Cofunctions and the Unit Circle

Trigonometric functions (six of them, for the ) are functions of Sine and Cosine, as the segments of the Unit Circle's Outscribed Triangles are in fact familiar functions of the Inscribed Triangles.

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. There are other connotations of a a complimentary dual object to a function, such as its inverse, but convention reserves its usage to when it is helpful. 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, 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 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 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. The Unit Circle, in Figure 3, gives the geometric reason why

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 below in Figure 3. The unit circle can be considered as a set of points with coordinates $(x,y)$ that satisfy the equation $x^2 + y^2 = r^2=1$, where $r$ is the hypotenuse which is now configured to be the Terminal Ray as variable ray.

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. The Terminal Ray is the same as the Initial Ray when the angle is any number of full turns.

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 circle, which is equivalent to 360 degrees.

The unit circle is also used to visualize the relationships between the trigonometric functions and the segments they represent. Consider the inscribed segments, which are the Cofunctions, in Figure 3.

Points on the circle are naturally described in terms of the radius, $r=1$ here, and angle, $\theta$ extending into the first quadrant for the $\theta=28.8^\circ=0.502$. This diagram entails all of the elements of the plane polar coordinate system, where $\Phi:(\theta, r)\mapsto (x,y)$ is the abstract map from polar coordinates to cartesian coordinates. The Plane Polar coordinates are given by the point, $P$:

$$P=(r\cos(\theta), r\sin(\theta))$$

When the radius $r$ is one, the points described are on the Unit Circle.

They have characteristics as a direct result of the right triangle properties, as can be seen in the Inscribed Unit Circle, in Figure 3.

The Angle Addition Formulas for the two are proven elsewhere, using Euler's Formula. The formulas demonstrate that the two are related by a phase difference of $90^\circ$—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, see Figure 2, where we see a positive offset in abscissa corresponds to a leftward shift by that much.

Tangent Triangle Segments

Just kissing the Unit Circle at the Terminal Ray's intersection, there is a triangle, $\triangle{OQS}$ that we will now consider. Zooming in on Quadrant 1 in Figure 4, we see the tangent lengthed segment as $\tan(\theta)=\overline{PS}$. 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 big triangle is divided between tangent and cotangent sided right-triangles, with the bisection angle being $\theta=45^\circ=\pi/4$ at which cotangent=tangent=1. Whatever the angle, $\overline{QS}=\tan(\theta)+\cot(\theta)$.

Tangent and Secant Segments

Using Figure 4, 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 Right Triangle Rule on the Inscribed Unit Circle:

$$\begin{equation}\label{sine_sqr_cosine_sqr} \sin^2(\theta)+\cos^2(\theta) = 1 \end{equation}$$

Which leaves the secant as a function of one cofunction:

$$\begin{equation}\label{secant_eq} \text{secant} = \sec(\theta)=\frac{1}{\cos(\theta)} \end{equation}$$

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 with the cotangent as hypotenuse:

$$ \overline{WQ} = \text{cosecant} - \sin(\theta), \quad \overline{WP} = \cos(\theta), \quad \overline{QP}=\text{cotangent} $$

Plugging into the Right Triangle Rule:

$$\begin{equation}\label{cotangent_as_hyp} \text{cotangent}^2 = (\text{cosecant} - \sin(\theta))^2 + \cos^2(\theta) \end{equation} $$

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 an equal distance from the origin on the other side is called symmetric, $f_s(x)=f_s(-x)$.
And antisymmetric is any function whose value for a given argument is exactly the negative of its value for an equal distance from the origin on the other side, $f_a(-x)=-f_a(x)$.

Odd and Even

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 both symmetric and antisymmetric.

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.

The Harmonic Addition Theorem

Let's consider the sum of the Sine and Cosine functions, in the special case they have the same argument, to derive some of its general properties.

$$ f(\theta) = A\cos(\theta) + B\sin(\theta) $$

Quoting The Angle Addition Formula for Cosine:

$$ \cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y) $$