### Trigonometric equation

In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, these are identities involving certain functions of one or more angles. They are distinct from triangle identities, which are identities involving both angles and side lengths of a triangle. Only the former are covered in this article.

These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is the integration of non-trigonometric functions: a common technique involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.

## Notation

### Angles

This article uses Greek letters such as alpha (α), beta (β), gamma (γ), and theta (θ) to represent angles. Several different units of angle measure are widely used, including degrees, radians, and grads:

1 full circle  = 360 degrees = 2$\pi$ radians  =  400 grads.

The following table shows the conversions for some common angles:

 Degrees Radians Grads Degrees Radians Grads 30° 60° 120° 150° 210° 240° 300° 330° $\frac\pi6\!$ $\frac\pi3\!$ $\frac\left\{2\pi\right\}3\!$ $\frac\left\{5\pi\right\}6\!$ $\frac\left\{7\pi\right\}6\!$ $\frac\left\{4\pi\right\}3\!$ $\frac\left\{5\pi\right\}3\!$ $\frac\left\{11\pi\right\}6\!$ 33⅓ grad 66⅔ grad 133⅓ grad 166⅔ grad 233⅓ grad 266⅔ grad 333⅓ grad 366⅔ grad 45° 90° 135° 180° 225° 270° 315° 360° $\frac\pi4\!$ $\frac\pi2\!$ $\frac\left\{3\pi\right\}4\!$ $\pi\!$ $\frac\left\{5\pi\right\}4\!$ $\frac\left\{3\pi\right\}2\!$ $\frac\left\{7\pi\right\}4\!$ $2\pi\!$ 50 grad 100 grad 150 grad 200 grad 250 grad 300 grad 350 grad 400 grad

Unless otherwise specified, all angles in this article are assumed to be in radians, but angles ending in a degree symbol (°) are in degrees. Per Niven's theorem multiples of 30° are the only rational angles with rational sin/cos, which may account for their popularity in examples.[1]

### Trigonometric functions

The primary trigonometric functions are the sine and cosine of an angle. These are sometimes abbreviated sin(θ) and cos(θ), respectively, where θ is the angle, but the parentheses around the angle are often omitted, e.g., sin θ and cos θ.

The tangent (tan) of an angle is the ratio of the sine to the cosine:

$\tan\theta = \frac\left\{\sin\theta\right\}\left\{\cos\theta\right\}.$

Finally, the reciprocal functions secant (sec), cosecant (csc), and cotangent (cot) are the reciprocals of the cosine, sine, and tangent:

$\sec\theta = \frac\left\{1\right\}\left\{\cos\theta\right\},\quad\csc\theta = \frac\left\{1\right\}\left\{\sin\theta\right\},\quad\cot\theta=\frac\left\{1\right\}\left\{\tan\theta\right\}=\frac\left\{\cos\theta\right\}\left\{\sin\theta\right\}.$

These definitions are sometimes referred to as ratio identities.

## Inverse functions

The inverse trigonometric functions are partial inverse functions for the trigonometric functions. For example, the inverse function for the sine, known as the inverse sine (sin−1) or arcsine (arcsin or asin), satisfies

$\sin\left(\arcsin x\right) = x\quad\text\left\{for\right\} \quad |x| \leq 1$

and

$\arcsin\left(\sin x\right) = x\quad\text\left\{for\right\} \quad |x| \leq \pi/2.$

 Function Inverse sin cos tangent sec csc cot arcsin arccos arctan arcsec arccsc arccot

## Pythagorean identity

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

$\cos^2\theta + \sin^2\theta = 1\!$

where cos2 θ means (cos(θ))2 and sin2 θ means (sin(θ))2.

This can be viewed as a version of the Pythagorean theorem, and follows from the equation x2 + y2 = 1 for the unit circle. This equation can be solved for either the sine or the cosine:

$\sin\theta = \pm \sqrt\left\{1-\cos^2\theta\right\} \quad \text\left\{and\right\} \quad \cos\theta = \pm \sqrt\left\{1 - \sin^2\theta\right\}. \,$

### Related identities

Dividing the Pythagorean identity through by either cos2 θ or sin2 θ yields two other identities:

$1 + \tan^2\theta = \sec^2\theta\quad\text\left\{and\right\}\quad 1 + \cot^2\theta = \csc^2\theta.\!$

