List of trigonometric identities

Reflections, shifts, and periodicity

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

Reflections

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 ${\displaystyle {\theta }^{\mathrm{\prime }}}$ of this reflected line (vector) has the value $${\displaystyle {\theta }^{\mathrm{\prime }}=2\alpha -\theta .}$$

The values of the trigonometric functions of these angles ${\displaystyle \theta ,{\theta }^{\mathrm{\prime }}}$ 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.

Shifts and periodicity

Signs

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

$${\displaystyle \begin{array}{cc}\mathrm{sgn}(\sin \theta )=\mathrm{sgn}(\csc \theta )& =\{\begin{array}{ll}+1& \text{if}0<\theta <\pi \\ -1& \text{if}-\pi <\theta <0\\ 0& \text{if}\theta \in \{0,\pi \}\end{array}\\ \mathrm{sgn}(\cos \theta )=\mathrm{sgn}(\sec \theta )& =\{\begin{array}{ll}+1& \text{if}-\frac{\pi }{2}<\theta <\frac{\pi }{2}\\ -1& \text{if}-\pi <\theta <-\frac{\pi }{2}\text{or}\frac{\pi }{2}<\theta <\pi \\ 0& \text{if}\theta \in \{-\frac{\pi }{2},\frac{\pi }{2}\}\end{array}\\ \mathrm{sgn}(\tan \theta )=\mathrm{sgn}(\cot \theta )& =\{\begin{array}{ll}+1& \text{if}-\pi <\theta <-\frac{\pi }{2}\text{or}0<\theta <\frac{\pi }{2}\\ -1& \text{if}-\frac{\pi }{2}<\theta <0\text{or}\frac{\pi }{2}<\theta <\pi \\ 0& \text{if}\theta \in \{-\frac{\pi }{2},0,\frac{\pi }{2},\pi \}\end{array}\end{array}}$$

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). The sign of a sinusoid or cosinusoid can be used to define a normalized square wave. For example, the functions ${\displaystyle \mathrm{sgn}(\sin x)}$ and ${\displaystyle \mathrm{sgn}(\cos x)}$ take values ±1 and correspond to square waves with a phase shift of ⁠π/2⁠.

Angle sum and difference identities

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

The angle difference identities for ${\displaystyle \sin (\alpha -\beta )}$ and ${\displaystyle \cos (\alpha -\beta )}$ can be derived from the angle sum versions (and vice versa) 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. They can also be seen as expressing the dot product and cross product of two vectors in terms of the cosine and the sine of the angle between them.

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.

Sines and cosines of sums of infinitely many angles

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

$${\displaystyle \begin{array}{cc}\sin (\sum _{i=1}^{\mathrm{\infty }}{\theta }_i)& =\sum _{\text{odd}k\ge 1}(-1{)}^{\frac{k-1}{2}}\sum _{\begin{array}{c}A\subseteq \{1,2,3,\dots \}\\ \left|A\right|=k\end{array}}(\prod _{i\in A}\sin {\theta }_i\prod _{i\notin A}\cos {\theta }_i)\\ \cos (\sum _{i=1}^{\mathrm{\infty }}{\theta }_i)& =\sum _{\text{even}k\ge 0}(-1{)}^{\frac{k}{2}}\sum _{\begin{array}{c}A\subseteq \{1,2,3,\dots \}\\ \left|A\right|=k\end{array}}(\prod _{i\in A}\sin {\theta }_i\prod _{i\notin A}\cos {\theta }_i).\end{array}}$$

Because the series $\sum _{i=1}^{\mathrm{\infty }}{\theta }_i$ converges absolutely, it is necessarily the case that $\underset{i\to \mathrm{\infty }}{lim}{\theta }_i=0,$ $\underset{i\to \mathrm{\infty }}{lim}\sin {\theta }_i=0,$ and $\underset{i\to \mathrm{\infty }}{lim}\cos {\theta }_i=1.$ 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.

