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# Exact trigonometric constants

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Exact constant expressions for trigonometric expressions are sometimes useful, mainly for simplifying solutions into radical forms which allow further simplification.

All values of sine, cosine, and tangent of angles with 3° increments are derivable using identities: Half-angle, Double-angle, Addition/subtraction and values for 0°, 30°, 36°, and 45°. Note that 1° = $\frac{\pi}{180}$ radians.

This article is incomplete in at least two senses. First, it is always possible to apply a half-angle formula and find an exact expression for the cosine of 1/2 the smallest angle on the list. Second, this article exploits only the first two of five known Fermat primes: 3 and 5; and the trigonometric functions of other angles, such as $\frac{2\pi}{7}, \frac{2\pi}{9}$ (= 40°), and $\frac{2\pi}{13}$ (as well as the other constructible polygons, $\frac{2\pi}{17},$ $\frac{2\pi}{257}$, or $\frac{2\pi}{65537})$ are soluble by radicals. In practice, all values of sine, cosine, and tangent not found in this article are approximated using the techniques described at Generating trigonometric tables.

The values of sine, cosine, and tangent of angles with 1° increments can also be derived using the triple-angle identity. However, they can only be expressed with intermediate (and irreductible) complex numbers in the expression (see the formulae to compute the roots of a cubic equation), or by transforming them using hyperbolic real functions).

## Table of constants

Values outside [0°, 45°] angle range are trivially extracted from circle axis reflection symmetry from these values. (See Trigonometric identity)

### 0°: fundamental

$\sin 0=0\,$
$\cos 0=1\,$
$\tan 0=0\,$
$\cot 0=\mbox{undefined}\,$

### 3°: 60-sided polygon

$\sin\frac{\pi}{60}=\sin 3^\circ=\frac{2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)}{16}\,$
$\cos\frac{\pi}{60}=\cos 3^\circ=\frac{2(1+\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3-1)}{16}\,$
$\tan\frac{\pi}{60}=\tan 3^\circ=\frac{\left((2-\sqrt3)(3+\sqrt5)-2\right)\left(2-\sqrt{2(5-\sqrt5)}\right)}{4}\,$
$\cot\frac{\pi}{60}=\cot 3^\circ=\frac{\left((2+\sqrt3)(3+\sqrt5)-2\right)\left(2+\sqrt{2(5-\sqrt5)}\right)}{4}\,$

### 6°: 30-sided polygon

$\sin\frac{\pi}{30}=\sin 6^\circ=\frac{\sqrt{6(5-\sqrt5)}-\sqrt5-1}{8}\,$
$\cos\frac{\pi}{30}=\cos 6^\circ=\frac{\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)}{8}\,$
$\tan\frac{\pi}{30}=\tan 6^\circ=\frac{\sqrt{2(5-\sqrt5)}-\sqrt3(\sqrt5-1)}{2}\,$
$\cot\frac{\pi}{30}=\cot 6^\circ=\frac{\sqrt3(3+\sqrt5)+\sqrt{2(25+11\sqrt5)}}{2}\,$

### 9°: 20-sided polygon

$\sin\frac{\pi}{20}=\sin 9^\circ=\frac{\sqrt2(\sqrt5+1)-2\sqrt{5-\sqrt5}}{8}\,$
$\cos\frac{\pi}{20}=\cos 9^\circ=\frac{\sqrt2(\sqrt5+1)+2\sqrt{5-\sqrt5}}{8}\,$
$\tan\frac{\pi}{20}=\tan 9^\circ=\sqrt5+1-\sqrt{5+2\sqrt5}\,$
$\cot\frac{\pi}{20}=\cot 9^\circ=\sqrt5+1+\sqrt{5+2\sqrt5}\,$

### 12°: 15-sided polygon

$\sin\frac{\pi}{15}=\sin 12^\circ=\frac{\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)}{8}\,$
$\cos\frac{\pi}{15}=\cos 12^\circ=\frac{\sqrt{6(5+\sqrt5)}+\sqrt5-1}{8}\,$
$\tan\frac{\pi}{15}=\tan 12^\circ=\frac{\sqrt3(3-\sqrt5)-\sqrt{2(25-11\sqrt5)}}{2}\,$
$\cot\frac{\pi}{15}=\cot 12^\circ=\frac{\sqrt3(\sqrt5+1)+\sqrt{2(5+\sqrt5)}}{2}\,$