Using these identities together with the ratio 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 the other five.[2]
in terms of $\sin \theta\!$ $\cos \theta\!$ $\tan \theta\!$ $\csc \theta\!$ $\sec \theta\!$ $\cot \theta\!$
$\sin \theta =\!$ $\sin \theta\$ $\pm\sqrt\left\{1 - \cos^2 \theta\right\}\!$ $\pm\frac\left\{\tan \theta\right\}\left\{\sqrt\left\{1 + \tan^2 \theta\right\}\right\}\!$ $\frac\left\{1\right\}\left\{\csc \theta\right\}\!$ $\pm\frac\left\{\sqrt\left\{\sec^2 \theta - 1\right\}\right\}\left\{\sec \theta\right\}\!$ $\pm\frac\left\{1\right\}\left\{\sqrt\left\{1 + \cot^2 \theta\right\}\right\}\!$
$\cos \theta =\!$ $\pm\sqrt\left\{1 - \sin^2\theta\right\}\!$ $\cos \theta\!$ $\pm\frac\left\{1\right\}\left\{\sqrt\left\{1 + \tan^2 \theta\right\}\right\}\!$ $\pm\frac\left\{\sqrt\left\{\csc^2 \theta - 1\right\}\right\}\left\{\csc \theta\right\}\!$ $\frac\left\{1\right\}\left\{\sec \theta\right\}\!$ $\pm\frac\left\{\cot \theta\right\}\left\{\sqrt\left\{1 + \cot^2 \theta\right\}\right\}\!$
$\tan \theta =\!$ $\pm\frac\left\{\sin \theta\right\}\left\{\sqrt\left\{1 - \sin^2 \theta\right\}\right\}\!$ $\pm\frac\left\{\sqrt\left\{1 - \cos^2 \theta\right\}\right\}\left\{\cos \theta\right\}\!$ $\tan \theta\!$ $\pm\frac\left\{1\right\}\left\{\sqrt\left\{\csc^2 \theta - 1\right\}\right\}\!$ $\pm\sqrt\left\{\sec^2 \theta - 1\right\}\!$ $\frac\left\{1\right\}\left\{\cot \theta\right\}\!$
$\csc \theta =\!$ $\frac\left\{1\right\}\left\{\sin \theta\right\}\!$ $\pm\frac\left\{1\right\}\left\{\sqrt\left\{1 - \cos^2 \theta\right\}\right\}\!$ $\pm\frac\left\{\sqrt\left\{1 + \tan^2 \theta\right\}\right\}\left\{\tan \theta\right\}\!$ $\csc \theta\!$ $\pm\frac\left\{\sec \theta\right\}\left\{\sqrt\left\{\sec^2 \theta - 1\right\}\right\}\!$ $\pm\sqrt\left\{1 + \cot^2 \theta\right\}\!$
$\sec \theta =\!$ $\pm\frac\left\{1\right\}\left\{\sqrt\left\{1 - \sin^2 \theta\right\}\right\}\!$
$\frac\left\{1\right\}\left\{\cos \theta\right\}\!$ $\pm\sqrt\left\{1 + \tan^2 \theta\right\}\!$ $\pm\frac\left\{\csc \theta\right\}\left\{\sqrt\left\{\csc^2 \theta - 1\right\}\right\}\!$ $\sec \theta\!$ $\pm\frac\left\{\sqrt\left\{1 + \cot^2 \theta\right\}\right\}\left\{\cot \theta\right\}\!$
$\cot \theta =\!$ $\pm\frac\left\{\sqrt\left\{1 - \sin^2 \theta\right\}\right\}\left\{\sin \theta\right\}\!$ $\pm\frac\left\{\cos \theta\right\}\left\{\sqrt\left\{1 - \cos^2 \theta\right\}\right\}\!$ $\frac\left\{1\right\}\left\{\tan \theta\right\}\!$ $\pm\sqrt\left\{\csc^2 \theta - 1\right\}\!$ $\pm\frac\left\{1\right\}\left\{\sqrt\left\{\sec^2 \theta - 1\right\}\right\}\!$ $\cot \theta\!$

## Historic shorthands

The versine, coversine, haversine, and exsecant were used in navigation. For example the haversine formula was used to calculate the distance between two points on a sphere. They are rarely used today.

Name(s) Abbreviation(s) Value[3]
versed sine, versine $\operatorname\left\{versin\right\}\left(\theta\right)$
$\operatorname\left\{vers\right\}\left(\theta\right)$
$\operatorname\left\{ver\right\}\left(\theta\right)$
$1 - \cos \left(\theta\right)$
versed cosine, vercosine $\operatorname\left\{vercosin\right\}\left(\theta\right)$ $1 + \cos \left(\theta\right)$
coversed sine, coversine $\operatorname\left\{coversin\right\}\left(\theta\right)$
$\operatorname\left\{cvs\right\}\left(\theta\right)$
$1 - \sin\left(\theta\right)$
coversed cosine, covercosine $\operatorname\left\{covercosin\right\}\left(\theta\right)$ $1 + \sin\left(\theta\right)$
half versed sine, haversine $\operatorname\left\{haversin\right\}\left(\theta\right)$ $\frac\left\{1 - \cos \left(\theta\right)\right\}\left\{2\right\}$
half versed cosine, havercosine $\operatorname\left\{havercosin\right\}\left(\theta\right)$ $\frac\left\{1 + \cos \left(\theta\right)\right\}\left\{2\right\}$
half coversed sine, hacoversine
cohaversine
$\operatorname\left\{hacoversin\right\}\left(\theta\right)$ $\frac\left\{1 - \sin \left(\theta\right)\right\}\left\{2\right\}$
half coversed cosine, hacovercosine
cohavercosine
$\operatorname\left\{hacovercosin\right\}\left(\theta\right)$ $\frac\left\{1 + \sin \left(\theta\right)\right\}\left\{2\right\}$
exterior secant, exsecant $\operatorname\left\{exsec\right\}\left(\theta\right)$ $\sec\left(\theta\right) - 1$
exterior cosecant, excosecant $\operatorname\left\{excsc\right\}\left(\theta\right)$ $\csc\left(\theta\right) - 1$
chord $\operatorname\left\{crd\right\}\left(\theta\right)$ $2\sin\left\left(\frac\left\{\theta\right\}\left\{2\right\}\right\right)$

Ancient Indian mathematicians used Sanskrit terms Jyā, koti-jyā and utkrama-jyā, based on the resemblance of the chord, arc, and radius to the shape of a bow and bowstring drawn back.

## Symmetry, shifts, and periodicity

By examining the unit circle, the following properties of the trigonometric functions can be established.

### Symmetry

When the trigonometric functions are reflected from certain angles, the result is often one of the other trigonometric functions. This leads to the following identities:

Reflected in $\theta=0$[4] Reflected in $\theta= \pi/2$
(co-function identities)[5]
Reflected in $\theta= \pi$


\begin{align} \sin(-\theta) &= -\sin \theta \\ \cos(-\theta) &= +\cos \theta \\ \tan(-\theta) &= -\tan \theta \\ \csc(-\theta) &= -\csc \theta \\ \sec(-\theta) &= +\sec \theta \\ \cot(-\theta) &= -\cot \theta \end{align}



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



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

### Shifts and periodicity

By shifting the function round by certain angles, it is often possible to find different trigonometric functions that express particular results more simply. Some examples of this are shown by shifting functions round by π/2, π and 2π radians. Because the periods of these functions are either π or 2π, there are cases where the new function is exactly the same as the old function without the shift.

Shift by π/2 Shift by π
Period for tan and cot[6]
Shift by 2π
Period for sin, cos, csc and sec[7]


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



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



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

## Angle sum and difference identities

These are also known as the addition and subtraction theorems or formulae. They were originally established by the 10th century Persian mathematician Abū al-Wafā' Būzjānī. One method of proving these identities is to apply Euler's formula. The use of the symbols $\pm$ and $\mp$ is described in the article plus-minus sign.