Tangents and cotangents of sums

Let ${\displaystyle e_k}$ (for ${\displaystyle k=0,1,2,3,\dots }$) be the kth-degree elementary symmetric polynomial in the variables $${\displaystyle x_i=\tan {\theta }_i}$$ for ${\displaystyle i=0,1,2,3,\dots ,}$ that is,

$${\displaystyle \begin{array}{cc}{e}_0& =1\\ e_1& =\sum _i{x}_i& & =\sum _i\tan {\theta }_i\\ e_2& =\sum _{i<j}{x}_i{x}_j& & =\sum _{i<j}\tan {\theta }_i\tan {\theta }_j\\ e_3& =\sum _{i<j<k}{x}_i{x}_j{x}_k& & =\sum _{i<j<k}\tan {\theta }_i\tan {\theta }_j\tan {\theta }_k\\ & ⋮& & ⋮\end{array}}$$

Then

$${\displaystyle \tan (\sum _i{\theta }_i)=\frac{e_1-e_3+e_5-\cdots }{e_0-e_2+e_4-\cdots }.}$$ This can be shown by using the sine and cosine sum formulae above: $${\displaystyle \begin{array}{cc}\tan (\sum _i{\theta }_i)& =\frac{\sin (\sum _i{\theta }_i)/\prod _i\cos {\theta }_i}{\cos (\sum _i{\theta }_i)/\prod _i\cos {\theta }_i}\\ & =\frac{{\displaystyle \sum _{\text{odd}k\ge 1}(-1{)}^{\frac{k-1}{2}}\sum _{\begin{array}{c}A\subseteq \{1,2,3,\dots \}\\ \left|A\right|=k\end{array}}\prod _{i\in A}\tan {\theta }_i}}{{\displaystyle \sum _{\text{even}k\ge 0}(-1{)}^{\frac{k}{2}}\sum _{\begin{array}{c}A\subseteq \{1,2,3,\dots \}\\ \left|A\right|=k\end{array}}\prod _{i\in A}\tan {\theta }_i}}=\frac{e_1-e_3+e_5-\cdots }{e_0-e_2+e_4-\cdots }\\ \cot (\sum _i{\theta }_i)& =\frac{e_0-e_2+e_4-\cdots }{e_1-e_3+e_5-\cdots }\end{array}}$$

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

For example: $${\displaystyle \begin{array}{cc}\tan ({\theta }_1+{\theta }_2)& =\frac{e_1}{e_0-e_2}=\frac{x_1+x_2}{1-x_1{x}_2}=\frac{\tan {\theta }_1+\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2},\\ \tan ({\theta }_1+{\theta }_2+{\theta }_3)& =\frac{e_1-e_3}{e_0-e_2}=\frac{(x_1+x_2+x_3)-(x_1{x}_2{x}_3)}{1-(x_1{x}_2+x_1{x}_3+x_2{x}_3)},\\ \tan ({\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4)& =\frac{e_1-e_3}{e_0-e_2+e_4}\\ & =\frac{(x_1+x_2+x_3+x_4)-(x_1{x}_2{x}_3+x_1{x}_2{x}_4+x_1{x}_3{x}_4+x_2{x}_3{x}_4)}{1-(x_1{x}_2+x_1{x}_3+x_1{x}_4+x_2{x}_3+x_2{x}_4+x_3{x}_4)+(x_1{x}_2{x}_3{x}_4)},\end{array}}$$

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

Linear fractional transformations of tangents, related to tangents of sums

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

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

and let $\phi$ be any number for which $\tan \phi =\frac{c}{d}.$ Suppose that $\frac{a}{c}\ne \frac{b}{d}$ so that the forgoing fraction cannot be ⁠0/0⁠. Then for all $\theta \in \mathbb{R}$

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

(In case the denominator of this fraction is 0, we take the value of the fraction to be $\mathrm{\infty }$, where the symbol $\mathrm{\infty }$ does not mean either $+\mathrm{\infty }$ or $-\mathrm{\infty }$, but is the $\mathrm{\infty }$ that is approached by going in either the positive or the negative direction, making the completion of the line $\mathbb{R}\cup \{\mathrm{\infty }\}$ 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 transformations, a result known since 1976.

Secants and cosecants of sums

$${\displaystyle \begin{array}{cc}\sec (\sum _i{\theta }_i)& =\frac{\prod _i\sec {\theta }_i}{e_0-e_2+e_4-\cdots }\\ \csc (\sum _i{\theta }_i)& =\frac{\prod _i\sec {\theta }_i}{e_1-e_3+e_5-\cdots }\end{array}}$$

where ${\displaystyle e_k}$ is the kth-degree elementary symmetric polynomial in the n variables ${\displaystyle x_i=\tan {\theta }_i,}$ ${\displaystyle i=1,\dots ,n,}$ 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. The case of only finitely many terms can be proved by mathematical induction on the number of such terms.

