Cofunctions and the Unit Circle
The Trigonometric functions are Sine and Cosine, and four more named functions of the two Cofunctions.
Sine and Cosine are very similar, and are uniquely called Cofunctions.
The prefix, co-, means together, and is used when an object is coordinated with or coincidental to the root word, here function of angle.
There are other connotations of a complimentary dual object to a function, such as its inverse, but convention reserves
its usage to when it is significant.
The Cotangent is a complimentary segment to the Tangent, and Cosecant is the compliment to the Secant segment, but none of these four are cofunctions themselves.
The Inscribed segments, which are the Cofunctions, are (non-linear) functions of their angle, they are bounded, and they have unique symmetric properties connected to drawing rays of a circle oriented by Cartesian axes. The Outscribed segments each have a factor of the inverse of a cofunction, thereby imbuing them with properties of the Hyperbola.
The right triangle forms the basis for the Sine and Cosine functions. The Sine function is first looked at as an area rule, where its definition is linked to the triangle's area.
SOH-CAH-TOA is the second most important thing in trigonometry—only after the quadratic relation of Pythagoras. The definitions in eq. \eqref{sohcahtoa} give the relationships of the sides of the right angle triangle to the three functions, Sine, Cosine, and Tangent in a cute mnemonic.
Let's derive the Sine Rule for Area using the Sine function and Euclidean geometry, then look at the functions graphically.
Sine Rule for Area
The Sine Rule for Area (SRA) states that the area of a triangle, $A_\triangle$, is equal to one-half the product of two adjacent side lengths, $a$ and $b$, times the Sine of the included angle, $\theta$.
In general, a triangle does not have a right angle, which is why the SRA is useful for finding its area as it introduces a cofunction to the . Using an overlay of inscribed right triangles, using the fact that you can always compose any triangle using inscribed right triangles, we can then use the tools of right angles.
$$ \theta = \angle{}OQP = \angle{}OQR$$
$$\overline{OR} = \overline{OQ}\sin(\theta) $$
$$\overline{UP} = \overline{PQ}\sin(\theta) $$
Where the SRA variables, $a$ and $b$, are represented by the lengths of the sides $\overline{OQ}$ and $\overline{PQ}$, and the angle $\theta=\angle{}OQP$ is the included angle between those two sides.
The area of the primary triangle, $\triangle{}OQP$, is equal to the sum of the areas of the two inscribed right triangles with shared side, $\overline{OR}$, $\triangle{}ORQ$ and $\triangle{}POR$. The area of the right triangle $\triangle{}ORQ$ is equal to one-half the product of its two shorter sides, $\overline{OR}$, and its height, $\overline{RQ}$, which is $A_{\triangle{}ORQ} = \frac{1}{2} \overline{OR}\ \overline{RQ}$. The area of the right triangle $\triangle{}POR$ is equal to one-half the product of its base,$\overline{PR}$ , and its height, $\overline{OR}$, which is $A_{\triangle{}POR} = \frac{1}{2} \overline{OR}\ \overline{PR}$. So the area of the triangle $\triangle{}OQP$ is equal to the sum of these two areas, $A_{\triangle{}OQP} = \frac{1}{2} \overline{OR}\ \overline{RQ} + \frac{1}{2} \overline{OR}\ \overline{PR}$, or simplifying, $A_{\triangle{}OQP} = \frac{1}{2} \overline{OR}(\overline{RQ} + \overline{PR})=\frac{1}{2} \overline{OR}\ \overline{PQ}$. And substituting for $\overline{OR}$, we have $A_{\triangle{}OQP} = \frac{1}{2} \overline{OQ}\ \overline{PQ}\sin(\theta)$, which is the SRA as stated in \eqref{SRA}.