 Sine $\sin\left(\alpha \pm \beta\right) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \!$[8][9] $\cos\left(\alpha \pm \beta\right) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta\,$[9][10] $\tan\left(\alpha \pm \beta\right) = \frac\left\{\tan \alpha \pm \tan \beta\right\}\left\{1 \mp \tan \alpha \tan \beta\right\}$[9][11] $\arcsin\alpha \pm \arcsin\beta = \arcsin\left\left(\alpha\sqrt\left\{1-\beta^2\right\} \pm \beta\sqrt\left\{1-\alpha^2\right\}\right\right)$[12] $\arccos\alpha \pm \arccos\beta = \arccos\left\left(\alpha\beta \mp \sqrt\left\{\left(1-\alpha^2\right)\left(1-\beta^2\right)\right\}\right\right)$[13] $\arctan\alpha \pm \arctan\beta = \arctan\left\left(\frac\left\{\alpha \pm \beta\right\}\left\{1 \mp \alpha\beta\right\}\right\right)$[14]

### Matrix form

The sum and difference formulae for sine and cosine can be written in matrix form as:



 \cos\alpha    & -\sin\alpha  \\
\sin\alpha & \cos\alpha


\end{array}\right) \left(\begin{array}{rr}

 \cos\beta    & -\sin\beta  \\
\sin\beta & \cos\beta


\end{array}\right) \\[12pt] & = \left(\begin{array}{rr}

 \cos\alpha\cos\beta - \sin\alpha\sin\beta & -\cos\alpha\sin\beta - \sin\alpha\cos\beta \\
\sin\alpha\cos\beta + \cos\alpha\sin\beta & -\sin\alpha\sin\beta + \cos\alpha\cos\beta


\end{array}\right) \\[12pt] & = \left(\begin{array}{rr}

 \cos(\alpha+\beta) & -\sin(\alpha+\beta) \\
\sin(\alpha+\beta) & \cos(\alpha+\beta)


\end{array}\right) \end{align} This shows that these matrices form a representation of the rotation group in the plane (technically, the special orthogonal group SO(2)), since the composition law is fulfilled: subsequent multiplications of a vector with these two matrices yields the same result as the rotation by the sum of the angles.

### Sines and cosines of sums of infinitely many terms

$\sin\left\left(\sum_\left\{i=1\right\}^\infty \theta_i\right\right)$

=\sum_{\text{odd}\ k \ge 1} (-1)^{(k-1)/2} \sum_{\begin{smallmatrix} A \subseteq \{\,1,2,3,\dots\,\} \\ \left|A\right| = k\end{smallmatrix}} \left(\prod_{i \in A} \sin\theta_i \prod_{i \not \in A} \cos\theta_i\right)

$\cos\left\left(\sum_\left\{i=1\right\}^\infty \theta_i\right\right)$

=\sum_{\text{even}\ k \ge 0} ~ (-1)^{k/2} ~~ \sum_{\begin{smallmatrix} A \subseteq \{\,1,2,3,\dots\,\} \\ \left|A\right| = k\end{smallmatrix}} \left(\prod_{i \in A} \sin\theta_i \prod_{i \not \in A} \cos\theta_i\right)

In these two identities an asymmetry appears that is not seen in the case of sums of finitely many terms: in each product, there are only finitely many sine factors and cofinitely many cosine factors.

If only finitely many of the terms θi are nonzero, then only finitely many of the terms on the right side will be nonzero because sine factors will vanish, and in each term, all but finitely many of the cosine factors will be unity.

### Tangents of sums

Let ek (for k = 0, 1, 2, 3, ...) be the kth-degree elementary symmetric polynomial in the variables

$x_i = \tan \theta_i\,$

for i = 0, 1, 2, 3, ..., i.e.,



\begin{align} 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{align}

Then

$\tan\left\left(\sum_i \theta_i\right\right) = \frac\left\{e_1 - e_3 + e_5 -\cdots\right\}\left\{e_0 - e_2 + e_4 - \cdots\right\}.\!$

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

For example:

\begin\left\{align\right\}

\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_1x_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{align}

and so on. The case of only finitely many terms can be proved by mathematical induction.[15]

### Secants and cosecants of sums



\begin{align} \sec\left(\sum_i \theta_i\right) & = \frac{\prod_i \sec\theta_i}{e_0 - e_2 + e_4 - \cdots} \\[8pt] \csc\left(\sum_i \theta_i \right) & = \frac{\prod_i \sec\theta_i }{e_1 - e_3 + e_5 - \cdots} \end{align}

where ek is the kth-degree elementary symmetric polynomial in the n variables xi = tan θi, i = 1, ..., 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. The convergence of the series in the denominators can be shown by writing the secant identity in the form

$e_0 - e_2 + e_4 - \cdots = \frac\left\{\prod_i \sec\theta_i\right\}\left\{\sec\left\left(\sum_i \theta_i\right\right)\right\}$

and then observing that the left side converges if the right side converges, and similarly for the cosecant identity.

For example,



\begin{align} \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{align}

## Multiple-angle formulae

 Tn is the nth Chebyshev polynomial $\cos n\theta =T_n \left(\cos \theta \right)\,$  [16] $\sin^2 n\theta = S_n \left(\sin^2\theta\right)\,$ $\cos n\theta +i\sin n\theta=\left(\cos\left(\theta\right)+i\sin\left(\theta\right)\right)^n \,$    [17]

### Double-angle, triple-angle, and half-angle formulae

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

Double-angle formulae[18][19]
\begin\left\{align\right\}

\sin 2\theta &= 2 \sin \theta \cos \theta \ \\ &= \frac{2 \tan \theta} {1 + \tan^2 \theta} \end{align}

\begin\left\{align\right\}

\cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ &= 2 \cos^2 \theta - 1 \\ &= 1 - 2 \sin^2 \theta \\ &= \frac{1 - \tan^2 \theta} {1 + \tan^2 \theta} \end{align}

$\tan 2\theta = \frac\left\{2 \tan \theta\right\} \left\{1 - \tan^2 \theta\right\}\!$ $\cot 2\theta = \frac\left\{\cot^2 \theta - 1\right\}\left\{2 \cot \theta\right\}\!$
Triple-angle formulae[16][20]
\begin\left\{align\right\}\sin 3\theta & = - \sin^3\theta + 3 \cos^2\theta \sin\theta\\

