Real Analysis/Trigonometry

Real Analysis

Prerequisite: Integrals; a bit of Fundamental Theorem of Calculus and Inverse Functions

Trigonometry is an ancient field of mathematics. Its relevance in the real world and its basic fundamentals – a series of rules – are enough to include it as a typical part of elementary mathematics. However, the main issue, like in everything one learns in elementary mathematics, is that the concepts in trigonometry were a series of rules and geometric images of circles and triangles. This section will correct this – through a rigorous redefinition of trigonometric functions from the ground up.

Construction

To redefine the trigonometric functions, we will first have to start off from some base. If we are to seriously remove all graphical interpretations and prove it using our current formalized mathematical concepts, we ought to clarify what we know and what we are aiming to show.

Axioms

Try as we might, our current scope of mathematics (that is real analysis plus some formalized algebra and a bit of formalized set theory) is inadequate to fully derive certain concepts we need in order to being our construction. Thus, we must accept the following as axioms.

List of Trigonometric Definitions (Axioms)

Description	Formula
The definition of a unit-circle: given a point at origin, this function contains two variables x and y such that the hypotenuse is always 1. Thus, this circle has radius 1.	$x^{2}+y^{2}=1$
The definition of "the area of a circle": The area of the circle is defined as follows	$A(r)=\pi r^{2}$

For our trigonometric functions, we also require the properties one learns in elementary mathematics such as the functions sine and cosine being periodic.

We will first begin by working with the equation $x^{2}+y^{2}=1$. It's not a function, but we will make it so, by restricting the equation.

${\begin{aligned}x^{2}+y^{2}&=1\\y^{2}&=1-x^{2}\\y&={\sqrt {1-x^{2}}}\\\end{aligned}}$

The function f will be defined as $f={\sqrt {1-x^{2}}}$. In elementary mathematics, this will usually be it; there are no more usual tools one has in order to further manipulate this function to do interesting things. However, we will show that calculus of all tools can crack the mystery of trigonometry.

Pi, and Area

The area for a unit circle is $A(1)=\pi (1)^{2}=\pi$. Special attention must be made that the area π is for the whole circle. Why is that mentioned? Well, we will attempt to align the concept of area for the circle with the concept of area for a function, using integrals.

$\int _{-1}^{1}{f}={\frac {\pi }{2}}$

As you can see, the fraction is because π represents the whole area, but our restriction on the original equation makes it so that we must halve the area. As an aside, understand that although we can use the variable τ (tau), which is precisely half pi and the value of that integral, we elected not to do so – simply because τ is a less familiar number and may appear more foreign to readers. The objective of this section is to illustrate how we can construct new concepts that fit preexisting concepts and more using calculus – not to baffle readers with new notation.

Returning to our earlier deceleration of what π ought to be, notice that we have not defined the integral of the function f either. In terms of definitions, it isn't good to define variables with more unknowns. So, we will not illustrate what the integral of the function f is.

To begin, we can make the definition of π more official looking:

The definition of π
$\pi =2\int _{-1}^{1}{\sqrt {1-x^{2}}}$

and we will spend the rest of this heading on defining the actual value of that integral.

We will first declare that we have not defined how to work with square roots. Thus, the following definition will actually rely of wishy-washy definitions of "area". Yes, this explanation might as well be conveyed as an axiom, even though there is a reason for it – it's just that the following reason may not resemble analysis but geometry. So, for those who are not geometrically-inclined, assume the equation at the very bottom as fact. In not, continue on.

The area of the function f is
$\Pi (x)={\frac {x{\sqrt {1-x}}}{2}}+\int _{x}^{1}{\sqrt {1-x^{2}}}$