Exact trigonometric constants

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The primary solution angles on the unit circle are at multiples of 30 and 45 degrees.

Trigonometry
History Usage Functions Inverse functions Further reading
Reference
List of identities Exact constants Generating trigonometric tables CORDIC
Euclidean theory
Law of sines Law of cosines Law of tangents Pythagorean theorem
Calculus
The Trigonometric integral Trigonometric substitution Integrals of functions Derivatives of functions Integrals of inverses

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° = π/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 one-half 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 2π/7, 2π/9 (= 40°), and 2π/13 (as well as the other constructible polygons, 2π/17, 2π/257, or 2π/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.

1 Table of constants
2 Notes
- 2.1 Uses for constants
- 2.2 Derivation triangles
3 Calculated trigonometric values for sine and cosine
4 Plans for simplifying
5 See also
6 References
7 External links

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

$\sin\frac{\pi}{10}=\sin 18^\circ=\frac{\sqrt5-1}{4}=\frac{\varphi-1}{2}=\frac{1}{2\varphi}\,$

$\cos\frac{\pi}{10}=\cos 18^\circ=\frac{\sqrt{2(5+\sqrt5)}}{4}\,$

$\tan\frac{\pi}{10}=\tan 18^\circ=\frac{\sqrt{5(5-2\sqrt5)}}{5}\,$

$\cot\frac{\pi}{10}=\cot 18^\circ=\sqrt{5+2\sqrt 5}\,$

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

$\sin\frac{\pi}{4}=\sin 45^\circ=\frac{\sqrt2}{2}\,$

$\cos\frac{\pi}{4}=\cos 45^\circ=\frac{\sqrt2}{2}\,$

$\tan\frac{\pi}{4}=\tan 45^\circ=1\,$

$\cot\frac{\pi}{4}=\cot 45^\circ=1\,$

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

Regular polygon (N-sided) and its fundamental right triangle. Angle: a=180/n °

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 {180/n, 90−180/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, ...
  - 30°-60°-90° triangle: triangle (3 sided)
  - 60°-30°-90° triangle: hexagon (6-sided)
  - 75°-15°-90° triangle: dodecagon (12-sided)
  - 82.5°-7.5°-90° triangle: icosikaitetragon (24-sided)
  - 86.25°-3.75°-90° triangle: tetracontakaioctagon (48-sided)
  - ...
- 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
  - 78°-12°-90° triangle: pentakaidecagon (15-sided)
  - 84°-6°-90° triangle: tricontagon (30-sided)
  - 87°-3°-90° triangle: hexacontagon (60-sided)
  - 88.5°-1.5°-90° triangle: hectoicosagon (120-sided)
  - 89.25°-0.75°-90° triangle: dihectotetracontagon (240-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.
- 9×2ⁿ-sided
  - 70°-20°-90° triangle: enneagon (9-sided)
  - 80°-10°-90° triangle: octakaidecagon (18-sided)
  - 85°-5°-90° triangle: triacontakaihexagon (36-sided)
  - 87.5°-2.5°-90° triangle: heptacontakaidigon (72-sided)
  - ...
- 45×2ⁿ-sided
  - 86°-4°-90° triangle: tetracontakaipentagon (45-sided)
  - 88°-2°-90° triangle: enneacontagon (90-sided)
  - 89°-1°-90° triangle: hectaoctacontagon (180-sided)
  - 89.5°-0.5°-90° triangle: trihectohexacontagon (360-sided)
  - ...

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/(5×2^m)

Chord(36°) = a/b = 1/f, from Ptolemy's theorem

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

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

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

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.