& = - 4\sin^3\theta + 3\sin\theta \end{align}

\begin\left\{align\right\}\cos 3\theta & = \cos^3\theta - 3 \sin^2 \theta\cos \theta \\

& = 4 \cos^3\theta - 3 \cos\theta\end{align}

$\tan 3\theta = \frac\left\{3 \tan\theta - \tan^3\theta\right\}\left\{1 - 3 \tan^2\theta\right\}\!$ $\cot 3\theta = \frac\left\{3 \cot\theta - \cot^3\theta\right\}\left\{1 - 3 \cot^2\theta\right\}\!$
Half-angle formulae[21][22]
\begin\left\{align\right\}&\sin \frac\left\{\theta\right\}\left\{2\right\} = \sgn \!\! \left\left( \!\! 2 \pi \! - \! \theta \! + \! 4 \pi \! \left\lfloor \! \frac\left\{\theta\right\}\left\{4\pi\right\} \! \right\rfloor \! \right\right) \!\! \sqrt\left\{\frac\left\{1 \! - \! \cos \theta\right\}\left\{2\right\}\right\} \\ \\

&\left(\mathrm{or}\,\,\sin^2\frac{\theta}{2}=\frac{1-\cos\theta}{2}\right)\end{align}

\begin\left\{align\right\}&\cos \frac\left\{\theta\right\}\left\{2\right\} = \sgn \!\! \left\left(\!\! \pi \! + \! \theta \! + \! 4 \pi \! \left\lfloor \! \frac\left\{\pi \! - \! \theta\right\}\left\{4\pi\right\} \! \right\rfloor \! \right\right) \!\! \sqrt\left\{\frac\left\{1 + \cos\theta\right\}\left\{2\right\}\right\} \\ \\

&\left(\mathrm{or}\,\,\cos^2\frac{\theta}{2}=\frac{1+\cos\theta}{2}\right)\end{align}

\begin\left\{align\right\} \tan \frac\left\{\theta\right\}\left\{2\right\} &= \csc \theta - \cot \theta \\ &= \pm\, \sqrt\left\{1 - \cos \theta \over 1 + \cos \theta\right\} \\\left[8pt\right] &= \frac\left\{\sin \theta\right\}\left\{1 + \cos \theta\right\} \\\left[8pt\right] &= \frac\left\{1-\cos \theta\right\}\left\{\sin \theta\right\} \\\left[10pt\right]

\tan\frac{\eta+\theta}{2} & = \frac{\sin\eta+\sin\theta}{\cos\eta+\cos\theta} \\[8pt] \tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right) & = \sec\theta + \tan\theta \\[8pt] \sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} & = \frac{1 - \tan(\theta/2)}{1 + \tan(\theta/2)} \\[8pt] \tan\tfrac{1}{2}\theta & = \frac{\tan\theta}{1 + \sqrt{1+\tan^2\theta}} \\ &\mbox{for}\quad \theta \in \left(-\tfrac{\pi}{2},\tfrac{\pi}{2} \right) \end{align}

\begin\left\{align\right\} \cot \frac\left\{\theta\right\}\left\{2\right\} &= \csc \theta + \cot \theta \\ &= \pm\, \sqrt\left\{1 + \cos \theta \over 1 - \cos \theta\right\} \\\left[8pt\right] &= \frac\left\{\sin \theta\right\}\left\{1 - \cos \theta\right\} \\\left[8pt\right] &= \frac\left\{1 + \cos \theta\right\}\left\{\sin \theta\right\} \end\left\{align\right\}

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 this is in general impossible using the given tools, by field theory.

A formula for computing the trigonometric identities for the third-angle exists, but it requires finding the zeroes of the cubic equation $x^3 - \frac\left\{3x+d\right\}\left\{4\right\}=0$, where x is the value of the sine function at some angle and d is the known value of the sine function at the triple angle. However, the discriminant of this equation is negative, so this equation has three real roots (of which only one is the solution within the correct third-circle) but none of these solutions is reducible to a real algebraic expression, as they use intermediate complex numbers under the cube roots, (which may be expressed in terms of real-only functions only if using hyperbolic functions).

### Sine, cosine, and tangent of multiple angles

For specific multiples, these follow from the angle addition formulas, while the general formula was given by 16th century French mathematician Vieta.

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

In each of these two equations, the first parenthesized term is a binomial coefficient, and the final trigonometric function equals one or minus one or zero so that half the entries in each of the sums are removed. tan  can be written in terms of tan θ using the recurrence relation:

$\tan\,\left(n\left\{+\right\}1\right)\theta = \frac\left\{\tan n\theta + \tan \theta\right\}\left\{1 - \tan n\theta\,\tan \theta\right\}.$

cot  can be written in terms of cot θ using the recurrence relation:

$\cot\,\left(n\left\{+\right\}1\right)\theta = \frac\left\{\cot n\theta\,\cot \theta - 1\right\}\left\{\cot n\theta + \cot \theta\right\}.$

### Chebyshev method

The Chebyshev method is a recursive algorithm for finding the nth multiple angle formula knowing the (n − 1)th and (n − 2)th formulae.[23]

The cosine for nx can be computed from the cosine of (n − 1)x and (n − 2)x as follows:

$\cos nx = 2 \cdot \cos x \cdot \cos \left(\left(n-1\right) x\right) - \cos \left(\left(n-2\right) x\right) \,$

Similarly sin(nx) can be computed from the sines of (n − 1)x and (n − 2)x

$\sin nx = 2 \cdot \cos x \cdot \sin \left(\left(n-1\right) x\right) - \sin \left(\left(n-2\right) x\right) \,$

For the tangent, we have:

$\tan nx = \frac\left\{H + K \tan x\right\}\left\{K- H \tan x\right\} \,$

where H/K = tan(n − 1)x.

### Tangent of an average

$\tan\left\left( \frac\left\{\alpha+\beta\right\}\left\{2\right\} \right\right)$

= \frac{\sin\alpha + \sin\beta}{\cos\alpha + \cos\beta} = -\,\frac{\cos\alpha - \cos\beta}{\sin\alpha - \sin\beta}

Setting either α or β to 0 gives the usual tangent half-angle formulæ.

