$	${\displaystyle \cot \left({\tfrac {3\pi }{2}}-\theta \right)=+\tan \theta }$	${\displaystyle \cot(2\pi -\theta )=-\cot(\theta )=\cot(-\theta )}$

Shift by one quarter period	Shift by one half period	Shift by full periods[4]	Period
${\displaystyle \sin(\theta \pm {\tfrac {\pi }{2}})=\pm \cos \theta }$	${\displaystyle \sin(\theta +\pi )=-\sin \theta }$	${\displaystyle \sin(\theta +k\cdot 2\pi )=+\sin \theta }$	${\displaystyle 2\pi }$
${\displaystyle \cos(\theta \pm {\tfrac {\pi }{2}})=\mp \sin \theta }$	${\displaystyle \cos(\theta +\pi )=-\cos \theta }$	${\displaystyle \cos(\theta +k\cdot 2\pi )=+\cos \theta }$	${\displaystyle 2\pi }$
${\displaystyle \csc(\theta \pm {\tfrac {\pi }{2}})=\pm \sec \theta }$	${\displaystyle \csc(\theta +\pi )=-\csc \theta }$	${\displaystyle \csc(\theta +k\cdot 2\pi )=+\csc \theta }$	${\displaystyle 2\pi }$
${\displaystyle \sec(\theta \pm {\tfrac {\pi }{2}})=\mp \csc \theta }$	${\displaystyle \sec(\theta +\pi )=-\sec \theta }$	${\displaystyle \sec(\theta +k\cdot 2\pi )=+\sec \theta }$	${\displaystyle 2\pi }$
${\displaystyle \tan(\theta \pm {\tfrac {\pi }{4}})={\tfrac {\tan \theta \pm 1}{1\mp \tan \theta }}}$	${\displaystyle \tan(\theta +{\tfrac {\pi }{2}})=-\cot \theta }$	${\displaystyle \tan(\theta +k\cdot \pi )=+\tan \theta }$	${\displaystyle \pi }$
${\displaystyle \cot(\theta \pm {\tfrac {\pi }{4}})={\tfrac {\cot \theta \mp 1}{1\pm \cot \theta }}}$	${\displaystyle \cot(\theta +{\tfrac {\pi }{2}})=-\tan \theta }$	${\displaystyle \cot(\theta +k\cdot \pi )=+\cot \theta }$	${\displaystyle \pi }$

The sign of trigonometric functions depends on quadrant of the angle. If ${\displaystyle {-\pi }<\theta \leq \pi }$ and sgn is the sign function,

$${\displaystyle {\begin{aligned}\operatorname {sgn}(\sin \theta )=\operatorname {sgn}(\csc \theta )&={\begin{cases}+1&{\text{if}}\ \ 0<\theta <\pi \\-1&{\text{if}}\ \ {-\pi }<\theta <0\\0&{\text{if}}\ \ \theta \in \{0,\pi \}\end{cases}}\\[5mu]\operatorname {sgn}(\cos \theta )=\operatorname {sgn}(\sec \theta )&={\begin{cases}+1&{\text{if}}\ \ {-{\tfrac {1}{2}}\pi }<\theta <{\tfrac {1}{2}}\pi \\-1&{\text{if}}\ \ {-\pi }<\theta <-{\tfrac {1}{2}}\pi \ \ {\text{or}}\ \ {\tfrac {1}{2}}\pi <\theta <\pi \\0&{\text{if}}\ \ \theta \in {\bigl \{}{-{\tfrac {1}{2}}\pi },{\tfrac {1}{2}}\pi {\bigr \}}\end{cases}}\\[5mu]\operatorname {sgn}(\tan \theta )=\operatorname {sgn}(\cot \theta )&={\begin{cases}+1&{\text{if}}\ \ {-\pi }<\theta <-{\tfrac {1}{2}}\pi \ \ {\text{or}}\ \ 0<\theta <{\tfrac {1}{2}}\pi \\-1&{\text{if}}\ \ {-{\tfrac {1}{2}}\pi }<\theta <0\ \ {\text{or}}\ \ {\tfrac {1}{2}}\pi <\theta <\pi \\0&{\text{if}}\ \ \theta \in {\bigl \{}{-{\tfrac {1}{2}}\pi },0,{\tfrac {1}{2}}\pi ,\pi {\bigr \}}\end{cases}}\end{aligned}}}$$

The trigonometric functions are periodic with common period ${\displaystyle 2\pi ,}$ so for values of θ outside the interval ${\displaystyle ({-\pi },\pi ],}$ they take repeating values (see § Shifts and periodicity above).