For example,

$${\displaystyle \begin{array}{cc}\sec (\alpha +\beta +\gamma )& =\frac{\sec \alpha \sec \beta \sec \gamma }{1-\tan \alpha \tan \beta -\tan \alpha \tan \gamma -\tan \beta \tan \gamma }\\ \csc (\alpha +\beta +\gamma )& =\frac{\sec \alpha \sec \beta \sec \gamma }{\tan \alpha +\tan \beta +\tan \gamma -\tan \alpha \tan \beta \tan \gamma }.\end{array}}$$

Ptolemy's theorem

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 (but expressed in terms of chords). 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. The relationship follows most easily when the circle is constructed to have a diameter of length one, as shown here.

By Thales's theorem, ${\displaystyle \mathrm{\angle }DAB}$ and ${\displaystyle \mathrm{\angle }DCB}$ 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 ${\displaystyle \mathrm{\angle }ADC}$, 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 $\frac{1}{2}$, so the length of ${\displaystyle \overline{AC}}$ is $2\times \frac{1}{2}\sin (\alpha +\beta )$, 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 ${\displaystyle \overline{BD}}$.

Multiple-angle and half-angle formulas

Multiple-angle formulae

Double-angle formulae

Triple-angle formulae

Formulae for triple angles.

$${\displaystyle \begin{array}{cccc}\sin (3\theta )& =3\sin \theta -4{\sin }^3\theta & & =4\sin \theta \sin \left(\frac{\pi }{3}-\theta \right)\sin \left(\frac{\pi }{3}+\theta \right)\\ \cos (3\theta )& =4{\cos }^3\theta -3\cos \theta & & =4\cos \theta \cos \left(\frac{\pi }{3}-\theta \right)\cos \left(\frac{\pi }{3}+\theta \right)\\ \tan (3\theta )& =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }& & =\tan \theta \tan \left(\frac{\pi }{3}-\theta \right)\tan \left(\frac{\pi }{3}+\theta \right)\\ \cot (3\theta )& =\frac{3\cot \theta -{\cot }^3\theta }{1-3{\cot }^2\theta }\\ \sec (3\theta )& =\frac{{\sec }^3\theta }{4-3{\sec }^2\theta }\\ \csc (3\theta )& =\frac{{\csc }^3\theta }{3{\csc }^2\theta -4}\end{array}}$$

Multiple-angle formulae

Formulae for multiple angles.

$${\displaystyle \begin{array}{cc}\sin (n\theta )& =\sum _{k\in {\mathbb{O}}^{+}}^n(-1{)}^{\frac{k-1}{2}}(\genfrac{}{}{0ex}{}{n}{k}){\cos }^{n-k}\theta {\sin }^k\theta =\sin \theta \sum _{i=0}^{\frac{n+1}{2}}\sum _{j=0}^i(-1{)}^{i-j}(\genfrac{}{}{0ex}{}{n}{2i+1})(\genfrac{}{}{0ex}{}{i}{j}){\cos }^{n-2(i-j)-1}\theta \\ & =\sin (\theta )\sum _{k=0}^{\lfloor \frac{n-1}{2}\rfloor }(-1{)}^k(2\cos (\theta ){)}^{n-2k-1}(\genfrac{}{}{0ex}{}{n-k-1}{k})\\ & =2^{(n-1)}\prod _{k=0}^{n-1}\sin \left(\frac{k\pi }{n}+\theta \right)\\ \cos (n\theta )& =\sum _{k\in {\mathbb{E}}_0^{+}}^n(-1{)}^{\frac{k}{2}}(\genfrac{}{}{0ex}{}{n}{k}){\cos }^{n-k}\theta {\sin }^k\theta =\sum _{i=0}^{\frac{n}{2}}\sum _{j=0}^i(-1{)}^{i-j}(\genfrac{}{}{0ex}{}{n}{2i})(\genfrac{}{}{0ex}{}{i}{j}){\cos }^{n-2(i-j)}\theta \\ & =\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }(-1{)}^k{(2\cos (\theta ))}^{n-2k}(\genfrac{}{}{0ex}{}{n-k}{k})\frac{n}{2n-2k}\\ \cos ((2n+1)\theta )& =(-1{)}^n{2}^{2n}\prod _{k=0}^{2n}\cos \left(\frac{k\pi }{2n+1}-\theta \right)\\ \cos (2n\theta )& =(-1{)}^n{2}^{2n-1}\prod _{k=0}^{2n-1}\cos \left(\frac{(1+2k)\pi }{4n}-\theta \right)\\ \tan (n\theta )& =\frac{{\displaystyle \sum _{k\in {\mathbb{O}}^{+}}^n(-1{)}^{\frac{k-1}{2}}(\genfrac{}{}{0ex}{}{n}{k}){\tan }^k\theta }}{{\displaystyle \sum _{k\in {\mathbb{E}}_0^{+}}^n(-1{)}^{\frac{k}{2}}(\genfrac{}{}{0ex}{}{n}{k}){\tan }^k\theta }}\\ {\mathbb{O}}^{+}& =\text{Positive odd integers}\\ {\mathbb{E}}_0^{+}& =\text{Non-negative even integers}\end{array}}$$