### Viète's infinite product

$\cos\left\left(\left\{\theta \over 2\right\}\right\right) \cdot \cos\left\left(\left\{\theta \over 4\right\}\right\right)$

\cdot \cos\left({\theta \over 8}\right)\cdots = \prod_{n=1}^\infty \cos\left({\theta \over 2^n}\right) = {\sin(\theta)\over \theta} = \operatorname{sinc}\,\theta.

## Power-reduction formula

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

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

and in general terms of powers of sin θ or cos θ the following is true, and can be deduced using De Moivre's formula, Euler's formula and binomial theorem.

Cosine Sine
$\text\left\{if \right\}n\text\left\{ is odd\right\}$ $\cos^n\theta = \frac\left\{2\right\}\left\{2^n\right\} \sum_\left\{k=0\right\}^\left\{\frac\left\{n-1\right\}\left\{2\right\}\right\} \binom\left\{n\right\}\left\{k\right\} \cos\left\{\left(\left(n-2k\right)\theta\right)\right\}$ $\sin^n\theta = \frac\left\{2\right\}\left\{2^n\right\} \sum_\left\{k=0\right\}^\left\{\frac\left\{n-1\right\}\left\{2\right\}\right\} \left(-1\right)^\left\{\left(\frac\left\{n-1\right\}\left\{2\right\}-k\right)\right\} \binom\left\{n\right\}\left\{k\right\} \sin\left\{\left(\left(n-2k\right)\theta\right)\right\}$
$\text\left\{if \right\}n\text\left\{ is even\right\}$ $\cos^n\theta = \frac\left\{1\right\}\left\{2^n\right\} \binom\left\{n\right\}\left\{\frac\left\{n\right\}\left\{2\right\}\right\} + \frac\left\{2\right\}\left\{2^n\right\} \sum_\left\{k=0\right\}^\left\{\frac\left\{n\right\}\left\{2\right\}-1\right\} \binom\left\{n\right\}\left\{k\right\} \cos\left\{\left(\left(n-2k\right)\theta\right)\right\}$ $\sin^n\theta = \frac\left\{1\right\}\left\{2^n\right\} \binom\left\{n\right\}\left\{\frac\left\{n\right\}\left\{2\right\}\right\} + \frac\left\{2\right\}\left\{2^n\right\} \sum_\left\{k=0\right\}^\left\{\frac\left\{n\right\}\left\{2\right\}-1\right\} \left(-1\right)^\left\{\left(\frac\left\{n\right\}\left\{2\right\}-k\right)\right\} \binom\left\{n\right\}\left\{k\right\} \cos\left\{\left(\left(n-2k\right)\theta\right)\right\}$

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

The product-to-sum identities or prosthaphaeresis formulas can be proven by expanding their right-hand sides using the angle addition theorems. See beat (acoustics) and phase detector for applications of the sum-to-product formulæ.

Product-to-sum[24]
$\cos \theta \cos \varphi = \cot\left(a_k - a_j\right)$

(in particular, A1,1, being an empty product, is 1). Then

$\cot\left(z - a_1\right)\cdots\cot\left(z - a_n\right) = \cos\frac\left\{n\pi\right\}\left\{2\right\} + \sum_\left\{k=1\right\}^n A_\left\{n,k\right\} \cot\left(z - a_k\right).$

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

$\cot\left(z - a_1\right)\cot\left(z - a_2\right) = -1 + \cot\left(a_1 - a_2\right)\cot\left(z - a_1\right) + \cot\left(a_2 - a_1\right)\cot\left(z - a_2\right).$

### Ptolemy's theorem

$\text\left\{If \right\}w + x + y + z = \pi = \text\left\{half circle,\right\} \,$
\begin\left\{align\right\} \text\left\{then \right\}

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

(The first three equalities are trivial; the fourth is the substance of this identity.) Essentially this is Ptolemy's theorem adapted to the language of modern trigonometry.

## 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. In the case of a non-zero linear combination of a sine and cosine wave[27] (which is just a sine wave with a phase shift of π/2), we have

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

where



\varphi = \begin{cases}\arcsin \left(\frac{b}{\sqrt{a^2+b^2}}\right) & \text{if }a \ge 0, \\ \pi-\arcsin \left(\frac{b}{\sqrt{a^2+b^2}}\right) & \text{if }a < 0, \end{cases}

or equivalently



\varphi = \text{sgn}(b)\arccos \left(\tfrac{a}{\sqrt{a^2+b^2}}\right)

or even



\varphi = \arctan \left(\frac{b}{a}\right) + \begin{cases} 0 & \text{if }a \ge 0, \\ \pi & \text{if }a < 0, \end{cases}

or using the atan2 function

$\varphi = \operatorname\left\{atan2\right\} \left\left( b, a \right\right).$

More generally, for an arbitrary phase shift, we have

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

where

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

and


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


0 & \text{if } a + b\cos \alpha \ge 0, \\ \pi & \text{if } a + b\cos \alpha < 0. \end{cases}

$\sum_i a_i \sin\left(x+\delta_i\right)= a \sin\left(x+\delta\right),$

where

$a^2=\sum_\left\{i,j\right\}a_i a_j \cos\left(\delta_i-\delta_j\right)$

and

$\tan \delta=\frac\left\{\sum_i a_i \sin\delta_i\right\}\left\{\sum_i a_i \cos\delta_i\right\}.$

## Lagrange's trigonometric identities

These identities, named after Joseph Louis Lagrange, are:[28][29]



\begin{align} \sum_{n=1}^N \sin n\theta & = \frac{1}{2}\cot\frac{\theta}{2}-\frac{\cos(N+\frac{1}{2})\theta}{2\sin\frac{1}{2}\theta}\\ \sum_{n=1}^N \cos n\theta & = -\frac{1}{2}+\frac{\sin(N+\frac{1}{2})\theta}{2\sin\frac{1}{2}\theta} \end{align}

A related function is the following function of x, called the Dirichlet kernel.

$1+2\cos\left(x\right) + 2\cos\left(2x\right) + 2\cos\left(3x\right) + \cdots + 2\cos\left(nx\right)$

= \frac{\sin\left(\left(n +\frac{1}{2}\right)x\right)}{\sin(x/2)}.

