Historically, a multitude of functions that can be derived from sine and cosine
were provided in tables along with their logarithms for early global navigators.
Today with the help of electronic computing, most of these are derived directly
from sine and cosine (ज्य "jyā" and कोज्य "kotijyā") when needed, or as complex exponentials e^{jθ}.

AX | = | sin | = | sine | AY | = | cos | = | cosine | AT | = | tan | = | tangent | AC | = | cot | = | cotangent |

CF | = | sec | = | secant | TF | = | csc | = | cosecant | ET | = | exs | = | exsecant | NC | = | exc | = | excosecant |

XE | = | ver | = | versine | YN | = | cvs | = | coversine | XW | = | vcs | = | vercosine | YS | = | cvc | = | covercosine |

HR | = | hav | = | haversine | HU | = | hcv | = | hacoversine | HL | = | hvc | = | havercosine | HD | = | hcc | = | hacovercosine |

E^{͡}A | = | arc | = | angle | AE | = | crd | = | chord | MG | = | sag | = | sagitta | MO | = | apo | = | apothem |

A hacked over cosine is a sine wave that oscillates between zero and one. Here it is on the unit circle and unit square with hcc and 19 of its trigonometric relatives. (Actually hacovercosine is short for halved-co-versed-co-sine, but hacked-over cosine seems to describe it more colorfully )

Sines and cosines marking the coordinate of a point on a unit circle corresponding to a given angle measured in radians (i.e. arclength on a unit circle) can be expressed through Taylor series expansion. When the imaginary number i is used to represent an imaginary unit orthogonal to one, (which when squared is equal to -1), the complex exponential emerges from Taylor expansion.

${e}^{i\theta}=\sum _{n=0}^{\mathrm{\infty}}\frac{{(i\theta )}^{n}}{n!}=1+i\frac{{\theta}^{1}}{1!}-\frac{{\theta}^{2}}{2!}-i\frac{{\theta}^{3}}{3!}+\frac{{\theta}^{4}}{4!}+i\frac{{\theta}^{5}}{5!}-\frac{{\theta}^{6}}{6!}-i\frac{{\theta}^{7}}{7!}+...=cos\theta +isin\theta $
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where
$$ {i}^{2}=-1$$with a repeating pattern that repeats in circles like:

${e}^{i0}={i}^{0}=1$

${e}^{i\mathrm{\pi /2}}={i}^{1}=i$

${e}^{i\mathrm{\pi 2/2}}={i}^{2}=-1$

${e}^{i\mathrm{\pi 3/2}}={i}^{3}=\mathrm{-i}$

${e}^{i\mathrm{\pi 4/2}}={i}^{4}=1$

${e}^{i\mathrm{\pi 5/2}}={i}^{5}=i$

${e}^{i\mathrm{\pi 6/2}}={i}^{6}=-1$

${e}^{i\mathrm{\pi 7/2}}={i}^{7}=\mathrm{-i}$

${e}^{i\mathrm{\pi 8/2}}={i}^{8}=1$

${e}^{i\mathrm{\pi 9/2}}={i}^{9}=i$

${e}^{i\mathrm{\pi 10/2}}={i}^{10}=-1$

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or if you're an engineer, avoid $i$ by using $j$, (since the letter
*i* is used for *intensité de courant* in engineering)

where ${j}^{2}=-1$

Just because both ${i}^{2}=-1$ and ${j}^{2}=-1$ does not mean $i$ and $j$ are the same imaginary number. In fact, we can multiply $i$ and $j$ together to get a different value than multiplying them together in reverse order.

$ij=-ji$So what is $ij$ ? It is yet another complex unit distinct from both $i$ and $j$. When you work with these complex numbers $i$, $j$, and $\mathrm{ij}$ together, they are hypercomplex numbers called quaternions.

$ijij=-jiij=-j\left({i}^{2}\right)j=-j(-1)j={j}^{2}=-1$
For convenience, *k* is defined as *ij*, and symetrically, it behaves exactly like *i* and *j*, mutatus mutandus.