### 15°: dodecagon

$\sin\frac{\pi}{12}=\sin 15^\circ=\frac{\sqrt2(\sqrt3-1)}{4}\,$
$\cos\frac{\pi}{12}=\cos 15^\circ=\frac{\sqrt2(\sqrt3+1)}{4}\,$
$\tan\frac{\pi}{12}=\tan 15^\circ=2-\sqrt3\,$
$\cot\frac{\pi}{12}=\cot 15^\circ=2+\sqrt3\,$

### 18°: decagon

$\sin\frac{\pi}{10}=\sin 18^\circ=\frac{\sqrt5-1}{4}=\frac{\varphi-1}{2}=\frac{1}{2\varphi}\,$
$\cos\frac{\pi}{10}=\cos 18^\circ=\frac{\sqrt{2(5+\sqrt5)}}{4}\,$
$\tan\frac{\pi}{10}=\tan 18^\circ=\sqrt{1-\sqrt\frac{4}{5}}\,$
$\cot\frac{\pi}{10}=\cot 18^\circ=\sqrt{5+2\sqrt 5}\,$

### 21°: sum 9° + 12°

$\sin\frac{7\pi}{60}=\sin 21^\circ=\frac{2(\sqrt3+1)\sqrt{5-\sqrt5}-\sqrt2(\sqrt3-1)(1+\sqrt5)}{16}\,$
$\cos\frac{7\pi}{60}=\cos 21^\circ=\frac{2(\sqrt3-1)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3+1)(1+\sqrt5)}{16}\,$
$\tan\frac{7\pi}{60}=\tan 21^\circ=\frac{\left(2-(2+\sqrt3)(3-\sqrt5)\right)\left(2-\sqrt{2(5+\sqrt5)}\right)}{4}\,$
$\cot\frac{7\pi}{60}=\cot 21^\circ=\frac{\left(2-(2-\sqrt3)(3-\sqrt5)\right)\left(2+\sqrt{10\sqrt5}\right)}{4}\,$

### 22.5°: octagon

$\sin\frac{\pi}{8}=\sin 22.5^\circ=\frac{\sqrt{2-\sqrt{2}}}{2}\,$
$\cos\frac{\pi}{8}=\cos 22.5^\circ=\frac{\sqrt{2+\sqrt{2}}}{2}\,$
$\tan\frac{\pi}{8}=\tan 22.5^\circ=\sqrt{2}-1\,$
$\cot\frac{\pi}{8}=\cot 22.5^\circ=\sqrt{2}+1\,$

### 24°: sum 12° + 12°

$\sin\frac{2\pi}{15}=\sin 24^\circ=\frac{\sqrt3(\sqrt5+1)-\sqrt2\sqrt{5-\sqrt5}}{8}\,$
$\cos\frac{2\pi}{15}=\cos 24^\circ=\frac{\sqrt6\sqrt{5-\sqrt5}+\sqrt5+1}{8}\,$
$\tan\frac{2\pi}{15}=\tan 24^\circ=\frac{\sqrt{50+22\sqrt5}-\sqrt3(3+\sqrt5)}{2}\,$
$\cot\frac{2\pi}{15}=\cot 24^\circ=\frac{\sqrt2\sqrt{5-\sqrt5}+\sqrt3(\sqrt5-1)}{2}\,$

### 27°: sum 12° + 15°

$\sin\frac{3\pi}{20}=\sin 27^\circ=\frac{(\sqrt5+1)\sqrt{5+\sqrt5}-\sqrt2(\sqrt5-1)}{8}\,$
$\cos\frac{3\pi}{20}=\cos 27^\circ=\frac{(\sqrt5+1)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)}{8}\,$
$\tan\frac{3\pi}{20}=\tan 27^\circ=\sqrt5-1-\sqrt{5-2\sqrt5}\,$
$\cot\frac{3\pi}{20}=\cot 27^\circ=\sqrt5-1+\sqrt{5-2\sqrt5}\,$

### 30°: hexagon

$\sin\frac{\pi}{6}=\sin 30^\circ=\frac{1}{2}\,$
$\cos\frac{\pi}{6}=\cos 30^\circ=\frac{\sqrt3}{2}\,$
$\tan\frac{\pi}{6}=\tan 30^\circ=\frac{\sqrt3}{3}\,$
$\cot\frac{\pi}{6}=\cot 30^\circ=\sqrt3\,$