## Other sums of trigonometric functions

Sum of sines and cosines with arguments in arithmetic progression:[30] if $\alpha\ne0$, then



\begin{align} & \sin{\varphi} + \sin{(\varphi + \alpha)} + \sin{(\varphi + 2\alpha)} + \cdots {} \\[8pt] & {} \qquad\qquad \cdots + \sin{(\varphi + n\alpha)} = \frac{\sin{\left(\frac{(n+1) \alpha}{2}\right)} \cdot \sin{(\varphi + \frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}} \quad\hbox{and}\\[10pt] & \cos{\varphi} + \cos{(\varphi + \alpha)} + \cos{(\varphi + 2\alpha)} + \cdots {} \\[8pt] & {} \qquad\qquad \cdots + \cos{(\varphi + n\alpha)} = \frac{\sin{\left(\frac{(n+1) \alpha}{2}\right)} \cdot \cos{(\varphi + \frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}}. \end{align}

For any a and b:

$a \cos\left(x\right) + b \sin\left(x\right) = \sqrt\left\{ a^2 + b^2 \right\} \cos\left(x - \operatorname\left\{atan2\right\}\,\left(b,a\right)\right) \;$

where atan2(y, x) is the generalization of arctan(y/x) that covers the entire circular range.

$\tan\left(x\right) + \sec\left(x\right) = \tan\left\left(\left\{x \over 2\right\} + \left\{\pi \over 4\right\}\right\right).$

The above identity is sometimes convenient to know when thinking about the Gudermannian function, which relates the circular and hyperbolic trigonometric functions without resorting to complex numbers.

If x, y, and z are the three angles of any triangle, i.e. if x + y + z = π, then

$\cot\left(x\right)\cot\left(y\right) + \cot\left(y\right)\cot\left(z\right) + \cot\left(z\right)\cot\left(x\right) = 1.\,$

## Certain linear fractional transformations

If ƒ(x) is given by the linear fractional transformation

$f\left(x\right) = \frac\left\{\left(\cos\alpha\right)x - \sin\alpha\right\}\left\{\left(\sin\alpha\right)x + \cos\alpha\right\},$

and similarly

$g\left(x\right) = \frac\left\{\left(\cos\beta\right)x - \sin\beta\right\}\left\{\left(\sin\beta\right)x + \cos\beta\right\},$

then

$f\left(g\left(x\right)\right) = g\left(f\left(x\right)\right)$

= \frac{(\cos(\alpha+\beta))x - \sin(\alpha+\beta)}{(\sin(\alpha+\beta))x + \cos(\alpha+\beta)}.

More tersely stated, if for all α we let ƒα be what we called ƒ above, then

$f_\alpha \circ f_\beta = f_\left\{\alpha+\beta\right\}. \,$

If x is the slope of a line, then ƒ(x) is the slope of its rotation through an angle of −α.

## Inverse trigonometric functions

$\arcsin\left(x\right)+\arccos\left(x\right)=\pi/2\;$
$\arctan\left(x\right)+\arccot\left(x\right)=\pi/2.\;$
$\arctan\left(x\right)+\arctan\left(1/x\right)=\left\\left\{\begin\left\{matrix\right\} \pi/2, & \mbox\left\{if \right\}x > 0 \\ -\pi/2, & \mbox\left\{if \right\}x < 0 \end\left\{matrix\right\}\right.$

### Compositions of trig and inverse trig functions

 $\sin\left[\arccos\left(x\right)\right]=\sqrt\left\{1-x^2\right\} \,$ $\tan\left[\arcsin \left(x\right)\right]=\frac\left\{x\right\}\left\{\sqrt\left\{1 - x^2\right\}\right\}$ $\sin\left[\arctan\left(x\right)\right]=\frac\left\{x\right\}\left\{\sqrt\left\{1+x^2\right\}\right\}$ $\tan\left[\arccos \left(x\right)\right]=\frac\left\{\sqrt\left\{1 - x^2\right\}\right\}\left\{x\right\}$ $\cos\left[\arctan\left(x\right)\right]=\frac\left\{1\right\}\left\{\sqrt\left\{1+x^2\right\}\right\}$ $\cot\left[\arcsin \left(x\right)\right]=\frac\left\{\sqrt\left\{1 - x^2\right\}\right\}\left\{x\right\}$ $\cos\left[\arcsin\left(x\right)\right]=\sqrt\left\{1-x^2\right\} \,$ $\cot\left[\arccos \left(x\right)\right]=\frac\left\{x\right\}\left\{\sqrt\left\{1 - x^2\right\}\right\}$

## Relation to the complex exponential function

$e^\left\{ix\right\} = \cos\left(x\right) +$

i\sin(x)\, [31] (Euler's formula),

$e^\left\{-ix\right\} = \cos\left(-x\right) + i\sin\left(-x\right) = \cos\left(x\right) - i\sin\left(x\right)$
$e^\left\{i\pi\right\} = -1$ (Euler's identity),
$\cos\left(x\right) = \frac\left\{e^\left\{ix\right\} + e^\left\{-ix\right\}\right\}\left\{2\right\}$[32]
$\sin\left(x\right) = \frac\left\{e^\left\{ix\right\} - e^\left\{-ix\right\}\right\}\left\{2i\right\}$[33]

and hence the corollary:

$\tan\left(x\right) = \frac\left\{\sin\left(x\right)\right\}\left\{\cos\left(x\right)\right\}= \frac\left\{e^\left\{ix\right\} - e^\left\{-ix\right\}\right\}\left\{i\left(\left\{e^\left\{ix\right\} + e^\left\{-ix\right\}\right\}\right)\right\}$



where $i^2 = -1$.

