Exact trigonometric constants |
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: Trigonometric_identity#Half-angle_formulas, Trigonometric_identity#Double-angle_formulas, Trigonometric_identity#Angle_sum_and_difference_identities and values for 0°, 30°, 36° and 45°. Note that 1° = π/180 radians.
= Table of constants =
Values outside 0° ... 45° angle range are trivially extracted from circle axis Coordinate_rotations_and_reflections symmetry from these values. (See Trigonometric identity#Periodicity, symmetry, and shifts)
== 0° Fundamental ==
:sin 0^circ = 0 :cos 0^circ = 1 : an 0^circ = 0
== 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}
: an frac {pi}{60} = an 3^circ = frac{ left( (2 - sqrt3) (3 + sqrt5) - 2 ight) left(2 - sqrt{2 (5 - sqrt5)} ight) }{4}
== 6° - 30-sided polygon ==
: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} : an frac {pi}{30} = an 6^circ = frac{(sqrt2) sqrt{5 - sqrt5} - sqrt3(sqrt5 - 1)}{2} :cot frac {pi}{30} = cot 6^circ = frac{sqrt3 (3 + sqrt5) + sqrt{50 + 22 sqrt5}}{2}
== 9° - 20-sided polygon ==
:sin frac {pi}{20} = sin 9^circ = frac{sqrt2(sqrt5 + 1) - 2sqrt{5 - sqrt5}}{8} :cos frac {pi}{20} = cos 9^circ = frac{sqrt2(sqrt5 + 1) + 2sqrt{5 - sqrt5}}{8} : an frac {pi}{20} = an 9^circ = sqrt5 + 1 - sqrt{5 + 2sqrt5} :cot frac {pi}{20} = cot 9^circ = sqrt5 + 1 + sqrt{5 + 2sqrt5}
== 12° - 15-sided polygon ==
: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} : an frac{pi}{15} = an 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}
== 15° - 12-sided polygon ==
:sin frac{pi}{12} = sin 15^circ = frac{sqrt 2 left(sqrt 3 - 1 ight)}{4}
:cos frac{pi}{12} = cos 15^circ = frac{sqrt 2 left(sqrt 3 + 1 ight)}{4}
: an frac{pi}{12} = an 15^circ = 2 - sqrt 3
:cot frac{pi}{12} = cot 15^circ = 2 + sqrt 3
== 18° - 10-sided polygon ==
: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} : an frac{pi}{10} = an 18^circ = frac{sqrt{5(5 - 2 sqrt 5)}}{5} :cot frac{pi}{10} = cot 18^circ = sqrt{5 + 2 sqrt 5}
== 21° - Sum 9° + 12° ==
:sin frac{7pi}{60} = sin 21^circ = frac{2(sqrt 3 + 1) sqrt{5 - sqrt 5} - sqrt 2 (sqrt 3 - 1) (1 + sqrt 5)} {16} :cos frac{7pi}{60} = cos 21^circ = frac{2 (sqrt 3 - 1) sqrt{5 - sqrt 5} + sqrt 2 (sqrt 3 + 1) (1 + sqrt 5)} {16} : an frac{7pi}{60} = an 21^circ = frac{(1 + 2 sqrt 3 - sqrt 5) sqrt{5 - 2 sqrt 5} + (2 + sqrt 3)(sqrt 5 - 3) + 2} {2}
== 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} : an frac {pi}{8} = an 22.5^circ = sqrt{2}-1 :cot frac {pi}{8} = cot 22.5^circ = sqrt{2}+1
== 24° - Sum 12° + 12° ==
:sin frac {2pi}{15} = sin 24^circ = frac{sqrt3(sqrt5 + 1) - sqrt2 sqrt{5 - sqrt5}}{8} :cos frac {2pi}{15} = cos 24^circ = frac{sqrt6 sqrt{5 - sqrt5} + sqrt5 + 1}{8} : an frac {2pi}{15} = an 24^circ = frac{sqrt{50 + 22 sqrt5} - sqrt3 (3 + sqrt5)}{2} :cot frac {2pi}{15} = cot 24^circ = frac{sqrt2 sqrt{5 - sqrt5} + sqrt3(sqrt5 - 1)}{2}
== 27° - Sum 12° + 15° ==
:sin frac {3pi}{20} = sin 27^circ = frac{2 sqrt{5 + sqrt5} - sqrt2(sqrt5 - 1)}{8} :cos frac {3pi}{20} = cos 27^circ = frac{2 sqrt{5 + sqrt5} + sqrt2(sqrt5 - 1)}{8} : an frac {3pi}{20} = an 27^circ = sqrt5 - 1 - sqrt{5 - 2 sqrt5} :cot frac {3pi}{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{sqrt 3}{2} : an frac{pi}{6} = an 30^circ = frac{sqrt 3}{3} :cot frac{pi}{6} = cot 30^circ = frac{3}{sqrt 3} = sqrt 3
== 33° - Sum 15° + 18° ==
