trigonometry

daamath maintains the full lattice of trigonometric functions. we also expose their relationship directly from [hypercomplex] algebra, believe it or not

introduction

the trigonometric functions are derived from the exponential series

ι⁰/0! + ι¹/1! + ι²/2! + ι³/3! + …

where if you set ι = ix or εx or jx, the even and odd terms will give you cos and sin for the elliptic, parabolic, hyperbolic geometries respectively

totally we have 2 ⋅ 3 ⋅ 3P2 = 36 functions. 2 from the normal trig function and its inverse 3 from the three geometries 3P2 from a triangle of three sides, from which we take two sides

unfortunately, parabolic trigonometry is trivial, because . thus daamath excludes parabolic trigonometry and maintains 24 instead.

lastly, you may notice that the word 'angle' is never mentioned. that is because, while angles are convenient in circular geometry, in hyperbolic functions, we dont take the angle of anything anymore. we take the semiarea of the hyperbola with the x = y line. in fact, in circular geometry, what we are really describing when we say 'the angle' is the semiarea subtended by the section. this idea of semiarea extends to parabolic geometry as well.

API implementation

sin

given the semiarea, return the ratio of opposite to hypotenuse in circular geometry

cos

given the semiarea, return the ratio of adjacent to hypotenuse in circular geometry

tan

given the semiarea, return the ratio of opposite to adjacent in circular geometry

cot

given the semiarea, return the ratio of adjacent to opposite in circular geometry

sec

given the semiarea, return the ratio of hypotenuse to adjacent in circular geometry

csc

given the semiarea, return the ratio of hypotenuse to opposite in circular geometry

asin

given the ratio of opposite to hypotenuse, return the semiarea in circular geometry

acos

given the ratio of adjacent to hypotenuse, return the semiarea in circular geometry

atan

given the ratio of opposite to adjacent, return the semiarea in circular geometry

acot

given the ratio of adjacent to opposite, return the semiarea in circular geometry

asec

given the ratio of hypotenuse to adjacent, return the semiarea in circular geometry

acsc

given the ratio of hypotenuse to opposite, return the semiarea in circular geometry

sinh

given the semiarea, return the ratio of opposite to hypotenuse in hyperbolic geometry

cosh

given the semiarea, return the ratio of adjacent to hypotenuse in hyperbolic geometry

tanh

given the semiarea, return the ratio of opposite to adjacent in hyperbolic geometry

coth

given the semiarea, return the ratio of adjacent to opposite in hyperbolic geometry

sech

given the semiarea, return the ratio of hypotenuse to adjacent in hyperbolic geometry

csch

given the semiarea, return the ratio of hypotenuse to opposite in hyperbolic geometry

asinh

given the ratio of opposite to hypotenuse, return the semiarea in hyperbolic geometry

acosh

given the ratio of adjacent to hypotenuse, return the semiarea in hyperbolic geometry

atanh

given the ratio of opposite to adjacent, return the semiarea in hyperbolic geometry

acoth

given the ratio of adjacent to opposite, return the semiarea in hyperbolic geometry

asech

given the ratio of hypotenuse to adjacent, return the semiarea in hyperbolic geometry

acsch

given the ratio of hypotenuse to opposite, return the semiarea in hyperbolic geometry

rant

why do trigonometric functions have anything to do with hypercomplex numbers? because the hypercomplex numbers encode geometries, which are beautifully expressed through these trig functions we have. there are three 2-dimensional unital algebrae: complex numbers, split-complex numbers, dual numbers. they correspond to circular, hyperbolic, and parabolic geometry.

okay im gonna solve why i use semiarea. for elliptic geometry, its trivial. 2π radians ∝ π area. 2 semiarea ∝ 1 area. so 2π radians ∝ 2π semiarea. so radian = semiarea already. but lets prove by geometry too.

ELLIPTIC GEOMETRY:

ellipse: x² + y² = r² line: y = tx intersection: x² + t²x² = r² x² = r² / (1 + t²) y² = r²t² / (1 + t²) (x = r(1 + t²)⁻⁰·⁵, y = rt(1 + t²)⁻⁰·⁵)

area between ellipse and line:

full area = abπr²

sector area = t

DUAL LINE GEOMETRY:

dual line: x² = r² line: y = tx intersection: (x = ±1, y = t/2)

HYPERBOLIC GEOMETRY: x² - y² = r²