## Infinite product formulae

For applications to special functions, the following infinite product formulae for trigonometric functions are useful:[34][35]

## Identities without variables

$\cos 20^\circ\cdot\cos 40^\circ\cdot\cos 80^\circ=\frac\left\{1\right\}\left\{8\right\}$

is a special case of an identity that contains one variable:

$\prod_\left\{j=0\right\}^\left\{k-1\right\}\cos\left(2^j x\right)=\frac\left\{\sin\left(2^k x\right)\right\}\left\{2^k\sin\left(x\right)\right\}.$

Similarly:

$\sin 20^\circ\cdot\sin 40^\circ\cdot\sin 80^\circ=\frac\left\{\sqrt\left\{3\right\}\right\}\left\{8\right\}.$

The same cosine identity in radians is

$\cos\frac\left\{\pi\right\}\left\{9\right\}\cos\frac\left\{2\pi\right\}\left\{9\right\}\cos\frac\left\{4\pi\right\}\left\{9\right\} = \frac\left\{1\right\}\left\{8\right\},$

Similarly:

$\tan 50^\circ\cdot\tan 60^\circ\cdot\tan 70^\circ=\tan 80^\circ.$
$\tan 40^\circ\cdot\tan 30^\circ\cdot\tan 20^\circ=\tan 10^\circ.$

The following is perhaps not as readily generalized to an identity containing variables (but see explanation below):

$\cos 24^\circ+\cos 48^\circ+\cos 96^\circ+\cos 168^\circ=\frac\left\{1\right\}\left\{2\right\}.$

Degree measure ceases to be more felicitous than radian measure when we consider this identity with 21 in the denominators:



\begin{align} & \cos\left( \frac{2\pi}{21}\right)

 + \cos\left(2\cdot\frac{2\pi}{21}\right)
+ \cos\left(4\cdot\frac{2\pi}{21}\right) \\[10pt]


& {} \qquad {} + \cos\left( 5\cdot\frac{2\pi}{21}\right)

 + \cos\left( 8\cdot\frac{2\pi}{21}\right)
+ \cos\left(10\cdot\frac{2\pi}{21}\right)=\frac{1}{2}.


\end{align}

The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear: they are those integers less than 21/2 that are relatively prime to (or have no prime factors in common with) 21. The last several examples are corollaries of a basic fact about the irreducible cyclotomic polynomials: the cosines are the real parts of the zeroes of those polynomials; the sum of the zeroes is the Möbius function evaluated at (in the very last case above) 21; only half of the zeroes are present above. The two identities preceding this last one arise in the same fashion with 21 replaced by 10 and 15, respectively.

Many of those curious identities stem from more general facts like the following:[36]

$\prod_\left\{k=1\right\}^\left\{n-1\right\} \sin\left\left(\frac\left\{k\pi\right\}\left\{n\right\}\right\right) = \frac\left\{n\right\}\left\{2^\left\{n-1\right\}\right\}$

and

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

Combining these gives us

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

If n is an odd number (n = 2m + 1) we can make use of the symmetries to get

$\prod_\left\{k=1\right\}^\left\{m\right\} \tan\left\left(\frac\left\{k\pi\right\}\left\{2m+1\right\}\right\right) = \sqrt\left\{2m+1\right\}$

The transfer function of the Butterworth low pass filter can be expressed in terms of polynomial and poles. By setting the frequency as the cutoff frequency, the following identity can be proved:

$\prod_\left\{k=1\right\}^\left\{n\right\} \sin\left\left(\frac\left\{\left\left(2k-1\right\right)\pi\right\}\left\{4n\right\}\right\right) = \prod_\left\{k=1\right\}^\left\{n\right\} \cos\left\left(\frac\left\{\left\left(2k-1\right\right)\pi\right\}\left\{4n\right\}\right\right) = \frac\left\{\sqrt\left\{2\right\}\right\}\left\{2^\left\{n\right\}\right\}$

### Computing π

An efficient way to compute π is based on the following identity without variables, due to Machin:

$\frac\left\{\pi\right\}\left\{4\right\} = 4 \arctan\frac\left\{1\right\}\left\{5\right\} - \arctan\frac\left\{1\right\}\left\{239\right\}$

or, alternatively, by using an identity of Leonhard Euler:

$\frac\left\{\pi\right\}\left\{4\right\} = 5 \arctan\frac\left\{1\right\}\left\{7\right\} + 2 \arctan\frac\left\{3\right\}\left\{79\right\}.$

### A useful mnemonic for certain values of sines and cosines

For certain simple angles, the sines and cosines take the form $\scriptstyle\sqrt\left\{n\right\}/2$ for 0 ≤ n ≤ 4, which makes them easy to remember.



\begin{matrix} \sin 0 & = & \sin 0^\circ & = & \sqrt{0}/2 & = & \cos 90^\circ & = & \cos \left( \frac {\pi} {2} \right) \\ \\ \sin \left( \frac {\pi} {6} \right) & = & \sin 30^\circ & = & \sqrt{1}/2 & = & \cos 60^\circ & = & \cos \left( \frac {\pi} {3} \right) \\ \\ \sin \left( \frac {\pi} {4} \right) & = & \sin 45^\circ & = & \sqrt{2}/2 & = & \cos 45^\circ & = & \cos \left( \frac {\pi} {4} \right) \\ \\ \sin \left( \frac {\pi} {3} \right) & = & \sin 60^\circ & = & \sqrt{3}/2 & = & \cos 30^\circ & = & \cos \left( \frac {\pi} {6} \right)\\ \\ \sin \left( \frac {\pi} {2} \right) & = & \sin 90^\circ & = & \sqrt{4}/2 & = & \cos 0^\circ & = & \cos 0 \end{matrix}

### Miscellany

With the golden ratio φ:

$\cos \left\left( \frac \left\{\pi\right\} \left\{5\right\} \right\right) = \cos 36^\circ=\left\{\sqrt\left\{5\right\}+1 \over 4\right\} = \frac\left\{\varphi \right\}\left\{2\right\}$

$\sin \left\left( \frac \left\{\pi\right\} \left\{10\right\} \right\right) = \sin 18^\circ = \left\{\sqrt\left\{5\right\}-1 \over 4\right\} = \left\{\varphi - 1 \over 2\right\} = \left\{1 \over 2\varphi\right\}$

Also see exact trigonometric constants.

### An identity of Euclid

Euclid showed in Book XIII, Proposition 10 of his Elements that the area of the square on the side of a regular pentagon inscribed in a circle is equal to the sum of the areas of the squares on the sides of the regular hexagon and the regular decagon inscribed in the same circle. In the language of modern trigonometry, this says:

$\sin^2\left(18^\circ\right)+\sin^2\left(30^\circ\right)=\sin^2\left(36^\circ\right). \,$

Ptolemy used this proposition to compute some angles in his table of chords.