These are also known as the angle addition and subtraction theorems (or formulae). $${\displaystyle {\begin{aligned}\sin(\alpha +\beta )&=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\\sin(\alpha -\beta )&=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\\cos(\alpha +\beta )&=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\\cos(\alpha -\beta )&=\cos \alpha \cos \beta +\sin \alpha \sin \beta \end{aligned}}}$$

The angle difference identities for ${\displaystyle \sin(\alpha -\beta )}$ and ${\displaystyle \cos(\alpha -\beta )}$ can be derived from the angle sum versions by substituting ${\displaystyle -\beta }$ for ${\displaystyle \beta }$ and using the facts that ${\displaystyle \sin(-\beta )=-\sin(\beta )}$ and ${\displaystyle \cos(-\beta )=\cos(\beta )}$. They can also be derived by using a slightly modified version of the figure for the angle sum identities, both of which are shown here.

These identities are summarized in the first two rows of the following table, which also includes sum and difference identities for the other trigonometric functions.

Sine	${\displaystyle \sin(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle \sin \alpha \cos \beta \pm \cos \alpha \sin \beta }$[5][6]
Cosine	${\displaystyle \cos(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle \cos \alpha \cos \beta \mp \sin \alpha \sin \beta }$[6][7]
Tangent	${\displaystyle \tan(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle {\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}}$[6][8]
Cosecant	${\displaystyle \csc(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle {\frac {\sec \alpha \sec \beta \csc \alpha \csc \beta }{\sec \alpha \csc \beta \pm \csc \alpha \sec \beta }}}$[9]
Secant	${\displaystyle \sec(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle {\frac {\sec \alpha \sec \beta \csc \alpha \csc \beta }{\csc \alpha \csc \beta \mp \sec \alpha \sec \beta }}}$[9]
Cotangent	${\displaystyle \cot(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle {\frac {\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha }}}$[6][10]
Arcsine	${\displaystyle \arcsin x\pm \arcsin y}$			${\displaystyle =}$	${\displaystyle \arcsin \left(x{\sqrt {1-y^{2}}}\pm y{\sqrt {1-x^{2}{\vphantom {y}}}}\right)}$[11]
Arccosine	${\displaystyle \arccos x\pm \arccos y}$			${\displaystyle =}$	${\displaystyle \arccos \left(xy\mp {\sqrt {\left(1-x^{2}\right)\left(1-y^{2}\right)}}\right)}$[12]
Arctangent	${\displaystyle \arctan x\pm \arctan y}$			${\displaystyle =}$	${\displaystyle \arctan \left({\frac {x\pm y}{1\mp xy}}\right)}$[13]
Arccotangent	${\displaystyle \operatorname {arccot} x\pm \operatorname {arccot} y}$			${\displaystyle =}$	${\displaystyle \operatorname {arccot} \left({\frac {xy\mp 1}{y\pm x}}\right)}$

When the series ${\textstyle \sum _{i=1}^{\infty }\theta _{i}}$ converges absolutely then

$${\displaystyle {\begin{aligned}{\sin }{\biggl (}\sum _{i=1}^{\infty }\theta _{i}{\biggl )}&=\sum _{{\text{odd}}\ k\geq 1}(-1)^{\frac {k-1}{2}}\!\!\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}{\biggl (}\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}{\biggr )}\\{\cos }{\biggl (}\sum _{i=1}^{\infty }\theta _{i}{\biggr )}&=\sum _{{\text{even}}\ k\geq 0}(-1)^{\frac {k}{2}}\,\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}{\biggl (}\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}{\biggr )}.\end{aligned}}}$$

Because the series ${\textstyle \sum _{i=1}^{\infty }\theta _{i}}$ converges absolutely, it is necessarily the case that ${\textstyle \lim _{i\to \infty }\theta _{i}=0,}$ ${\textstyle \lim _{i\to \infty }\sin \theta _{i}=0,}$ and ${\textstyle \lim _{i\to \infty }\cos \theta _{i}=1.}$ Particularly, in these two identities, an asymmetry appears that is not seen in the case of sums of finitely many angles: in each product, there are only finitely many sine factors but there are cofinitely many cosine factors. Terms with infinitely many sine factors would necessarily be equal to zero.

