# Approximating Trig Functions with Polynomials

I was thinking about the graphs of different trig functions and noticed that most of them are of a similar shape to some different types of polynomials. For example:

• Higher degree polynomials create a wave like sin or cos
• $x^3$ looks like one repetition of tan, and could be flipped and shifted to look like cot
• Each repetition of sec and csc looks like two quadratic parabolas

While obviously the polynomials aren't going to be an exact approximation, are there a set of coefficients that create a reasonably close (to a few decimal places) approximation of one period of the trig functions?

If so, is this useful? Or are there other, better, post Pre Calculus approximations of the trig functions?

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You will learn about Taylor series, which can be used to approximate functions using polynomials in higher calculus courses – user130512 Feb 25 '14 at 18:30

Seems to me that you are getting ready for Taylor series of trig functions. I would suggest to google this and you are getting lots of answers

http://en.wikipedia.org/wiki/Taylor_series would do but there are many many other great sites.

As far as usefullness, that can't be even described in one sentence. I appreciate you being inquisitive. That approach is very good, therefore (+1)

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My teacher mentioned those when I asked her. Am I right in saying that it is a kind of Fourier transform with polynomials instead of waves? It tells you what coefficients you need for the best fitting polynomial of a given degree? – Linuxios Feb 25 '14 at 18:35
Fourier used an (infinite) sum of trig terms to describe certain types of waves. That's not quite what Taylor series is about, but the idea is similar in the sense that a sum of infinite terms "models" a particular curve – imranfat Feb 25 '14 at 19:43
That's more what I meant. Awesome! Thanks for the quick answer, and I finally understand why precalc is spending so much time on curves (polynomials, e^x, log, ln, sin, etc.). – Linuxios Feb 25 '14 at 19:44
@Linuxios If you intend to go into the Calculus sequence, consider your precalc course as its foundation. Make it strong. Lots of hard calc problems are considered "hard" because people are not solid in their algebra that is so much needed. Notwithstanding the fact that Calc isn't to be taken easy, having a solid precalc background really, REALLY puts you at an advantage. Good luck – imranfat Feb 25 '14 at 20:16
Thanks. I'm just excited. I love math :). – Linuxios Feb 25 '14 at 20:16

First of all, I'm not native English, so sorry for my bad english. Although matematically imranfant was right, I will share with you some knowledge I made so an arduino could calculate "arc tan" of some given value with 2 decimal points, and having a surprising accurate result.

This only works for angles between 0 and 180 (It was the only thing I needed, probably you can do something equivalent to angles between 180 and 360) In the program I made, I had a vector (x,y), so I calculated the tan of that vector: tan(x,y)=y/x (in my program I could only use up to 2 decimals) Once I have the tangent, I needed to know in wich octant my angle was. This was easy. If x>0, then its between 0º and 90º, and if x<0, its between 90º and 180º. That gives you the quadrant, but to know wich octant you were: if |y|>|x| then the angle is between 45º and 135º. So by knowing these two things you know where the angle is.

Then depending on the octant, you use one of these equations (polinomical aproximations I found to the function f(tan(a),a)): We will call "a" the angle, and we will use the X and Y coordinates of the vector.

if "a" is between 0º and 45º: a= -16.343*(y/x)^2 + 61.701(y/x) - 0.2593

if "a" is between 45º and 90º: a = 16.254*(x/y)^2 - 61.634(x/y) + 90.252

if "a" is between 90º and 135º: a = -16.343*(x/y)^2 - 61.701(x/y) + 89.741

if "a" is between 135º and 180º: a = 16.343*(y/x)^2 + 61.701(y/x) + 180.26

Let's say you want to know the angle of the vector (3,2): First of all: y=3>2=x so its between 45º and 135º. Second, x>0 so its between 0º and 90º, so now we know its between 45º and 90º. So we have to use this: a = 16.254*(x/y)^2 - 61.634(x/y) + 90.252

And the answer to that being x=2 and y=3 is:

a=56.3867

The real angle of (3,2) is arctan(3/2)=56.3099 so the estimation has a mistake of slightly less than 0.1 of a degree.

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