Chebyshev method

The Chebyshev method is a recursive algorithm for finding the nth multiple angle formula knowing the ${\displaystyle (n-1)}$th and ${\displaystyle (n-2)}$th values.

${\displaystyle \cos (nx)}$ can be computed from ${\displaystyle \cos ((n-1)x)}$, ${\displaystyle \cos ((n-2)x)}$, and ${\displaystyle \cos (x)}$ with

$${\displaystyle \cos (nx)=2\cos x\cos ((n-1)x)-\cos ((n-2)x).}$$

This can be proved by adding together the formulae

$${\displaystyle \begin{array}{cc}\cos ((n-1)x+x)& =\cos ((n-1)x)\cos x-\sin ((n-1)x)\sin x\\ \cos ((n-1)x-x)& =\cos ((n-1)x)\cos x+\sin ((n-1)x)\sin x\end{array}}$$

It follows by induction that ${\displaystyle \cos (nx)}$ is a polynomial of ⁠${\displaystyle \cos x}$⁠, the so-called Chebyshev polynomial of the first kind, T_n ; thus, $${\displaystyle \cos (n\theta )=T_n(\cos \theta )}$$

Similarly, ${\displaystyle \sin (nx)}$ can be computed from ⁠${\displaystyle \sin ((n-1)x)}$⁠, ⁠${\displaystyle \sin ((n-2)x)}$⁠, and ${\displaystyle \cos x}$ with $${\displaystyle \sin (nx)=2\cos x\sin ((n-1)x)-\sin ((n-2)x)}$$

This can be proved by adding formulae for ${\displaystyle \sin ((n-1)x+x)}$ and ${\displaystyle \sin ((n-1)x-x).}$

Serving a purpose similar to that of the Chebyshev method, for the tangent we can write:

$${\displaystyle \tan (nx)=\frac{\tan ((n-1)x)+\tan x}{1-\tan ((n-1)x)\tan x}.}$$

Half-angle formulas

$${\displaystyle \begin{array}{cc}\sin \frac{\theta }{2}& =\mathrm{sgn}\left(\sin \frac{\theta }{2}\right)\sqrt{\frac{1-\cos \theta }{2}}\\ \cos \frac{\theta }{2}& =\mathrm{sgn}\left(\cos \frac{\theta }{2}\right)\sqrt{\frac{1+\cos \theta }{2}}\\ \tan \frac{\theta }{2}& =\frac{1-\cos \theta }{\sin \theta }=\frac{\sin \theta }{1+\cos \theta }=\csc \theta -\cot \theta =\frac{\tan \theta }{1+\sec \theta }\\ & =\mathrm{sgn}(\sin \theta )\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}=\frac{-1+\mathrm{sgn}(\cos \theta )\sqrt{1+{\tan }^2\theta }}{\tan \theta }\\ \cot \frac{\theta }{2}& =\frac{1+\cos \theta }{\sin \theta }=\frac{\sin \theta }{1-\cos \theta }=\csc \theta +\cot \theta =\mathrm{sgn}(\sin \theta )\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}\\ \sec \frac{\theta }{2}& =\mathrm{sgn}\left(\cos \frac{\theta }{2}\right)\sqrt{\frac{2}{1+\cos \theta }}\\ \csc \frac{\theta }{2}& =\mathrm{sgn}\left(\sin \frac{\theta }{2}\right)\sqrt{\frac{2}{1-\cos \theta }}\end{array}}$$