### 33°: sum 15° + 18°

$\sin\frac{11\pi}{60}=\sin 33^\circ=\frac{2(\sqrt3-1)\sqrt{5+\sqrt5}+\sqrt2(1+\sqrt3)(\sqrt5-1)}{16}\,$
$\cos\frac{11\pi}{60}=\cos 33^\circ=\frac{2(\sqrt3+1)\sqrt{5+\sqrt5}+\sqrt2(1-\sqrt3)(\sqrt5-1)}{16}\,$
$\tan\frac{11\pi}{60}=\tan 33^\circ=\frac{\left(2-(2-\sqrt3)(3+\sqrt5)\right)\left(2+\sqrt{2(5-\sqrt5)}\right)}{4}\,$
$\cot\frac{11\pi}{60}=\cot 33^\circ=\frac{\left(2-(2+\sqrt3)(3+\sqrt5)\right)\left(2-\sqrt{2(5-\sqrt5)}\right)}{4}\,$

### 36°: pentagon

$\sin\frac{\pi}{5}=\sin 36^\circ=\frac{\sqrt{2(5-\sqrt5)}}{4}\,$
$\cos\frac{\pi}{5}=\cos 36^\circ=\frac{1+\sqrt5}{4}=\frac{\varphi}{2}\,$
$\tan\frac{\pi}{5}=\tan 36^\circ=\sqrt{5-2\sqrt5}\,$
$\cot\frac{\pi}{5}=\cot 36^\circ=\frac{\sqrt{5(5+2\sqrt5)}}{5}\,$

### 39°: sum 18° + 21°

$\sin\frac{13\pi}{60}=\sin 39^\circ=\frac{2(1-\sqrt3)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3+1)(\sqrt5+1)}{16}\,$
$\cos\frac{13\pi}{60}=\cos 39^\circ=\frac{2(1+\sqrt3)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3-1)(\sqrt5+1)}{16}\,$
$\tan\frac{13\pi}{60}=\tan 39^\circ=\frac{\left((2-\sqrt3)(3-\sqrt5)-2\right)\left(2-\sqrt{2(5+\sqrt5)}\right)}{4}\,$
$\cot\frac{13\pi}{60}=\cot 39^\circ=\frac{\left((2+\sqrt3)(3-\sqrt5)-2\right)\left(2+\sqrt{2(5+\sqrt5)}\right)}{4}\,$

### 42°: sum 21° + 21°

$\sin\frac{7\pi}{30}=\sin 42^\circ=\frac{\sqrt6\sqrt{5+\sqrt5}-\sqrt5+1}{8}\,$
$\cos\frac{7\pi}{30}=\cos 42^\circ=\frac{\sqrt2\sqrt{5+\sqrt5}+\sqrt3(\sqrt5-1)}{8}\,$
$\tan\frac{7\pi}{30}=\tan 42^\circ=\frac{\sqrt3(\sqrt5+1)-\sqrt2\sqrt{5+\sqrt5}}{2}\,$
$\cot\frac{7\pi}{30}=\cot 42^\circ=\frac{\sqrt{2(25-11\sqrt5)}+\sqrt3(3-\sqrt5)}{2}\,$

### 45°: square

$\sin\frac{\pi}{4}=\sin 45^\circ=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}\,$
$\cos\frac{\pi}{4}=\cos 45^\circ=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}\,$
$\tan\frac{\pi}{4}=\tan 45^\circ=1\,$
$\cot\frac{\pi}{4}=\cot 45^\circ=1\,$

### 60°: triangle

$\sin\frac{\pi}{3}=\sin 60^\circ=\frac{\sqrt3}{2}\,$
$\cos\frac{\pi}{3}=\cos 60^\circ=\frac{1}{2}\,$
$\tan\frac{\pi}{3}=\tan 60^\circ=\sqrt3\,$
$\cot\frac{\pi}{3}=\cot 60^\circ=\frac{{\sqrt3}}{3}\,$

## Notes

### Uses for constants

As an example of the use of these constants, consider a dodecahedron with the following volume, where a is the length of an edge:

$V=\frac{5a^3\cos{36^\circ}}{\tan^2{36^\circ}}$

Using

$\cos 36^\circ=\frac{\sqrt5+1}{4}\,$
$\tan 36^\circ=\sqrt{5-2\sqrt5}\,$

this can be simplified to:

$V=\frac{a^3(15+7\sqrt5)}{4}\,$

### Derivation triangles

The derivation of sine, cosine, and tangent constants into radial forms is based upon the constructability of right triangles.

