# Taking trigonometry series-ly

### Part 1: Numerical

A mathematician friend mentioned to me that for small values of $x$ we might closely approximate $\sin(x)$ and $\cos(x)$ by cubic polynomials. Was she correct? To answer this question, use your calculator in degree mode to try to find the coefficients $a, \dots, h$ in the following suggested polynomial approximations: \[\sin(x) = a +bx+cx^2+dx^3\quad\quad \cos(x) = e+fx+gx^2+hx^3\quad\quad -0.1\leq x\leq 0.1\] Do you find that there is a best answer to finding these coefficients? How do the coefficients change when you try out the problem in radian mode on the calculator? Can you find a good choice of the coefficients of a possible fourth order term? Don't forget that you already possibly know certain values of the trig functions which might give you a good starting point for a search for these functions.

### Part 2: Using calculus

My mathematician friend now tells me that she thinks of trigonometrical functions in terms of solutions to differential equations, and not in terms of triangles. She says that $\sin(x)$ and $\cos(x)$ are both solutions to the second order differential equation: \[\frac{d^2 f}{dx^2}+f =0\;.\] Using this approach, what values for the coefficients $a, \dots, g$ emerge for the polynomial approximations from the first part? Does this correspond to your degree or radians expression? Or something different?

If the polynomial approximation continued to an arbitrarily high order, what would the coefficients be?

If you find the approximation to the sixth power of $x$ you can now estimate trigonometrical values without using the $\sin$ or $\cos$ button on your calculator. Test the accuracy of your series for various values of $x$ between $0$ and $\pi$/2.

*Discussion points: Do you think that your calculator stores values of sin and cos, or works them out on demand? Would your series provide an efficient way of evaluating the numerical values of sin(x) and cos(x)?*

We know that $\sin(x)$ and $\cos(x)$ both satisfy the equation, so try substituting $f(x) = \sin(x)$ into the differential equation.

Part1:

There are various approaches to this part of the problem, although the answers in radians should look something close to $$ \sin(x)\approx x-\frac{1}{6}x^3\quad\quad \cos(x) = 1-\frac{1}{2}x^2 $$

One approach would be to start using well known properties of trig functions. For example, we know that $a$ must be $0$ and $e$ must be exactly 1 because $\sin(0)=0$ and $\cos(0) = 1$.

We can also use some shortcuts because we know that $\sin(x) = -\sin(-x)$ and $\cos(x)=+\cos(-x)$. So, we can say that

$$

\sin(x)\approx bx +dx^3\quad\quad \cos(x) \approx 1+fx^2

$$

Then look at a very small value of $x$, say $0.0000001$. Unless the other coefficients are very large, we can ignore then. This tells us that, approximately

$$

b\approx \sin(0.0000001) \quad f\approx \sqrt{\frac{\cos(0.0000001)-1}{0.0000001^2}}

$$

We could find a value of $d$ by trying some larger values of $x$.

It should be clear that the numbers in radians are much smaller (and nicer) than in degrees.

Part 2:

$ \frac{d^2 f}{dx^2}+x =0 $

Substitute $ f(x) = sinx, f ' (x) = cosx , f ''(x) = -sinx $

$ \frac{d^2 f}{dx^2} =-x $

$\frac{df}{dx}= cosx =\int -x = -\frac{x^2}{\frac{x^5}{5!}2} + c$ (sub cos0 = 1 to find c=1)

$f(x)=sinx = \int cosx = \int -\frac{x^2}{2} + 1 = -\frac{x^3}{3!} + x + c $ (sub sin(0) = 0 to find c=0)

$\int sinx = -cosx = \int-\frac{x^3}{6!} + x = -\frac{x^4}{4!} + \frac{x^2}{2!} + c$ (-cos(0) = c =-1)

$\int -cos = -sinx = \int -\frac{x^4}{4!} + \frac{x^2}{2!} -1 = -\frac{x^5}{5!} + \frac{x^3}{3!} -x$

$\int -sinx = cosx = \int -\frac{x^5}{5!} + \frac{x^3}{3!} -x = -\frac{x^6}{6!} + \frac{x^4}{4!}-\frac{x^2}{2!} + 1 $

Hence cosx = $1 -\frac{x^2}{2!}+ \frac{x^4}{4!} -\frac{x^6}{6!}$

And Sinx = $x - \frac{x^3}{3!} +\frac{x^5}{5!}$

Extension:

Infinite power series of sinx =$x - \frac{x^3}{3!} +\frac{x^5}{5!} + ... + \frac{x^n}{n!}$ for odd n

Infinite series of cosx =$1 -\frac{x^2}{2!}+ \frac{x^4}{4!} -\frac{x^6}{6!} +... \frac{x^n}{n!} $for even n

Here we make a comparison of the evaluation of \frac{\pi}{2} by the series approxomations and the actuall functions.

$sin(\frac{\pi}{2}) = 1$

approximation :$sin(\frac{\pi}{2}) = 1.004524856$

$cos (\frac{\pi}{2}) =0 $

appromimation: $cos (\frac{\pi}{2}) =-8.945230096 x 10^{-4}$

Calculators and computers do not store a list of results for each and every possible function. A much more efficient method is to store a polynomial expression for each funtion using the taylor/maclaurin series.