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.

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

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

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

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

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

6°: 30-sided polygon

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

$cosfrac\{pi\}\{30\}=cos\; 6^circ=frac\{sqrt\{2(5-sqrt5)\}+sqrt3(sqrt5+1)\}\{8\},$

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

$cotfrac\{pi\}\{30\}=cot\; 6^circ=frac\{sqrt3(3+sqrt5)+sqrt\{2(25+11sqrt5)\}\}\{2\},$

9°: 20-sided polygon

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

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

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

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

12°: 15-sided polygon

$sinfrac\{pi\}\{15\}=sin\; 12^circ=frac\{sqrt\{2(5+sqrt5)\}-sqrt3(sqrt5-1)\}\{8\},$

$cosfrac\{pi\}\{15\}=cos\; 12^circ=frac\{sqrt\{6(5+sqrt5)\}+sqrt5-1\}\{8\},$

$tanfrac\{pi\}\{15\}=tan\; 12^circ=frac\{sqrt3(3-sqrt5)-sqrt\{2(25-11sqrt5)\}\}\{2\},$

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

15°: dodecagon

$sinfrac\{pi\}\{12\}=sin\; 15^circ=frac\{sqrt2(sqrt3-1)\}\{4\},$

$cosfrac\{pi\}\{12\}=cos\; 15^circ=frac\{sqrt2(sqrt3+1)\}\{4\},$

$tanfrac\{pi\}\{12\}=tan\; 15^circ=2-sqrt3,$

$cotfrac\{pi\}\{12\}=cot\; 15^circ=2+sqrt3,$

18°: decagon

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

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

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

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

21°: sum 9° + 12°

$sinfrac\{7pi\}\{60\}=sin\; 21^circ=frac\{2(sqrt3+1)sqrt\{5-sqrt5\}-sqrt2(sqrt3-1)(1+sqrt5)\}\{16\},$

$cosfrac\{7pi\}\{60\}=cos\; 21^circ=frac\{2(sqrt3-1)sqrt\{5-sqrt5\}+sqrt2(sqrt3+1)(1+sqrt5)\}\{16\},$

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

$cotfrac\{7pi\}\{60\}=cot\; 21^circ=frac\{left(2-(2-sqrt3)(3-sqrt5)right)left(2+sqrt\{10sqrt5\}right)\}\{4\},$

22.5°: octagon

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

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

$tanfrac\{pi\}\{8\}=tan\; 22.5^circ=sqrt\{2\}-1,$

$cotfrac\{pi\}\{8\}=cot\; 22.5^circ=sqrt\{2\}+1,$

24°: sum 12° + 12°

$sinfrac\{2pi\}\{15\}=sin\; 24^circ=frac\{sqrt3(sqrt5+1)-sqrt2sqrt\{5-sqrt5\}\}\{8\},$

$cosfrac\{2pi\}\{15\}=cos\; 24^circ=frac\{sqrt6sqrt\{5-sqrt5\}+sqrt5+1\}\{8\},$

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

$cotfrac\{2pi\}\{15\}=cot\; 24^circ=frac\{sqrt2sqrt\{5-sqrt5\}+sqrt3(sqrt5-1)\}\{2\},$

27°: sum 12° + 15°

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

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

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

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

30°: hexagon

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

$cosfrac\{pi\}\{6\}=cos\; 30^circ=frac\{sqrt3\}\{2\},$

$tanfrac\{pi\}\{6\}=tan\; 30^circ=frac\{sqrt3\}\{3\},$

$cotfrac\{pi\}\{6\}=cot\; 30^circ=sqrt3,$

33°: sum 15° + 18°

$sinfrac\{11pi\}\{60\}=sin\; 33^circ=frac\{2(sqrt3-1)sqrt\{5+sqrt5\}+sqrt2(1+sqrt3)(sqrt5-1)\}\{16\},$

$cosfrac\{11pi\}\{60\}=cos\; 33^circ=frac\{2(sqrt3+1)sqrt\{5+sqrt5\}+sqrt2(1-sqrt3)(sqrt5-1)\}\{16\},$