$$ A*{\triangle{}UPQ} = A*{\triangle{}OPQ} + A_{\triangle{}UPO} $$
To consider the other choice of the hypotenuse for angle $\theta$, we consider the triangle $\triangle{}UPQ$, where angle $\angle{}PUQ = 90^\circ$ as indicated with the small square at vertex $U$. Now the side $\overline{PQ}$ is the hypotenuse for angle $\theta$, and the side $\overline{UQ}$ is the short side adjacent to angle $\theta$, to use the right angle triangle area formula. So the area of triangle $\triangle{}UPQ$ is equal to the sum of the areas of triangles $\triangle{}OPQ$ and $\triangle{}UPO$, or rearranging, $A_{\triangle{}OPQ} = A_{\triangle{}UPQ} - A_{\triangle{}UPO} $, which is the SRA as stated in \eqref{SRA} since $\overline{OQ} = \overline{UQ} - \overline{UO}$, or:
$$ A_{\triangle{}OPQ} = A_{\triangle{}UPQ} - A_{\triangle{}UPO} = \frac{1}{2} \overline{UQ}\ \overline{PQ}\sin(\theta) - \frac{1}{2} \overline{UO}\ \overline{PQ}\sin(\theta) = \frac{1}{2} \overline{OQ}\ \overline{PQ}\sin(\theta)$$.
Which is what we set out to prove.
The notation of overlined points as segments is from the proofs of the early landmark work, Principia Mathematica (Isaac Newton, 1685).
Archimedean Area of the Unit Circle
Area of a circle: $ \pi r^2 $ where $r$ stands for the radius, or distance from the center of the circle to the edge. And note that the area of the square which is $2r$ on the side, is $4r^2$ and that the circle covers roughly 3/4 of the square. Pi was first studied by Archimedes, who bounded the number $\pi$ by the sum of sides of a smaller polygon (inscribed) and a larger one (circumscribed). So the circumferance of the circle ($2\pi r$, or $\pi$ times the diameter) is between the inner polygon side length (less than) and the outer polygon side length (greater than). $ Nl_{inner} \lt 2\pi r \lt Nl_{outer} $ where $l$ is the distance between adjacent polygon vertices, and $N$ is the number of sides (Archimedes reportedly made this calculation for $N=96$).
Sine and Cosine
The bounded Trigonometric functions are Sine and Cosine, meaning their values stay on the closed interval, $[-1,1]$. The Unit Circle, seen in Figure 3 (below) with the position of the Terminal Ray's intersection with the circle given in terms of the cofunctions. The right triangle formed with unit hypotenuse gives the geometric reason why we have the Pythagorean Identity:
$$\begin{equation}\label{sine_sqr_cosine_sqr} \sin^2(\theta)+\cos^2(\theta) = 1 \end{equation}$$
Observing the graph of these Cofunctions, in Figure 2 (below), plotted over the argument in Radians, shows they have the signature characteristics of an ideal wave, be it sound pressure vibrations or light in a lens (Dutch scientists Huygens, Snell, 17th century).
Radians
In the Unit Circle, an angle is measured in Radians, which is a unit of measure that is based on the radius of the circle. One radian is defined as the angle that subtends an arc length equal to the radius of the circle. Since the circumference of a circle is $2\pi\approx 6.28$ times its radius, where $\pi$ is a constant convention for the ratio of the circumference to the diameter of a circle. There are $2\pi$ Radians in a full turn, which is equivalent to 360 degrees.
Phase Difference of Cofunctions
The Angle Addition Formulas for the two Cofunctions, seen here: using Euler's Formula, demonstrates that one is the other with an argument offset. They are related by a phase difference of $90^\circ=\frac{\pi}{2}$, which is what makes them the cofunctions:
$$\sin(\theta + \frac{\pi}{2})=\cos(\theta)$$
And,
$$\cos(\theta + \frac{\pi}{2})=-\sin(\theta)$$
Which you can convince yourself of graphically, in Figure 2 (above), where we see a positive offset in argument corresponds to a leftward shift by that much.
The Unit Circle
The Unit Circle is often used to define the Trigonometric functions, such as Sine, Cosine, and Tangent, which are based on the coordinates of points on the circle. The Unit Circle is a fundamental concept in trigonometry and geometry, and it is defined as a circle with a radius of one, centered at the origin of a Cartesian coordinate system, where a point on the circle is identified with an angle, $\angle POQ$ in the diagram at Figure 3 (below). The Unit Circle can be considered as a set of points with coordinates $(x,y)$ that satisfy the equation of the circle in the plane:
$$\begin{equation}\label{circle_eq} x^2 + y^2 = r^2=1 \end{equation}$$
Where $r$ is the hypotenuse, also serving as the Terminal Ray in the Plane Polar coordinate system.