Also $${\displaystyle \begin{array}{cc}\tan \frac{\eta \pm \theta }{2}& =\frac{\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }\\ \tan \left(\frac{\theta }{2}+\frac{\pi }{4}\right)& =\sec \theta +\tan \theta \\ \sqrt{\frac{1-\sin \theta }{1+\sin \theta }}& =\frac{\left|1-\tan \frac{\theta }{2}\right|}{\left|1+\tan \frac{\theta }{2}\right|}\end{array}}$$

Table

These can be shown by using either the sum and difference identities or the multiple-angle formulae.

The fact that the triple-angle formula for sine and cosine only involves powers of a single function allows one to relate the geometric problem of a compass and straightedge construction of angle trisection to the algebraic problem of solving a cubic equation, which allows one to prove that trisection is in general impossible using the given tools.

A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x³ − 3x + d = 0, where x is the value of the cosine function at the one-third angle and d is the known value of the cosine function at the full angle. However, the discriminant of this equation is positive, so this equation has three real roots (of which only one is the solution for the cosine of the one-third angle). None of these solutions are reducible to a real algebraic expression, as they use intermediate complex numbers under the cube roots.

Power-reduction formula

Obtained by solving the second and third versions of the cosine double-angle formula.

In general terms of powers of ${\displaystyle \sin \theta }$ or ${\displaystyle \cos \theta }$ the following is true, and can be deduced using De Moivre's formula, Euler's formula and the binomial theorem.

Product-to-sum and sum-to-product identities

The product-to-sum identities or prosthaphaeresis formulae can be proven by expanding their right-hand sides using the angle addition theorems. Historically, the first four of these were known as Werner's formulas, after Johannes Werner who used them for astronomical calculations. See amplitude modulation for an application of the product-to-sum formulae, and beat (acoustics) and phase detector for applications of the sum-to-product formulae.

Product-to-sum identities

The product of two sines or cosines of different angles can be converted to a sum of trigonometric functions of a sum and difference of those angles:

$${\displaystyle \begin{array}{cc}\cos \theta \cos \phi & =\frac{1}{2}(\cos (\theta -\phi )+\cos (\theta +\phi )),\\ \sin \theta \sin \phi & =\frac{1}{2}(\cos (\theta -\phi )-\cos (\theta +\phi )),\\ \sin \theta \cos \phi & =\frac{1}{2}(\sin (\theta +\phi )+\sin (\theta -\phi )),\\ \cos \theta \sin \phi & =\frac{1}{2}(\sin (\theta +\phi )-\sin (\theta -\phi )).\end{array}}$$ As a corollary, the product or quotient of tangents can be converted to a quotient of sums of cosines or sines, respectively, $${\displaystyle \begin{array}{cc}\tan \theta \tan \phi & =\frac{\cos (\theta -\phi )-\cos (\theta +\phi )}{\cos (\theta -\phi )+\cos (\theta +\phi )},\\ \frac{\tan \theta }{\tan \phi }& =\frac{\sin (\theta +\phi )+\sin (\theta -\phi )}{\sin (\theta +\phi )-\sin (\theta -\phi )}.\end{array}}$$

More generally, for a product of any number of sines or cosines, $${\displaystyle \begin{array}{cc}\prod _{k=1}^n\cos {\theta }_k& =\frac{1}{2^n}\sum _{e\in S}\cos (e_1{\theta }_1+\cdots +e_n{\theta }_n)\\ & \text{where }e=(e_1,\dots ,e_n)\in S=\{1,-1{\}}^n,\\ \prod _{k=1}^n\sin {\theta }_k& =\frac{(-1{)}^{\lfloor \frac{n}{2}\rfloor }}{2^n}\{\begin{array}{l}{\displaystyle \sum _{e\in S}\cos (e_1{\theta }_1+\cdots +e_n{\theta }_n)\prod _{j=1}^n{e}_j\text{if}n\text{is even},}\\ {\displaystyle \sum _{e\in S}\sin (e_1{\theta }_1+\cdots +e_n{\theta }_n)\prod _{j=1}^n{e}_j\text{if}n\text{is odd}.}\end{array}\end{array}}$$