Here are right triangles made from symmetry sections of regular polygons are used to calculate fundamental trigonometric ratios. Each right triangle represents three points in a regular polygon: a vertex, an edge center containing that vertex, and the polygon center. A n-gon can be divided into 2n right triangle with angles of {$\frac{180}{n},$ $90 - \frac{180}{n},$ 90} degrees, for n in 3, 4, 5, ...

Constructibility of 3, 4, 5, and 15 sided polygons are the basis, and angle bisectors allow multiples of two to also be derived.

• Constructible
• 3×2n-sided regular polygons, for n in 0, 1, 2, 3, ...
• 4×2n-sided
• 45°-45°-90° triangle: square (4-sided)
• 67.5°-22.5°-90° triangle: octagon (8-sided)
• 78.75°-11.25°-90° triangle: hexakaidecagon (16-sided)
• ...
• 5×2n-sided
• 54°-36°-90° triangle: pentagon (5-sided)
• 72°-18°-90° triangle: decagon (10-sided)
• 81°-9°-90° triangle: icosagon (20-sided)
• 85.5°-4.5°-90° triangle: tetracontagon (40-sided)
• 87.75°-2.25°-90° triangle: octacontagon (80-sided)
• ...
• 15×2n-sided
• ... (Higher constructible regular polygons don't make whole degree angles: 17, 51, 85, 255, 257...)
• Nonconstructable (with whole or half degree angles) - No finite radical expressions involving real numbers for these triangle edge ratios are possible, therefore its multiples of two are also not possible.

## Calculated trigonometric values for sine and cosine

### The trivial ones

In degree format: 0, 90, 45, 30 and 60 can be calculated from their triangles, using the Pythagorean theorem.

### n × π/(5×2m)

#### Geometrical method

Applying Ptolemy's theorem to the cyclic quadrilateral ABCD defined by four successive vertices of the pentagon, we can find that:

$\mathrm{crd}\ {36^\circ}=\mathrm{crd}\left(\angle\mathrm{ADB}\right)=\frac{a}{b}=\frac{2}{1+\sqrt{5}},$

which is the reciprocal 1/f of the golden ratio. Crd is the Chord function,

$\mathrm{crd}\ {\theta}=2\sin{\frac{\theta}{2}}\,$

Thus

$\sin{18^\circ}=\frac{1}{1+\sqrt{5}}.$

(Alternatively, without using Ptolemy's theorem, label as X the intersection of AC and BD, and note by considering angles that triangle AXB is isosceles, so AX=AB=a. Triangles AXD and CXB are similar, because AD is parallel to BC. So XC=a.(a/b). But AX+XC=AC, so a+a2/b=b. Solving this gives a/b=1/f, as above).

Similarly

$\mathrm{crd}\ 108^\circ=\mathrm{crd}(\angle\mathrm{ABC})=\frac{b}{a}=\frac{1+\sqrt{5}}{2},$

so

$\sin 54^\circ=\cos 36^\circ=\frac{1+\sqrt{5}}{4}.$

#### Algebraic method

The multiple angle formulas for functions of $5x\,$, where $x\in\{18,36,54,72,90\}\,$ and $5x\in\{90,180,270,360,450\}\,$, can be solved for the functions of $x$, since we know the function values of $5x\,$. The multiple angle formulas are:

$\sin{5x}=16\sin^5 x-20\sin^3 x+5\sin x\,$,
$\cos{5x}=16\cos^5 x-20\cos^3 x+5\cos x\,$.
• When $\sin{5x}=0\,$ or $\cos{5x}=0\,$, we let $y=\sin x\,$ or $y=\cos x\,$ and solve for $y\,$:
$16y^5-20y^3+5y=0\,$.
One solution is zero, and the resulting 4th degree equation can be solved as a quadratic in $y^2\,$.
• When $\sin{5x}=1\,$ or $\cos{5x}=1\,$, we again let $y=\sin x\,$ or $y=\cos x\,$ and solve for $y\,$:
$16y^5-20y^3+5y-1=0\,$,
which factors into:
$(y-1)(4y^2+2y-1)^2=0\,$.

### n × π/20

9° is 45-36, and 27° is 45−18; so we use the subtraction formulas for sine and cosine.