The unit disk is the set of points which includes all circles with radius less than unity. The intersection of the unit disk, $D={(x,y)\in \mathbb{R}^2 : x^2 + y^2\le 1}$, and the positive $x$-axis is the Initial Ray, or the angular origin, being line segment $\overline{OS}$. The Terminal Ray is the same as the Initial Ray when the angle is any number of full turns.
Inscribed Segments
Consider the Inscribed segments $x$ and $y$, in Figure 3 (below); they obey the equation of the circle, eq. \eqref{circle_eq}. These are the coordinates of a point $P$ on the Unit Circle, indicating the position of the Terminal Ray, with the inter-system mapping as follows:
$$ x=\cos(\theta), \quad y=\sin(\theta) $$
Points on the circle are naturally described in terms of the length of the rays and the angle between the two. The Unit Circle Inscribed diagram entails all of the elements of the plane polar coordinate system and Cartesian coordinates.
Tangent Triangle Segments
Just kissing the Unit Circle, called an osculation, at the Terminal point, there is the vertex of the ray and the adjacent side of an outscribed triangle, $\triangle{OQS}$ that we will now consider. Zooming in on Quadrant 1 in Figure 5 (below), we see the Tangent lengthed segment as being $\tan(\theta)=\overline{PS}$. The left hand side is the line segment, $\overline{OQ}=\csc(\theta)$. The intersection of the Tangent segment with the $x$-axis, at point $S=(\sec(\theta), 0)$, defines the third side, $\overline{OS}=\sec(\theta)$. The Terminal Ray defines how the spanning outscribed triangle is divided between Tangent and Cotangent sided right-triangles. When the bisection angle is $\theta=45^\circ=\pi/4$ then the $\text{cotangent}=\text{tangent}=1$. Whatever the angle, $\overline{QS}=\tan(\theta)+\cot(\theta)\ge 2$.
Now lets calculate these Outscribed segments using right-triangles.
Tangent and Secant Segments
Using Figure 5 (above), we'll solve for Tangent and Secant segments using the Right Triangle Rule applied to two overlapping triangles.
Triangles With Shared Verteces
The two triangles we use share the Tangent segment and therefore the points $P$ and $S$, as follows.
$\triangle{OPS}$: This is the Tangent triangle with the Secant as hypotenuse:
$$ \overline{OS} = \text{secant}, \quad \overline{OP} = r = 1, \quad \overline{PS}=\text{tangent} $$
Using the fact that $\overline{OP}\perp \overline{PS}$, we employ the Pythagorean Theorem:
$$\begin{equation}\label{secant_as_hyp} \text{secant}^2 = 1^2 + \text{tangent}^2 \end{equation} $$
Eq. \eqref{secant_as_hyp} defines each outscribed segment in terms of the other. To obtain the dependency on quantities we know, we use the next triangle.
$\triangle{TPS}$: The triangle with the Tangent segment as hypotenuse:
$$ \overline{TP} = \sin(\theta), \quad \overline{TS} = \text{secant} - \cos(\theta), \quad \overline{PS}=\text{tangent} $$
$$\begin{equation}\label{tangent_as_hyp} \text{tangent}^2 = (\text{secant}-\cos(\theta))^2 + \sin^2(\theta) \end{equation} $$
We can either equate the two definitions of Tangent, or Secant, to isolate the other one with the cofunctions, so let's isolate Tangent. Combining the two Pythagorean applications, eqs. \eqref{secant_as_hyp} & \eqref{tangent_as_hyp}, by equating expressions for $\text{tangent}^2$, we isolate the segment we called Secant:
$$ \text{secant}^2 -1 = \sin^2(\theta) + (\text{secant}-\cos(\theta))^2 $$
Expanding the binomial:
$$ \text{secant}^2 -1 = \sin^2(\theta) + \text{secant}^2-2\text{secant}\cos(\theta) + \cos^2(\theta) $$
Using the Pythagorean Identity we are left with the Secant as a function of Cosine:
$$\begin{equation}\label{secant_eq} \text{secant} = \sec(\theta)=\frac{1}{\cos(\theta)} \end{equation}$$
Plugging back into the Secant as hypotenuse equation, eq. \eqref{secant_as_hyp}, we have for Tangent:
$$\begin{equation}\label{tangent_eq} \text{tangent} = \tan(\theta)= \frac{\sin(\theta)}{\cos(\theta)} \end{equation}$$
Cotangent Triangles
The Cotangent segment forms side and hypotenuse to two triangles, comparable to the Tangent triangles, involving the Cosecant segment instead of the Secant.