Sum-to-product identities

The sum of sines or cosines of two angles can be converted to a product of sines or cosines of the mean and half the difference of the angles:

$${\displaystyle \begin{array}{cc}\sin \theta +\sin \phi & =2\sin \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2},\\ \sin \theta -\sin \phi & =2\cos \frac{\theta +\phi }{2}\sin \frac{\theta -\phi }{2},\\ \cos \theta +\cos \phi & =2\cos \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2},\\ \cos \theta -\cos \phi & =-2\sin \frac{\theta +\phi }{2}\sin \frac{\theta -\phi }{2}.\end{array}}$$

The sum of the tangent of two angles can be converted to a quotient of the sine of angles divided by the product of the cosines: $${\displaystyle \tan \theta \pm \tan \phi =\frac{\sin (\theta \pm \phi )}{\cos \theta \cos \phi }.}$$

Hermite's cotangent identity

Charles Hermite demonstrated the following identity. Suppose ${\displaystyle a_1,\dots ,a_n}$ are complex numbers, no two of which differ by an integer multiple of π. Let

$${\displaystyle A_{n,k}=\prod _{\begin{array}{c}1\le j\le n\\ j\ne k\end{array}}\cot (a_k-a_j)}$$

(in particular, ${\displaystyle A_{1,1},}$ being an empty product, is 1). Then

$${\displaystyle \cot (z-a_1)\cdots \cot (z-a_n)=\cos \frac{n\pi }{2}+\sum _{k=1}^n{A}_{n,k}\cot (z-a_k).}$$

The simplest non-trivial example is the case n = 2:

$${\displaystyle \cot (z-a_1)\cot (z-a_2)=-1+\cot (a_1-a_2)\cot (z-a_1)+\cot (a_2-a_1)\cot (z-a_2).}$$

Finite products of trigonometric functions

For coprime integers n, m

$${\displaystyle \prod _{k=1}^n\left(2a+2\cos \left(\frac{2\pi km}{n}+x\right)\right)=2\left(T_n(a)+{(-1)}^{n+m}\cos (nx)\right)}$$

where T_n is the Chebyshev polynomial.

The following relationship holds for the sine function

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

More generally for an integer n > 0

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

or written in terms of the chord function $\mathrm{crd}x\equiv 2\sin \frac{1}{2}x$,

$${\displaystyle \mathrm{crd}(nx)=\prod _{k=1}^n\mathrm{crd}\left(\frac{2k\pi }{n}-x\right).}$$

This comes from the factorization of the polynomial $z^n-1$ into linear factors (cf. root of unity): For any complex z and an integer n > 0,

$${\displaystyle z^n-1=\prod _{k=1}^n\left(z-\exp \frac{2ki\pi }{n}\right).}$$

Linear combinations

For some purposes it is important to know that any linear combination of sine waves of the same period or frequency but different phase shifts is also a sine wave with the same period or frequency, but a different phase shift. This is useful in sinusoid data fitting, because the measured or observed data are linearly related to the a and b unknowns of the in-phase and quadrature components basis below, resulting in a simpler Jacobian, compared to that of ${\displaystyle c}$ and ${\displaystyle \phi }$.

Sine and cosine

The linear combination, or harmonic addition, of sine and cosine waves is equivalent to a single sine wave with a phase shift and scaled amplitude,

$${\displaystyle a\cos x+b\sin x=c\cos (x+\phi )}$$

where ${\displaystyle c}$ and ${\displaystyle \phi }$ are defined as so:

$${\displaystyle \begin{array}{cc}c& =\mathrm{sgn}(a)\sqrt{a^2+b^2},\\ \phi & =\arctan \left(-\frac{b}{a}\right),\end{array}}$$

given that ${\displaystyle a\ne 0.}$

Arbitrary phase shift

More generally, for arbitrary phase shifts, we have

$${\displaystyle a\sin (x+{\theta }_a)+b\sin (x+{\theta }_b)=c\sin (x+\phi )}$$

where ${\displaystyle c}$ and ${\displaystyle \phi }$ satisfy:

$${\displaystyle \begin{array}{cc}{c}^2& =a^2+b^2+2ab\cos \left({\theta }_a-{\theta }_b\right),\\ \tan \phi & =\frac{a\sin {\theta }_a+b\sin {\theta }_b}{a\cos {\theta }_a+b\cos {\theta }_b}.\end{array}}$$