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

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

36°: pentagon

$sinfrac\{pi\}\{5\}=sin\; 36^circ=frac\{sqrt\{2(5-sqrt5)\}\}\{4\},$

$cosfrac\{pi\}\{5\}=cos\; 36^circ=frac\{1+sqrt5\}\{4\}=frac\{varphi\}\{2\},$

$tanfrac\{pi\}\{5\}=tan\; 36^circ=sqrt\{5-2sqrt5\},$

$cotfrac\{pi\}\{5\}=cot\; 36^circ=frac\{sqrt\{5(5+2sqrt5)\}\}\{5\},$

39°: sum 18°+ 21°

$sinfrac\{13pi\}\{60\}=sin\; 39^circ=frac\{2(1-sqrt3)sqrt\{5-sqrt5\}+sqrt2(sqrt3+1)(sqrt5+1)\}\{16\},$

$cosfrac\{13pi\}\{60\}=cos\; 39^circ=frac\{2(1+sqrt3)sqrt\{5-sqrt5\}+sqrt2(sqrt3-1)(sqrt5+1)\}\{16\},$

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

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

42°: sum 21° + 21°

$sinfrac\{7pi\}\{30\}=sin\; 42^circ=frac\{sqrt6sqrt\{5+sqrt5\}-sqrt5+1\}\{8\},$

$cosfrac\{7pi\}\{30\}=cos\; 42^circ=frac\{sqrt2sqrt\{5+sqrt5\}+sqrt3(sqrt5-1)\}\{8\},$

$tanfrac\{7pi\}\{30\}=tan\; 42^circ=frac\{sqrt3(sqrt5+1)-sqrt2sqrt\{5+sqrt5\}\}\{2\},$

$cotfrac\{7pi\}\{30\}=cot\; 42^circ=frac\{sqrt\{2(25-11sqrt5)\}+sqrt3(3-sqrt5)\}\{2\},$

45°: square

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

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

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

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

60°: triangle

$sinfrac\{pi\}\{3\}=sin\; 60^circ=frac\{sqrt3\}\{2\},$

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

$tanfrac\{pi\}\{3\}=tan\; 60^circ=sqrt3,$

$cotfrac\{pi\}\{3\}=cot\; 60^circ=frac\{3\},$

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:

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

Using

$cos\; 36^circ=frac\{sqrt5+1\}\{4\},$

$tan\; 36^circ=sqrt\{5-2sqrt5\},$

this can be simplified to:

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

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 {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)

Geometrical method

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

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

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

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

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

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

so

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

Algebraic method

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

$sin\{5x\}=16sin^5\; x-20sin^3\; x+5sin\; x,$,

$cos\{5x\}=16cos^5\; x-20cos^3\; x+5cos\; x,$.

• When $sin\{5x\}=0,$ or $cos\{5x\}=0,$, we let $y=sin\; x,$ or $y=cos\; x,$ and solve for $y,$:

$16y^5-20y^3+5y=0,$.

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 $sqrt\{a+bsqrt\; c\},$,

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

and both

$d=pmsqrt\{frac\{apm\; R\}\{2\}\},$ and

$e=pmsqrt\{frac\{apm\; R\}\{2c\}\},$

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

$sqrt\{a+bsqrt\; c\}=d+esqrt\; c,$

Example:

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

See also

• Trigonometric function
• Trigonometric identity
• Constructible polygon
• Trigonometric number
• 17-gonal construction

References

• π/3 (60°) — π/6 (30°) — π/12 (15°) — π/24 (7.5°)
• π/4 (45°) — π/8 (22.5°) — π/16 (11.25°) — π/32 (6.625°)
• π/5 (36°) — π/10 (18°) — π/20 (9°)
• π/7 — π/14
• π/9 (20°) — π/18 (10°)
• π/11
• π/13
• π/15 (12°) — π/30 (6°)
• π/17
• π/19
• π/23

External links

• Constructible Regular Polygons
• Naming polygons