## Composition of trigonometric functions

This identity involves a trigonometric function of a trigonometric function:

$\cos\left(t \sin\left(x\right)\right) = J_0\left(t\right) + 2 \sum_\left\{k=1\right\}^\infty J_\left\{2k\right\}\left(t\right) \cos\left(2kx\right)$

where J0 and J2k are Bessel functions.

## Calculus

In calculus the relations stated below require angles to be measured in radians; the relations would become more complicated if angles were measured in another unit such as degrees. If the trigonometric functions are defined in terms of geometry, along with the definitions of arc length and area, their derivatives can be found by verifying two limits. The first is:

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

verified using the unit circle and squeeze theorem. The second limit is:

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

verified using the identity tan(x/2) = (1 − cos x)/sin x. Having established these two limits, one can use the limit definition of the derivative and the addition theorems to show that (sin x)′ = cos x and (cos x)′ = −sin x. If the sine and cosine functions are defined by their Taylor series, then the derivatives can be found by differentiating the power series term-by-term.

$\left\{d \over dx\right\}\sin x = \cos x$

The rest of the trigonometric functions can be differentiated using the above identities and the rules of differentiation:[37][38][39]



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

The integral identities can be found in "list of integrals of trigonometric functions". Some generic forms are listed below.

$\int\left\{\frac\left\{du\right\}\left\{\sqrt\left\{a^\left\{2\right\}-u^\left\{2\right\}\right\}\right\}\right\}=\sin ^\left\{-1\right\}\left\left( \frac\left\{u\right\}\left\{a\right\} \right\right)+C$
$\int\left\{\frac\left\{du\right\}\left\{a^\left\{2\right\}+u^\left\{2\right\}\right\}\right\}=\frac\left\{1\right\}\left\{a\right\}\tan ^\left\{-1\right\}\left\left( \frac\left\{u\right\}\left\{a\right\} \right\right)+C$
$\int\left\{\frac\left\{du\right\}\left\{u\sqrt\left\{u^\left\{2\right\}-a^\left\{2\right\}\right\}\right\}\right\}=\frac\left\{1\right\}\left\{a\right\}\sec ^\left\{-1\right\}\left| \frac\left\{u\right\}\left\{a\right\} \right|+C$

### Implications

The fact that the differentiation of trigonometric functions (sine and cosine) results in linear combinations of the same two functions is of fundamental importance to many fields of mathematics, including differential equations and Fourier transforms.

## Exponential definitions

Function Inverse function[40]
$\sin \theta = \frac\left\{e^\left\{i\theta\right\} - e^\left\{-i\theta\right\}\right\}\left\{2i\right\} \,$ $\arcsin x = -i \ln \left\left(ix + \sqrt\left\{1 - x^2\right\}\right\right) \,$
$\cos \theta = \frac\left\{e^\left\{i\theta\right\} + e^\left\{-i\theta\right\}\right\}\left\{2\right\} \,$ $\arccos x = i\,\ln\left\left(x-i\,\sqrt\left\{1-x^2\right\}\right\right) \,$
$\tan \theta = \frac\left\{e^\left\{i\theta\right\} - e^\left\{-i\theta\right\}\right\}\left\{i\left(e^\left\{i\theta\right\} + e^\left\{-i\theta\right\}\right)\right\} \,$ $\arctan x = \frac\left\{i\right\}\left\{2\right\} \ln \left\left(\frac\left\{i + x\right\}\left\{i - x\right\}\right\right) \,$
$\csc \theta = \frac\left\{2i\right\}\left\{e^\left\{i\theta\right\} - e^\left\{-i\theta\right\}\right\} \,$ $\arccsc x = -i \ln \left\left(\tfrac\left\{i\right\}\left\{x\right\} + \sqrt\left\{1 - \tfrac\left\{1\right\}\left\{x^2\right\}\right\}\right\right) \,$
$\sec \theta = \frac\left\{2\right\}\left\{e^\left\{i\theta\right\} + e^\left\{-i\theta\right\}\right\} \,$ $\arcsec x = -i \ln \left\left(\tfrac\left\{1\right\}\left\{x\right\} + \sqrt\left\{1 - \tfrac\left\{i\right\}\left\{x^2\right\}\right\}\right\right) \,$
$\cot \theta = \frac\left\{i\left(e^\left\{i\theta\right\} + e^\left\{-i\theta\right\}\right)\right\}\left\{e^\left\{i\theta\right\} - e^\left\{-i\theta\right\}\right\} \,$ $\arccot x = \frac\left\{i\right\}\left\{2\right\} \ln \left\left(\frac\left\{x - i\right\}\left\{x + i\right\}\right\right) \,$
$\operatorname\left\{cis\right\} \, \theta = e^\left\{i\theta\right\} \,$ $\operatorname\left\{arccis\right\} \, x = \frac\left\{\ln x\right\}\left\{i\right\} = -i \ln x = \operatorname\left\{arg\right\} \, x \,$

## Miscellaneous

### Dirichlet kernel

The Dirichlet kernel Dn(x) is the function occurring on both sides of the next identity:

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

The convolution of any integrable function of period 2π with the Dirichlet kernel coincides with the function's nth-degree Fourier approximation. The same holds for any measure or generalized function.

### Tangent half-angle substitution

Main article: Tangent half-angle substitution

If we set

$t = \tan\left\left(\frac\left\{x\right\}\left\{2\right\}\right\right),$

then[41]

$\sin\left(x\right) = \frac\left\{2t\right\}\left\{1 + t^2\right\}\text\left\{ and \right\}\cos\left(x\right) = \frac\left\{1 - t^2\right\}\left\{1 + t^2\right\}\text\left\{ and \right\}e^\left\{i x\right\} = \frac\left\{1 + i t\right\}\left\{1 - i t\right\}$

where eix = cos(x) + i sin(x), sometimes abbreviated to cis(x).

When this substitution of t for tan(x/2) is used in calculus, it follows that sin(x) is replaced by 2t/(1 + t2), cos(x) is replaced by (1 − t2)/(1 + t2) and the differential dx is replaced by (2 dt)/(1 + t2). Thereby one converts rational functions of sin(x) and cos(x) to rational functions of t in order to find their antiderivatives.