### n × π/30

6° is 36-30, 12° is 30−18, 24° is 54−30, and 42° is 60−18; so we use the subtraction formulas for sine and cosine.

### n × π/60

3° is 18−15, 21° is 36−15, 33° is 18+15, and 39° is 54−15, so we use the subtraction (or addition) formulas for sine and cosine.

## Plans for simplifying

### Rationalize the denominator

If the denominator is a square root, multiply the numerator and denominator by that radical.
If the denominator is the sum or difference of two terms, multiply the numerator and denominator by the conjugate of the denominator. The conjugate is the identical, except the sign between the terms is changed.
Sometimes you need to rationalize the denominator more than once.

### Split a fraction in two

Sometimes it helps to split the fraction into the sum of two fractions and then simplify both separately.

### Squaring and square rooting

If there is a complicated term, with only one kind of radical in a term, this plan may help. Square the term, combine like terms, and take the square root. This may leave a big radical with a smaller radical inside, but it is often better than the original.

### Simplification of nested radical expressions

In general nested radicals cannot be reduced.

But if for $\sqrt{a+b\sqrt c}\,$,

$R=\sqrt{a^2-b^2c}\,$ is rational,

and both

$d=\pm\sqrt{\frac{a\pm R}{2}}\,$ and
$e=\pm\sqrt{\frac{a\pm R}{2c}}\,$

are rational, with the appropriate choice of the four $\pm$signs, then

$\sqrt{a+b\sqrt c}=d+e\sqrt c\,$

Example:

$4\sin{18^\circ}=\sqrt{6-2\sqrt5}=\sqrt5-1\,$


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 Krzysztof R. Apt Computer Science , 2011, Abstract: This is a short introduction to the subject of strategic games. We focus on the concepts of best response, Nash equilibrium, strict and weak dominance, and mixed strategies, and study the relation between these concepts in the context of the iterated elimination of strategies. Also, we discuss some variants of the original definition of a strategic game. Finally, we introduce the basics of mechanism design and use pre-Bayesian games to explain it.
 Burkhard C. Schipper Computer Science , 2015, Abstract: It is known that there are uncoupled learning heuristics leading to Nash equilibrium in all finite games. Why should players use such learning heuristics and where could they come from? We show that there is no uncoupled learning heuristic leading to Nash equilibrium in all finite games that a player has an incentive to adopt, that would be evolutionary stable or that could "learn itself". Rather, a player has an incentive to strategically teach such a learning opponent in order secure at least the Stackelberg leader payoff. The impossibility result remains intact when restricted to the classes of generic games, two-player games, potential games, games with strategic complements or 2x2 games, in which learning is known to be "nice". More generally, it also applies to uncoupled learning heuristics leading to correlated equilibria, rationalizable outcomes, iterated admissible outcomes, or minimal curb sets. A possibility result restricted to "strategically trivial" games fails if some generic games outside this class are considered as well.
 Computer Science , 2014, DOI: 10.4204/EPTCS.144.6 Abstract: In this paper we describe an approach to resolve strategic games in which players can assume different types along the game. Our goal is to infer which type the opponent is adopting at each moment so that we can increase the player's odds. To achieve that we use Markov games combined with hidden Markov model. We discuss a hypothetical example of a tennis game whose solution can be applied to any game with similar characteristics.
 Computer Science , 2014, DOI: 10.4204/EPTCS.146.2 Abstract: We introduce a new class of games where each player's aim is to randomise her strategic choices in order to affect the other players' expectations aside from her own. The way each player intends to exert this influence is expressed through a Boolean combination of polynomial equalities and inequalities with rational coefficients. We offer a logical representation of these games as well as a computational study of the existence of equilibria.
 Computer Science , 2011, Abstract: We introduce a new measure of the discrepancy in strategic games between the social welfare in a Nash equilibrium and in a social optimum, that we call selfishness level. It is the smallest fraction of the social welfare that needs to be offered to each player to achieve that a social optimum is realized in a pure Nash equilibrium. The selfishness level is unrelated to the price of stability and the price of anarchy and is invariant under positive linear transformations of the payoff functions. Also, it naturally applies to other solution concepts and other forms of games. We study the selfishness level of several well-known strategic games. This allows us to quantify the implicit tension within a game between players' individual interests and the impact of their decisions on the society as a whole. Our analyses reveal that the selfishness level often provides a deeper understanding of the characteristics of the underlying game that influence the players' willingness to cooperate. In particular, the selfishness level of finite ordinal potential games is finite, while that of weakly acyclic games can be infinite. We derive explicit bounds on the selfishness level of fair cost sharing games and linear congestion games, which depend on specific parameters of the underlying game but are independent of the number of players. Further, we show that the selfishness level of the $n$-players Prisoner's Dilemma is $c/(b(n-1)-c)$, where $b$ and $c$ are the benefit and cost for cooperation, respectively, that of the $n$-players public goods game is $(1-\frac{c}{n})/(c-1)$, where $c$ is the public good multiplier, and that of the Traveler's Dilemma game is $\frac{1}{2}(b-1)$, where $b$ is the bonus. Finally, the selfishness level of Cournot competition (an example of an infinite ordinal potential game, Tragedy of the Commons, and Bertrand competition is infinite.
 Martin Hoefer Computer Science , 2010, Abstract: In this paper we consider strategic cost sharing games with so-called arbitrary sharing based on various combinatorial optimization problems, such as vertex and set cover, facility location, and network design problems. We concentrate on the existence and computational complexity of strong equilibria, in which no coalition can improve the cost of each of its members. Our main result reveals a connection between strong equilibrium in strategic games and the core in traditional coalitional cost sharing games studied in economics. For set cover and facility location games this results in a tight characterization of the existence of strong equilibrium using the integrality gap of suitable linear programming formulations. Furthermore, it allows to derive all existing results for strong equilibria in network design cost sharing games with arbitrary sharing via a unified approach. In addition, we are able to show that in general the strong price of anarchy is always 1. This should be contrasted with the price of anarchy of \Theta(n) for Nash equilibria. Finally, we indicate that the LP-approach can also be used to compute near-optimal and near-stable approximate strong equilibria.
 Games , 2010, DOI: 10.3390/g1040381 Abstract: It is well-known that in finite strategic games true common belief (or common knowledge) of rationality implies that the players will choose only strategies that survive the iterated elimination of strictly dominated strategies. We establish a general theorem that deals with monotonic rationality notions and arbitrary strategic games and allows to strengthen the above result to arbitrary games, other rationality notions, and transfinite iterations of the elimination process. We also clarify what conclusions one can draw for the customary dominance notions that are not monotonic. The main tool is Tarski’s Fixpoint Theorem.
 Computer Science , 2010, Abstract: In [Van Benthem 2007] the concept of a public announcement is used to study the effect of the iterated elimination of strictly dominated strategies. We offer a simple generalisation of this approach to cover arbitrary strategic games and many optimality notions. We distinguish between announcements of optimality and announcements of rationality.
 Krzysztof R. Apt Computer Science , 2007, Abstract: We provide here an epistemic analysis of arbitrary strategic games based on the possibility correspondences. Such an analysis calls for the use of transfinite iterations of the corresponding operators. Our approach is based on Tarski's Fixpoint Theorem and applies both to the notions of rationalizability and the iterated elimination of strictly dominated strategies.
 Computer Science , 2012, Abstract: We present the first general bounds on the mixing time of the Markov chain associated to the logit dynamics for wide classes of strategic games. The logit dynamics with inverse noise beta describes the behavior of a complex system whose individual components act selfishly and keep responding according to some partial ("noisy") knowledge of the system, where the capacity of the agent to know the system and compute her best move is measured by the inverse of the parameter beta. In particular, we prove nearly tight bounds for potential games and games with dominant strategies. Our results show that, for potential games, the mixing time is upper and lower bounded by an exponential in the inverse of the noise and in the maximum potential difference. Instead, for games with dominant strategies, the mixing time cannot grow arbitrarily with the inverse of the noise. Finally, we refine our analysis for a subclass of potential games called graphical coordination games, a class of games that have been previously studied in Physics and, more recently, in Computer Science in the context of diffusion of new technologies. We give evidence that the mixing time of the logit dynamics for these games strongly depends on the structure of the underlying graph. We prove that the mixing time of the logit dynamics for these games can be upper bounded by a function that is exponential in the cutwidth of the underlying graph and in the inverse of noise. Moreover, we consider two specific and popular network topologies, the clique and the ring. For games played on a clique we prove an almost matching lower bound on the mixing time of the logit dynamics that is exponential in the inverse of the noise and in the maximum potential difference, while for games played on a ring we prove that the time of convergence of the logit dynamics to its stationary distribution is significantly shorter.
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