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° = 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. One could in principle write down formulae involving the angles 2π/17, 2π/257, or 2π/65537, but they would be too unwieldy for most applications. 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 How can the trig values for sin and cos be calculated?
4 How can the trig values for tan and cot be calculated?
5 Plans for simplifying
6 See also
7 External links

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

$\sin 0^\circ = 0$

$\cos 0^\circ = 1$

$\tan 0^\circ = 0$

$\sin \frac {\pi}{60} = \sin 3^\circ = \frac{ 2 (1 - \sqrt3) \sqrt{5 + \sqrt5} + \sqrt2 (\sqrt5 - 1) (\sqrt3 + 1) }{16}$

$\cos \frac {\pi}{60} = \cos 3^\circ = \frac{ 2 (1 + \sqrt3) \sqrt{5 + \sqrt5} + \sqrt2 (\sqrt5 - 1) (\sqrt3 - 1) }{16}$

$\tan \frac {\pi}{60} = \tan 3^\circ = \frac{ \left( (2 - \sqrt3) (3 + \sqrt5) - 2 \right) \left(2 - \sqrt{2 (5 - \sqrt5)}\right) }{4}$

$\cot \frac {\pi}{60} = \cot 3^\circ = \frac{ \left( (2 + \sqrt3) (3 + \sqrt5) - 2 \right) \left(2 + \sqrt{2 (5 - \sqrt5)}\right) }{4}$

$\sin \frac {\pi}{30} = \sin 6^\circ = \frac{(\sqrt6) \sqrt{5 - \sqrt5} - (\sqrt5 + 1)}{8}$

$\cos \frac {\pi}{30} = \cos 6^\circ = \frac{(\sqrt2) \sqrt{5- \sqrt5} + \sqrt3( \sqrt5+1)}{8}$

$\tan \frac {\pi}{30} = \tan 6^\circ = \frac{(\sqrt2) \sqrt{5 - \sqrt5} - \sqrt3(\sqrt5 - 1)}{2}$

$\cot \frac {\pi}{30} = \cot 6^\circ = \frac{\sqrt3 (3 + \sqrt5) + \sqrt{51 + 22 \sqrt5}}{2}$

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

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

$\tan \frac {\pi}{20} = \tan 9^\circ = \sqrt5 + 1 - \sqrt{5 + 2\sqrt5}$

$\cot \frac {\pi}{20} = \cot 9^\circ = \sqrt5 + 1 + \sqrt{5 + 2\sqrt5}$

$\sin \frac{\pi}{15} = \sin 12^\circ = \frac{(\sqrt2) \sqrt{5 + \sqrt5} - \sqrt 3 (\sqrt 5 -1)}{8}$

$\cos \frac{\pi}{15} = \cos 12^\circ = \frac{(\sqrt6) \sqrt{5 + \sqrt5} + (\sqrt 5 - 1)}{8}$

$\tan \frac{\pi}{15} = \tan 12^\circ = \frac{(\sqrt3) (3 - \sqrt5 ) - \sqrt{50 - 22 \sqrt5}}{2}$

$\cot \frac{\pi}{15} = \cot 12^\circ = \frac{\sqrt3 (\sqrt5 + 1) + \sqrt2 \sqrt{5 + \sqrt5}}{2}$

$\sin \frac{\pi}{12} = \sin 15^\circ = \frac{\sqrt 2 \left(\sqrt 3 - 1\right)}{4}$

$\cos \frac{\pi}{12} = \cos 15^\circ = \frac{\sqrt 2 \left(\sqrt 3 + 1\right)}{4}$

$\tan \frac{\pi}{12} = \tan 15^\circ = 2 - \sqrt 3$

$\cot \frac{\pi}{12} = \cot 15^\circ = 2 + \sqrt 3$

$\sin \frac{\pi}{10} = \sin 18^\circ = \frac{\sqrt 5 - 1}{4}$

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

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

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

$\sin \frac{7\pi}{60} = \sin 21^\circ = \frac{2(\sqrt 3 + 1) \sqrt{5 - \sqrt 5} - \sqrt 2 (\sqrt 3 - 1) (1 + \sqrt 5)} {16}$