When only finitely many of the angles ${\displaystyle \theta _{i}}$ are nonzero then only finitely many of the terms on the right side are nonzero because all but finitely many sine factors vanish. Furthermore, in each term all but finitely many of the cosine factors are unity.

Let ${\displaystyle e_{k}}$ (for ${\displaystyle k=0,1,2,3,\ldots }$) be the kth-degree elementary symmetric polynomial in the variables $${\displaystyle x_{i}=\tan \theta _{i}}$$ for ${\displaystyle i=0,1,2,3,\ldots ,}$ that is,

$${\displaystyle {\begin{aligned}e_{0}&=1\\[6pt]e_{1}&=\sum _{i}x_{i}&&=\sum _{i}\tan \theta _{i}\\[6pt]e_{2}&=\sum _{i<j}x_{i}x_{j}&&=\sum _{i<j}\tan \theta _{i}\tan \theta _{j}\\[6pt]e_{3}&=\sum _{i<j<k}x_{i}x_{j}x_{k}&&=\sum _{i<j<k}\tan \theta _{i}\tan \theta _{j}\tan \theta _{k}\\&\ \ \vdots &&\ \ \vdots \end{aligned}}}$$

Then

$${\displaystyle \tan {\Bigl (}\sum _{i}\theta _{i}{\Bigr )}={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }}.}$$ This can be shown by using the sine and cosine sum formulae above: $${\displaystyle {\begin{aligned}\tan {\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {{\sin }{\bigl (}\sum _{i}\theta _{i}{\bigr )}/\prod _{i}\cos \theta _{i}}{{\cos }{\bigl (}\sum _{i}\theta _{i}{\bigr )}/\prod _{i}\cos \theta _{i}}}\\[10pt]&={\frac {\displaystyle \sum _{{\text{odd}}\ k\geq 1}(-1)^{\frac {k-1}{2}}\sum _{\begin{smallmatrix}A\subseteq \{1,2,3,\dots \}\\\left|A\right|=k\end{smallmatrix}}\prod _{i\in A}\tan \theta _{i}}{\displaystyle \sum _{{\text{even}}\ k\geq 0}~(-1)^{\frac {k}{2}}~~\sum _{\begin{smallmatrix}A\subseteq \{1,2,3,\dots \}\\\left|A\right|=k\end{smallmatrix}}\prod _{i\in A}\tan \theta _{i}}}={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }}\\[10pt]\cot {\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {e_{0}-e_{2}+e_{4}-\cdots }{e_{1}-e_{3}+e_{5}-\cdots }}\end{aligned}}}$$

The number of terms on the right side depends on the number of terms on the left side.

For example: $${\displaystyle {\begin{aligned}\tan(\theta _{1}+\theta _{2})&={\frac {e_{1}}{e_{0}-e_{2}}}={\frac {x_{1}+x_{2}}{1\ -\ x_{1}x_{2}}}={\frac {\tan \theta _{1}+\tan \theta _{2}}{1\ -\ \tan \theta _{1}\tan \theta _{2}}},\\[8pt]\tan(\theta _{1}+\theta _{2}+\theta _{3})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\[8pt]\tan(\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\[8pt]&={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}$$

and so on. The case of only finitely many terms can be proved by mathematical induction.^[14] The case of infinitely many terms can be proved by using some elementary inequalities.^[15]

Suppose ${\textstyle a,b,c,d,p,q\in \mathbb {R} }$ and ${\textstyle i={\sqrt {-1}}}$ and

${\displaystyle {\frac {ai+b}{ci+d}}=pi+q}$

and let ${\textstyle \varphi }$ be any number for which ${\textstyle \tan \varphi =c/d.}$ Suppose that ${\textstyle a/c\neq b/d}$ so that the forgoing fraction cannot be ${\textstyle 0/0}$. Then for all ${\textstyle \theta \in \mathbb {R} }$^[16]

${\displaystyle {\frac {a\tan \theta +b}{c\tan \theta +d}}=p\tan(\theta -\varphi )+q.}$

(In case the denominator of this fraction is 0, we take the value of the fraction to be ${\textstyle \infty }$, where the symbol ${\textstyle \infty }$ does not mean either ${\textstyle +\infty }$ or ${\textstyle -\infty }$, but is the ${\textstyle \infty }$ that is approached by going in either the positive or the negative direction, making the completion of the line ${\textstyle \mathbb {R} \cup \{\,\infty \,\}}$ topologically a circle.)