More than two sinusoids

The general case reads

$${\displaystyle \sum _i{a}_i\sin (x+{\theta }_i)=a\sin (x+\theta ),}$$ where $${\displaystyle a^2=\sum _{i,j}{a}_i{a}_j\cos ({\theta }_i-{\theta }_j)}$$ and $${\displaystyle \tan \theta =\frac{\sum _i{a}_i\sin {\theta }_i}{\sum _i{a}_i\cos {\theta }_i}.}$$

Lagrange's trigonometric identities

These identities, named after Joseph Louis Lagrange, are: $${\displaystyle \begin{array}{cc}\sum _{k=0}^n\sin k\theta & =\frac{\cos \frac{1}{2}\theta -\cos \left(\left(n+\frac{1}{2}\right)\theta \right)}{2\sin \frac{1}{2}\theta }\\ \sum _{k=1}^n\cos k\theta & =\frac{-\sin \frac{1}{2}\theta +\sin \left(\left(n+\frac{1}{2}\right)\theta \right)}{2\sin \frac{1}{2}\theta }\end{array}}$$ for ${\displaystyle \theta \not\equiv 0(\mathrm{mod}2\pi ).}$

A related function is the Dirichlet kernel:

$${\displaystyle D_n(\theta )=1+2\sum _{k=1}^n\cos k\theta =\frac{\sin \left(\left(n+\frac{1}{2}\right)\theta \right)}{\sin \frac{1}{2}\theta }.}$$

A similar identity is

$${\displaystyle \sum _{k=1}^n\cos (2k-1)\alpha =\frac{\sin (2n\alpha )}{2\sin \alpha }.}$$

The proof is the following. By using the angle sum and difference identities, $${\displaystyle \sin (A+B)-\sin (A-B)=2\cos A\sin B.}$$ Then let's examine the following formula,

$${\displaystyle 2\sin \alpha \sum _{k=1}^n\cos (2k-1)\alpha =2\sin \alpha \cos \alpha +2\sin \alpha \cos 3\alpha +2\sin \alpha \cos 5\alpha +\cdots +2\sin \alpha \cos (2n-1)\alpha }$$ and this formula can be written by using the above identity,

$${\displaystyle \begin{array}{cc}& 2\sin \alpha \sum _{k=1}^n\cos (2k-1)\alpha \\ & =\sum _{k=1}^n(\sin (2k\alpha )-\sin (2(k-1)\alpha ))\\ & =(\sin 2\alpha -\sin 0)+(\sin 4\alpha -\sin 2\alpha )+(\sin 6\alpha -\sin 4\alpha )+\cdots +(\sin (2n\alpha )-\sin (2(n-1)\alpha ))\\ & =\sin (2n\alpha ).\end{array}}$$

So, dividing this formula with ${\displaystyle 2\sin \alpha }$ completes the proof.

Certain linear fractional transformations

If ${\displaystyle f(x)}$ is given by the linear fractional transformation $${\displaystyle f(x)=\frac{(\cos \alpha )x-\sin \alpha }{(\sin \alpha )x+\cos \alpha },}$$ and similarly $${\displaystyle g(x)=\frac{(\cos \beta )x-\sin \beta }{(\sin \beta )x+\cos \beta },}$$ then $${\displaystyle f(g(x))=g(f(x))=\frac{(\cos (\alpha +\beta ))x-\sin (\alpha +\beta )}{(\sin (\alpha +\beta ))x+\cos (\alpha +\beta )}.}$$

More tersely stated, if for all ${\displaystyle \alpha }$ we let ${\displaystyle f_{\alpha }}$ be what we called ${\displaystyle f}$ above, then $${\displaystyle f_{\alpha }\circ f_{\beta }=f_{\alpha +\beta }.}$$

If ${\displaystyle x}$ is the slope of a line, then ${\displaystyle f(x)}$ is the slope of its rotation through an angle of ${\displaystyle -\alpha .}$