$\cos \frac{7\pi}{60} = \cos 21^\circ = \frac{2 (\sqrt 3 - 1) \sqrt{5 - \sqrt 5} + \sqrt 2 (\sqrt 3 + 1) (1 + \sqrt 5)} {16}$

$\tan \frac {7 \pi}{60} = \tan 21^\circ = \frac{ \left(2 - (2 + \sqrt3) (3 - \sqrt5) \right) \left(2 - \sqrt{2 (5 + \sqrt5)}\right) }{4}$

$\cot \frac {7 \pi}{60} = \cot 21^\circ = \frac{ \left(2 - (2 - \sqrt3) (3 - \sqrt5) \right) \left(2 + \sqrt{2 (5 +\sqrt5)}\right) }{4}$

$\sin \frac {\pi}{8} = \sin 22.5^\circ = \frac{\sqrt{2 - \sqrt{2}}}{2}$

$\cos \frac {\pi}{8} = \cos 22.5^\circ = \frac{\sqrt{2 + \sqrt{2}}}{2}$

$\tan \frac {\pi}{8} = \tan 22.5^\circ = \sqrt{2}-1$

$\cot \frac {\pi}{8} = \cot 22.5^\circ = \sqrt{2}+1$

$\sin \frac {2\pi}{15} = \sin 24^\circ = \frac{\sqrt3(\sqrt5 + 1) - \sqrt2 \sqrt{5 - \sqrt5}}{8}$

$\cos \frac {2\pi}{15} = \cos 24^\circ = \frac{\sqrt6 \sqrt{5 - \sqrt5} + \sqrt5 + 1}{8}$

$\tan \frac {2\pi}{15} = \tan 24^\circ = \frac{\sqrt{50 + 22 \sqrt5} - \sqrt3 (3 + \sqrt5)}{2}$

$\cot \frac {2\pi}{15} = \cot 24^\circ = \frac{\sqrt2 \sqrt{5 - \sqrt5} + \sqrt3(\sqrt5 - 1)}{2}$

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

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

$\tan \frac {3\pi}{20} = \tan 27^\circ = \sqrt5 - 1 - \sqrt{5 - 2 \sqrt5}$

$\cot \frac {3\pi}{20} = \cot 27^\circ = \sqrt5 - 1 + \sqrt{5 - 2 \sqrt5}$

$\sin \frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}$

$\cos \frac{\pi}{6} = \cos 30^\circ = \frac{\sqrt 3}{2}$

$\tan \frac{\pi}{6} = \tan 30^\circ = \frac{\sqrt 3}{3}$

$\cot \frac{\pi}{6} = \cot 30^\circ = \frac{3}{\sqrt 3} = \sqrt 3$

$\sin \frac{11\pi}{60} = \sin 33^\circ = \frac{2(\sqrt 3 - 1) \sqrt{5 + \sqrt 5} + \sqrt 2 (1 + \sqrt 3) (\sqrt 5 - 1)} {16}$

$\cos \frac{11\pi}{60} = \cos 33^\circ = \frac{2 (\sqrt 3 + 1) \sqrt{5 + \sqrt 5} + \sqrt 2 (1 - \sqrt 3) (\sqrt 5 - 1)} {16}$

$\tan \frac {11 \pi}{60} = \tan 33^\circ = \frac{ \left(2 - (2 - \sqrt3) (3 + \sqrt5) \right) \left(2 + \sqrt{2 (5 - \sqrt5)}\right) }{4}$

$\cot \frac {11 \pi}{60} = \cot 33^\circ = \frac{ \left(2 - (2 + \sqrt3) (3 + \sqrt5) \right) \left(2 - \sqrt{2 (5 - \sqrt5)}\right) }{4}$

$\sin \frac{\pi}{5} = \sin 36^\circ = \frac{\sqrt{2(5 - \sqrt 5)} }{4}$

$\cos \frac{\pi}{5} = \cos 36^\circ = \frac{\sqrt 5+1}{4}$

$\tan \frac{\pi}{5} = \tan 36^\circ = \sqrt{5 - 2\sqrt 5}$

$\cot \frac{\pi}{5} = \cot 36^\circ = \frac{ \sqrt{5(5 + 2\sqrt 5)}}{5}$

$\sin{\frac{13\pi}{60}} = \sin{39^\circ} = \frac{2(1-\sqrt 3)\sqrt{5-\sqrt 5} + \sqrt 2 (\sqrt 3 + 1)(\sqrt 5 + 1)}{16}$