From this identity it can be shown to follow quickly that the family of all Cauchy-distributed random variables is closed under linear fractional tranformations, a result known since 1976.^[17]

$${\displaystyle {\begin{aligned}{\sec }{\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {\prod _{i}\sec \theta _{i}}{e_{0}-e_{2}+e_{4}-\cdots }}\\[8pt]{\csc }{\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {\prod _{i}\sec \theta _{i}}{e_{1}-e_{3}+e_{5}-\cdots }}\end{aligned}}}$$

where ${\displaystyle e_{k}}$ is the kth-degree elementary symmetric polynomial in the n variables ${\displaystyle x_{i}=\tan \theta _{i},}$ ${\displaystyle i=1,\ldots ,n,}$ and the number of terms in the denominator and the number of factors in the product in the numerator depend on the number of terms in the sum on the left.^[18] The case of only finitely many terms can be proved by mathematical induction on the number of such terms.

For example,

$${\displaystyle {\begin{aligned}\sec(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{1-\tan \alpha \tan \beta -\tan \alpha \tan \gamma -\tan \beta \tan \gamma }}\\[8pt]\csc(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{\tan \alpha +\tan \beta +\tan \gamma -\tan \alpha \tan \beta \tan \gamma }}.\end{aligned}}}$$

Ptolemy's theorem is important in the history of trigonometric identities, as it is how results equivalent to the sum and difference formulas for sine and cosine were first proved. It states that in a cyclic quadrilateral ${\displaystyle ABCD}$, as shown in the accompanying figure, the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. In the special cases of one of the diagonals or sides being a diameter of the circle, this theorem gives rise directly to the angle sum and difference trigonometric identities.^[19] The relationship follows most easily when the circle is constructed to have a diameter of length one, as shown here.

By Thales's theorem, ${\displaystyle \angle DAB}$ and ${\displaystyle \angle DCB}$ are both right angles. The right-angled triangles ${\displaystyle DAB}$ and ${\displaystyle DCB}$ both share the hypotenuse ${\displaystyle {\overline {BD}}}$ of length 1. Thus, the side ${\displaystyle {\overline {AB}}=\sin \alpha }$, ${\displaystyle {\overline {AD}}=\cos \alpha }$, ${\displaystyle {\overline {BC}}=\sin \beta }$ and ${\displaystyle {\overline {CD}}=\cos \beta }$.

By the inscribed angle theorem, the central angle subtended by the chord ${\displaystyle {\overline {AC}}}$ at the circle's center is twice the angle ${\displaystyle \angle ADC}$, i.e. ${\displaystyle 2(\alpha +\beta )}$. Therefore, the symmetrical pair of red triangles each has the angle ${\displaystyle \alpha +\beta }$ at the center. Each of these triangles has a hypotenuse of length ${\textstyle {\frac {1}{2}}}$, so the length of ${\displaystyle {\overline {AC}}}$ is ${\textstyle 2\times {\frac {1}{2}}\sin(\alpha +\beta )}$, i.e. simply ${\displaystyle \sin(\alpha +\beta )}$. The quadrilateral's other diagonal is the diameter of length 1, so the product of the diagonals' lengths is also ${\displaystyle \sin(\alpha +\beta )}$.

When these values are substituted into the statement of Ptolemy's theorem that ${\displaystyle |{\overline {AC}}|\cdot |{\overline {BD}}|=|{\overline {AB}}|\cdot |{\overline {CD}}|+|{\overline {AD}}|\cdot |{\overline {BC}}|}$, this yields the angle sum trigonometric identity for sine: ${\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta }$. The angle difference formula for ${\displaystyle \sin(\alpha -\beta )}$ can be similarly derived by letting the side ${\displaystyle {\overline {CD}}}$ serve as a diameter instead of ${\displaystyle {\overline {BD}}}$.^[19]

Tn is the nth Chebyshev polynomial	${\displaystyle \cos(n\theta )=T_{n}(\cos \theta )}$[20]
de Moivre's formula, i is the imaginary unit	${\displaystyle \cos(n\theta )+i\sin(n\theta )=(\cos \theta +i\sin \theta )^{n}}$[21]

Formulae for twice an angle.^[22]

Formulae for triple angles.^[22]

Formulae for multiple angles.^[23]