:sin frac{11pi}{60} = sin 33^circ = frac{2(sqrt 3 - 1) sqrt{5 + sqrt 5} + sqrt 2 (sqrt 3 + 1) (sqrt 5 - 1)} {16} :cos frac{11pi}{60} = cos 33^circ = frac{2 (sqrt 3 + 1) sqrt{5 + sqrt 5} + sqrt 2 (1 - sqrt 3) (sqrt 5 - 1)} {16} : an frac{11pi}{60} = an 33^circ = frac {(-15 + 10sqrt 3 - 7sqrt 5 + 4sqrt{15} )sqrt{5(5 - 2 sqrt 5)} + 5(sqrt 3 - 2)(3 + sqrt 5) + 10}{10}
== 36° - Pentagon ==
: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} : an frac{pi}{5} = an 36^circ = sqrt{5 - 2sqrt 5} :cot frac{pi}{5} = cot 36^circ = frac{ sqrt{5(5 + 2sqrt 5)}}{5}
== 39° - Sum 18°+ 21° ==
:sin{frac{13pi}{60}} = sin{39^circ} = frac{2(1-sqrt 3)sqrt{5-sqrt 5} + sqrt 2 (1 + sqrt 3)(1 + sqrt 5)}{16} :cos{frac{13pi}{60}} = cos{39^circ} = frac{2 (1+sqrt 3)sqrt{5-sqrt 5} + sqrt2(sqrt 3 - 1)(sqrt 5 + 1)}{16} : an{frac{13pi}{60}} = cos{39^circ} = frac{left(sqrt{2(5+sqrt 5)}-2 ight)left((2-sqrt 3)(sqrt 5 - 3) + 2 ight)}{4}
== 42° - Sum 21° + 21° ==
:sin frac {7pi}{30} = sin 42^circ = frac{ sqrt6 sqrt{5 + sqrt5} - (sqrt5 + 1)}{8} :cos frac {7pi}{30} = cos 42^circ = frac{ sqrt2 sqrt{5 + sqrt5} + sqrt3(sqrt5 - 1)}{8} : an frac {7pi}{30} = an 42^circ = frac{ sqrt3(sqrt5 + 1) - sqrt2 sqrt{5 + sqrt5}}{2} :cot frac {7pi}{30} = cot 42^circ = frac{ sqrt{50 - 22 sqrt5} + sqrt3(3 - sqrt5)}{2}
== 45° - Square ==
:sin frac{pi}{4} = sin 45^circ = frac{sqrt 2}{2} :cos frac{pi}{4} = cos 45^circ = frac{sqrt 2}{2} : an frac{pi}{4} = an 45^circ = 1 :cot frac{pi}{4} = cot 45^circ = 1
= Notes =
== Uses for constants ==
As an example of the use of these constants, consider a dodecahedron with the following volume:
:V = frac{5e^3cos{36^circ}}{ an^2{36^circ}}
Using :cos{36^circ} = frac{sqrt 5 + 1}{4} : an{36^circ} = sqrt{5 - 2 sqrt 5}
this can be simplified to:
:V = frac{ e^3left(15 + 7 sqrt 5 ight)}{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 -agon can be divided into 2 N 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.
= How can the trig values for sin and cos be calculated =
== The trivial ones ==
:In degree format: 0, 90, 45, 30 and 60 can be calculated from their triangles, using the pythagorean theorem.
== n π over 10 ==
: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: ::sin{5x} = 16 sin^5 x - 20sin^3 x + 5 sin x ::cos{5x} = 16 cos^5 x - 20cos^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: ::16 y^5 - 20 y^3 + 5y = 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 + 5y - 1 = 0 :which factors into ::(4y^2 + 2y - 1)^2 = 0
== n π over 20 ==
:9° is 45 - 36, and 27° is 45 - 18; so we use the subtraction formulas for sin and cos.
== n π over 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 sin and cos.
== n π over 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 sin and cos.
= How can the trig values for tan and cot be calculated =
:Tangent is sine divided by cosine, and cotangent is cosine divided by sine. :Set up each fraction and simplify.
= 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==
: Main Article: Nested_radicals
In general nested radicals cannot be reduced.
But if for sqrt{a + bsqrt c} ,
: R = sqrt{a^2 - b^2 c} 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 pmsigns,
then sqrt{a + bsqrt c} = d + esqrt c
Example:
: 4sin 18^circ = sqrt{6 - 2 sqrt 5} = sqrt 5 - 1
= See also =
= External links =
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