$\cos{\frac{13\pi}{60}} = \cos{39^\circ} = \frac{2 (1+\sqrt 3)\sqrt{5-\sqrt 5} + \sqrt2(\sqrt 3 - 1)(\sqrt 5 + 1)}{16}$

$\tan \frac {13\pi}{60} = \tan 39^\circ = \frac{ \left( (2 - \sqrt3) (3 - \sqrt5) - 2 \right) \left(2 - \sqrt{2 (5 + \sqrt5)}\right) }{4}$

$\cot \frac {13\pi}{60} = \cot 39^\circ = \frac{ \left( (2 + \sqrt3) (3 - \sqrt5) - 2 \right) \left(2 + \sqrt{2 (5 + \sqrt5)}\right) }{4}$

$\sin \frac {7\pi}{30} = \sin 42^\circ = \frac{ \sqrt6 \sqrt{5 + \sqrt5} - (\sqrt5 + 1)}{8}$

$\cos \frac {7\pi}{30} = \cos 42^\circ = \frac{ \sqrt2 \sqrt{5 + \sqrt5} + \sqrt3(\sqrt5 - 1)}{8}$

$\tan \frac {7\pi}{30} = \tan 42^\circ = \frac{ \sqrt3(\sqrt5 + 1) - \sqrt2 \sqrt{5 + \sqrt5}}{2}$

$\cot \frac {7\pi}{30} = \cot 42^\circ = \frac{ \sqrt{50 - 22 \sqrt5} + \sqrt3(3 - \sqrt5)}{2}$

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

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

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

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

$\sin \frac{\pi}{3} = \sin 60^\circ = \frac{\sqrt 3}{2}$

$\cos \frac{\pi}{3} = \cos 60^\circ = \frac{1}{2}$

$\tan \frac{\pi}{3} = \tan 60^\circ = \sqrt 3$

$\cot \frac{\pi}{3} = \cot 60^\circ = \frac{\sqrt 3}{3}$

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

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

Using

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

$\tan{36^\circ} = \sqrt{5 - 2 \sqrt 5}$

this can be simplified to:

$V = \frac{ e^3\left(15 + 7 \sqrt 5 \right)}{4}$

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 = 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
- 3×2^X-sided regular polygons, X = 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)
  - ...
- 4×2^X-sided
  - 45°-45°-90° triangle - square (4-sided)
  - 67.5°-22.5°-90° triangle - octagon (8-sided)
  - 88.75°-11.25°-90° triangle - hexakaidecagon (16-sided)
  - ...
- 5×2^X-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)
  - ...
- 15×2^X-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 because of Casus Irreducibilis.
- 9×2^X-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)
  - ...
- 45×2^X-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)
  - ...

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

The multiple angle formulas for functions of 5x, where x = {18, 36, 54, 72, 90} and 5x = {90, 180, 270, 360, 540}, can be solved for the functions of x, since we know the function values of 5x. The multiple angle formulas are:

sin5 x = 16sin 5 x - 20sin 3 x + 5sin x

cos5 x = 16cos 5 x - 20cos 3 x + 5cos x

When sin 5x = 0 or cos 5x = 0, we let y = sin x or y = cos x and solve for y:

16 y 5 - 20 y 3 + 5 y = 0

One solution is zero, and the resulting 4th degree equation can be solved as a quadratic in y-squared.

When sin 5x = 1 or cos 5x = 1, we again let y = sin x or y = cos x and solve for y:

16 y 5 - 20 y 3 + 5 y - 1 = 0

which factors into

(y - 1)(4 y 2 + 2 y - 1) 2 = 0

9° is 45 - 36, and 27° is 45 - 18; so we use the subtraction formulas for sin and cos.

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

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 sin and cos.

Tangent is sine divided by cosine, and cotangent is cosine divided by sine.

Set up each fraction and simplify.

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.

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

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.

Main article: Nested radicals

In general nested radicals cannot be reduced.

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