Relation to the complex exponential function

Euler's formula states that, for any real number x: $${\displaystyle e^{ix}=\cos x+i\sin x,}$$ where i is the imaginary unit. Substituting −x for x gives us: $${\displaystyle e^{-ix}=\cos (-x)+i\sin (-x)=\cos x-i\sin x.}$$

These two equations can be used to solve for cosine and sine in terms of the exponential function. Specifically, $${\displaystyle \cos x=\frac{e^{ix}+e^{-ix}}{2}}$$ $${\displaystyle \sin x=\frac{e^{ix}-e^{-ix}}{2i}}$$

These formulae are useful for proving many other trigonometric identities. For example, that e^i(θ+φ) = e^iθ e^iφ means that

cos(θ + φ) + i sin(θ + φ) = (cos θ + i sin θ) (cos φ + i sin φ) = (cos θ cos φ − sin θ sin φ) + i (cos θ sin φ + sin θ cos φ).

That the real part of the left hand side equals the real part of the right hand side is an angle addition formula for cosine. The equality of the imaginary parts gives an angle addition formula for sine.

The following table expresses the trigonometric functions and their inverses in terms of the exponential function and the complex logarithm.

Relation to complex hyperbolic functions

Trigonometric functions may be deduced from hyperbolic functions with complex arguments. The formulae for the relations are shown below.$${\displaystyle \begin{array}{cc}\sin x& =-i\sinh (ix)\\ \cos x& =\cosh (ix)\\ \tan x& =-i\tanh (ix)\\ \cot x& =i\coth (ix)\\ \sec x& =\mathrm{sech}(ix)\\ \csc x& =i\mathrm{csch}(ix)\end{array}}$$

Series expansion

When using a power series expansion to define trigonometric functions, the following identities are obtained:

$${\displaystyle \begin{array}{cccc}\sin x& =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots & & =\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}\\ \cos x& =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots & & =\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\frac{x^{2n}}{(2n)!}\end{array}}$$

Infinite product formulae

For applications to special functions, the following infinite product formulae for trigonometric functions are useful:

$${\displaystyle \begin{array}{cccc}\sin x& =x\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{x^2}{{\pi }^2{n}^2}\right),& \cos x& =\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{x^2}{{\pi }^2\left(n-\frac{1}{2}\right){\phantom{)}}^2}\right),\\ \sinh x& =x\prod _{n=1}^{\mathrm{\infty }}\left(1+\frac{x^2}{{\pi }^2{n}^2}\right),& \cosh x& =\prod _{n=1}^{\mathrm{\infty }}\left(1+\frac{x^2}{{\pi }^2\left(n-\frac{1}{2}\right){\phantom{)}}^2}\right).\end{array}}$$

Inverse trigonometric functions

The following identities give the result of composing a trigonometric function with an inverse trigonometric function.

$${\displaystyle \begin{array}{cccccc}\sin (\arcsin x)& =x& \cos (\arcsin x)& =\sqrt{1-x^2}& \tan (\arcsin x)& =\frac{x}{\sqrt{1-x^2}}\\ \sin (\arccos x)& =\sqrt{1-x^2}& \cos (\arccos x)& =x& \tan (\arccos x)& =\frac{\sqrt{1-x^2}}{x}\\ \sin (\arctan x)& =\frac{x}{\sqrt{1+x^2}}& \cos (\arctan x)& =\frac{1}{\sqrt{1+x^2}}& \tan (\arctan x)& =x\\ \sin (\mathrm{arccsc}x)& =\frac{1}{x}& \cos (\mathrm{arccsc}x)& =\sqrt{1-\frac{1}{x^2}}& \tan (\mathrm{arccsc}x)& =\frac{1}{x\sqrt{1-\frac{1}{x^2}}}\\ \sin (\mathrm{arcsec}x)& =\sqrt{1-\frac{1}{x^2}}& \cos (\mathrm{arcsec}x)& =\frac{1}{x}& \tan (\mathrm{arcsec}x)& =x\sqrt{1-\frac{1}{x^2}}\\ \sin (\mathrm{arccot}x)& =\frac{1}{\sqrt{1+x^2}}& \cos (\mathrm{arccot}x)& =\frac{x}{\sqrt{1+x^2}}& \tan (\mathrm{arccot}x)& =\frac{1}{x}\end{array}}$$