Cotangent Triangles with Shared Verteces
The two triangles we use here share the Cotangent segment and therefore the points $P$ and $Q$, as follows.
$\triangle{OQP}$: This is the Cotangent triangle with the Cosecant as hypotenuse:
$$ \overline{OQ} = \text{cosecant}, \quad \overline{OP} = r = 1, \quad \overline{QP}=\text{cotangent} $$
Using the fact that $\overline{OP}\perp \overline{QP}$, we employ the Pythagorean Theorem:
$$\begin{equation}\label{cosecant_as_hyp} \text{cosecant}^2 = 1^2 + \text{cotangent}^2 \end{equation} $$
$\triangle{WQP}$: This is the Cotangent triangle which has the Cotangent as hypotenuse:
$$ \overline{WQ} = \text{cosecant} - \sin(\theta), \quad \overline{WP} = \cos(\theta), \quad \overline{QP}=\text{cotangent} $$
Plugging into the Pythagorean Theorem:
$$\begin{equation}\label{cotangent_as_hyp} \text{cotangent}^2 = (\text{cosecant} - \sin(\theta))^2 + \cos^2(\theta) \end{equation} $$
Isolating Cosecant:
$$ \text{cosecant}^2 - 1 = (\text{cosecant} - \sin(\theta))^2 + \cos^2(\theta) $$
Solving for Cosecant using eq. \eqref{sine_sqr_cosine_sqr}, we are left with:
$$\begin{equation}\label{cosecant_eq} \text{cosecant} = \csc(\theta)= \frac{1}{\sin(\theta)} \end{equation}$$
And plugging back into eq. \eqref{cosecant_as_hyp} to solve for Cotangent:
$$\begin{equation}\label{cotangent_eq} \text{cotangent} = \cot(\theta)= \frac{\cos(\theta)}{\sin(\theta)} \end{equation}$$
Symmetry and Antisymmetry
The Cofunctions have characteristics of symmetry.
Symmetry of a function is with respect to the origin. A function whose values mirror those of it at an equal distance from the origin on the other side is called symmetric, $f_s(x)=f_s(-x)$. A function whose values are similar except for the negation of one side of the equation exhibits the other kind of symmetry: antisymmetry. Antisymmetric functions have values exactly negative of its value for negative argument, $f_a(-x)=-f_a(x)$.
Sine is antisymmetric, while Cosine is symmetric, which effects the Principal Branch for purposes of finding the angle, $\theta$, as a function of the Cartesian Coordinates $(x,y)$ forming opposite and adjacent sides. To determine the symmetry of the outscribed segments (Tangent, Cotangent, Secant, and Cosecant), we recall that they exhibit reciprocals of the inscribed segments.
The Riemann Circle
The hyperbola is antisymmetric, and it doesn't go through zero at the origin since the magnitude of the ordinate for small argument is large. It's a little hard to see what's happening around inverse-zero, where the ordinate goes from negative large to positive large. The Real numbers have to be extended to accomodate large magnitude numbers, which are over and beyond that which we can fathom. Consider the pair of values plus or minus the largest computationally standard finite number ($\pm1.79\text{E}+308$)—or as the single point at infinity that the Real slice of the Complex Riemann Sphere exhibits, $g(\pi/2)=\infty$, or $NaN$ for Not a Number, which is always an approximation for the non-number Improper (complex) number. The domain of relevancy is extended to the complex plane by attributing an orthogonal dimension with the same mapping, it makes a sphere, a 2-sphere to be precise, and it's identical to the Real line, but orthogonal. This means that infinity is the point as a limit in the large complex limit, since the top of the sphere has one point as the circle does.
There is a mapping of the Real number line which exhibits the continuity at zero.