Using imaginary values *i*,*j*, and *k* individually with real numbers
constructs complex numbers. Using them together constructs hypercomplex numbers called quaternions.
Quaternions are nice for representating rotations in 3 dimensional space for robotics,
or virtual reality and other 3D rendering applications. They can also be used for
representing composite color. In either application, the "complex" parts are more
related to real-life quantities than the so-called "real" parts... so maybe they should
have been named the other way around.

I used quaternions for color object recognition that I use for feedback for a crane robot in simulation. It's fairly easy to calculate inverse kinematics to determine the length that you want the ropes to be to control the robotic crane, but the calculations are more involved when you want to determine the position of the platform from the lengths of the ropes. Fortunately that doesn't need to be calculated to control the robot, but I looked into how to do it anyway.

A horizontal beam runs s_{0} units between two points
P_{1} on the left, and P_{2} on the right.
String S_{1} of length s_{1} is attached at its upper end to P_{1}, and
string S_{2} of length s_{2} is attached at its upper end to P_{2}.
The two strings are connected together at their lower ends to
a heavy object at point to Z to keep them taut, and creating a swing.
The swing has radius r along a circle about point Q on the horizontal beam.

Brahmagupta's Formula calculates the area of a cyclic quadrilateral, given the side lengths:

${K}_{4}=\frac{\sqrt{{(\sum {s}^{2})}^{2}+8\prod s-2\sum {s}^{4}}}{4}$ or Heron's a for cyclic triangle ${K}_{3}=\frac{\sqrt{{(\sum {s}^{2})}^{2}-2\sum {s}^{4}}}{4}$The radius of the swing then is the height of the triangle

$r=\frac{2{K}_{3}}{{s}_{0}}=\frac{2\sqrt{{(\sum {s}^{2})}^{2}-2\sum {s}^{4}}}{4{s}_{0}}$The distance P_{1}Q and the distance P_{2}Q are

Three such swings, joined at the endpoints of their beams to
form a horizontal equilateral triangle P_{1}P_{2}P_{3}
mutatus mutandus to form swing traingles
P_{1}Z_{1}P_{2},
P_{2}Z_{2}P_{3}, and
P_{3}Z_{3}P_{1}.
The beams have a common size, s_{0}s, but the string lengths
s_{1} and s_{2} are independently variable, having
s_{11} and s_{21} for the first swing,
s_{12} and s_{22} for the second swing, and
s_{13} and s_{23} for the third swing.
The swinging points Z_{1},Z_{2},Z_{3} are constrained to be equadistant from one another,
connecting them with straight line beams of length one results in another equilateral triangle similar in configuration
to an inverted Stewart platform, or a NIST RoboCrane®.
Here is a diagram showing circles of swing of one such a robotic crane.

The forward kinematic problem is reduced to finding the set of three points
(x_{1},y_{1},z_{1}),
(x_{2},y_{2},z_{2}),
(x_{3},y_{3},z_{3})
lower than the support such that
the distance separating each pair of points one.
Then since
$x=T{x}_{0}$, and
${x}_{0}$ is known at the
zero-reference pose of the platform, then then

$X=\left(\begin{array}{ccc}{x}_{1}& {x}_{2}& {x}_{3}\\ {y}_{1}& {y}_{2}& {y}_{3}\\ {z}_{1}& {z}_{2}& {z}_{3}\end{array}\right)$

$X=T{X}_{0}$

$T=X{{X}_{0}}^{-1}$

The 8th order direct kinematic solution was published by Nanua, Waldron, and Murthy (1990). Yes, after deriving all of that, I found it was already done, for it is written: ט מַה-שֶּׁהָיָה, הוּא שֶׁיִּהְיֶה, וּמַה-שֶּׁנַּעֲשָׂה, הוּא שֶׁיֵּעָשֶׂה; וְאֵין כָּל-חָדָשׁ, תַּחַת הַשָּׁמֶשׁ. (What has been will be again, what has been done will be done again; there is nothing new under the sun.)