There is a quantum leap from any large magnitude point in the regular Real line to the point actually at infinity which is only microscopically measurable from the origin.
Convention holds that the observer's point is at the top,
so that the line of sight is drawn down and through the circle creating the arc as in case of Quadrant 1
and Quadrant 2 angles,
see line segment $\overline{NX_1}$ in Figure 6 (below).
The other points on the Real line are those small segments, selected by angles in Quadrant 3 and 4,
see line segment $\overline{NX_2}$.
Solving For The Map
Given any point on the $x$-axis there is one point on the circle where it intersects, in addition to passing through $N$. To solve for this point we need the equation of a line and the coordinates of a point on the Unit Circle, $(\cos(\theta),\sin(\theta))$:
$$ y=\left( \frac{\sin(\theta)-1}{\cos(\theta)-0} \right)x + 1=0 $$
We set the equation of the line equal to zero to find the $x$-intercept, and solve for $x$ as a function of angle, $g(\theta)$:
$$ g(\theta) = x = \frac{\cos(\theta)}{1 - \sin(\theta)} $$
So, for the two angles in Figure 6 (above), we have:
$$ g(\frac{\pi}{4}) = \frac{\frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} = 2.414 $$
And,
$$ g(-\frac{2\pi}{3}) = \frac{-\frac{1}{2}}{1 + 0.866} = -0.267 $$
The small points on the $x$-axis correspond to the angles in Quadrants 3 and 4, and the large points correspond to the top of the circle.
Odd and Even
Odd and even functions nomenclature comes from the characteristics of the odd and even integer powers. The powers of $x$ are either symmetric or antisymmetric. If the power is odd, then the function is antisymmetric (a.k.a. odd function), $(-x)^{2n+1} = -x^{2n+1}$ . If the power is even, then the function is symmetric (a.k.a. even function), $(-x)^{2n} = ((-1)^2)^nx^{2n}=x^{2n}$. Unlike the integers, functions are usually neither symmetric nor antisymmetric—if an arbitrary continuous function is constructed it will generally be asymmetric, meaning it will have no symmetry.
Every function is a sum of symmetric and antisymmetric parts, and if it's mixed then it's asymmetric.
$$\begin{equation}\label{symmetric_parts} f(x)=\frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}\end{equation} $$
The RHS of eq. \eqref{symmetric_parts} is separated into symmetric and antisymmetric functions, and when each part is nonzero the function is asymmetric.
The zero function, $f(x)=0$, is uniquely both symmetric and antisymmetric, because $-0=0$.
Additive Symmetry
The addition of functions with symmetry need to be pure to preserve symmetry.
$$ f_{a,1}(-x) + f_{a,2}(-x) = - (f_{a,1}(x) + f_{a,2}(x)) $$
The sum of two antisymmetric functions is also antisymmetric.
$$ f_{s,1}(-x) + f_{s,2}(-x) = f_{s,1}(x) + f_{s,2}(x) $$
The sume of two symmetric functions is also symmetric.
$$ f_a(-x) + f_s(-x) = -f_a(x) + f_s(x) \neq \pm (f_a(x) + f_s(x) ) $$
And, the sum of an antisymmetric and a symmetric function is asymmetric.
Product Symmetry
The product of two functions with symmetry is also symmetric.
$$ f_s(x)f_s(x)=f_s(-x)f_s(-x) $$
Which shows the product of even functions is just like the product of even numbers, and produces an even function.
$$ f_a(x)f_a(x)=(-f_a(-x))(-f_a(-x)) $$
Here we see that the product of an even number of antisymmetric functions produces a symmetric function.
$$ f_a(x)f_s(x)=-f_a(-x)f_s(-x) $$
And that the product of an odd function and an even one produce an odd function.
Composite Symmetry
If a function is a composition of functions with symmetry, the symmetry of the inner function dominates if it is even, otherwise the outer function determines the result.
$$ f_s(f_a(-x)) = f_s(-f_a(x)) = f_s(f_a(x)) $$
$$ f_a(f_s(-x)) = f_a(f_s(x)) $$
In both mixed cases, the result is a symmetric (even) function. Consequently, the squaring (an even function, $x^2$) of an odd function ($\sin(x)$) makes it even, so $\sin^2(\theta)$ is even.