The Idea Shop: Deriving the Trigonometric Functions Without Geometry

Deriving the Trigonometric Functions Without Geometry

Most people first meet sine and cosine in a basic geometry class. The lesson starts with a right triangle, and derives sine and cosine diagramtically using angles, lines and ratios. This is the "triangle view" of the trigonometric functions.

Most people are suprised to learn that's not the only way to derive them. While the triangle view of sine and cosine is intuitive, it actually conceals many interesting properties. But more importantly, the focus on lines and angles gives many people the impression that sine and cosine are "something from geometry" -- not objects with their own interesting properties and deep connections to other branches of math.

In this post, I'll explain a more elegant way of deriving sine and cosine that uses no geometry. This is the "real analysis" view. We'll start with a single assumption, and develop them analytically based on our earlier discussion of the Taylor series expansion (read here).

Starting Simple
Imagine a function with the following property. The function's second derivative is equal to the negative of the function itself. That is,

$f''=-f$

This is our starting assumption. It's just another way of saying that sine and cosine -- whatever they are -- need to have this one property. If we start with sine or cosine and take their derivative twice, we should end up back at the original function with a negative sign on it. Our goal is to show how to get from this assumption to actual algebraic functions for sine and cosine.

Let's take a closer look at our assumption. It relates the function to its second derivative -- a simple differential equation. But what about the other derivitives? To find them, let's take the next few derivatives of our assumption and see what happens.

Differentiating both sides and solving each time for f'', f''' and so on, we get the following

$f'''=-f'$

$f^{(4)}=-f''=f \Rightarrow f''=-f$

$f^{(5)}=-f'''=f' \Rightarrow f'''=-f'$

$f^{(6)}=-f^{(4)}=-f%20\Rightarrow%20f^{(4)}=f$

$f^{(7)}=-f^{(5)}=-f' \Rightarrow f^{(5)}=f'$

$\vdots$

We can already see the pattern here. As we take more and more derivatives, we always end up back at either f or f'. The only thing that changes is the positive or negative sign in front.

Now that we have n derivitives of f, we're ready to use our earlier finding about Taylor series expansions and write an expression for f.

Recall the Taylor expansion of f around the point c is given by

$f(x)=\sum_{j=0}^{\infty}\frac{f^{(j)}(c)}{j!}(x-c)^j$

Letting c = 0 -- which is technically a Maclaurin rather than a Taylor -- we get the following expansion

$f(x)=f(0)+f'(0)x+\frac{f''(0)x^{2}}{2!}+\frac{f'''(0)x^{3}}{3!}+\frac{f^{(4)}(0)x^{4}}{4!}+\hdots$

But as we've seen above, we already have expressions for the second, third and so on derivitives of f. They're all either f, -f, f' or -f'. Substituting these in, we get the following

$f(x)=f(0)+f'(0)x-\frac{f(0)x^{2}}{2!}-\frac{f'(0)x^{3}}{3!}+\frac{f(0)x^{4}}{4!}+\hdots$

Notice the pattern in the signs. The expansion starts with two positive terms, then has two negative terms, then has two positive terms, and so on. The pattern repeats forever, and each term involves only f or f'.

Since the function is really based two parameters -- f(0) and f'(0) -- we can rewrite this in a more useful way as follows. Label our two parameters c1 and c2. That is, let f(0) = c1, and let f'(0) = c2. Rewriting with this notation we get

$f(x)=c_1+c_2x-\frac{c_1x^2}{2!}-\frac{c_2x^{3}}{3!}+\frac{c_1x^{4}}{4!}+\hdots$

Now let's pick two sets of values for our parameters c1 and c2. These will define two distinct functions, both of which will satisfy our initial assumption.

First, let c1 = 0, and c2 = 1. Plugging into the above, we get the following function for sine

$f(x)=\sin(x)=0+(1)x-\frac{(0)x^2}{2!}-\frac{(1)x^{3}}{3!}+\frac{(0)x^{4}}{4!}+\hdots$

$\Rightarrow \sin(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\hdots$

This is the sine function we're looking for. When x = 0, the above function is zero. When x = pi/2 the function is one. And it works for all real values of x.

For our second case, let c1 = 1, and c2 = 0. Plugging this into our expansion, we get the following for cosine:

$f(x)=\cos(x)=1+(0)x-\frac{(1)x^{2}}{2!}-\frac{(0)x^{3}}{3!}+\frac{(1)x^{4}}{4!}+\hdots$

$\Rightarrow \cos(x)=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\hdots$

This is the cosine function we're looking for. When x = 0, the function is one. When x = pi/2, it's equal to zero. And like the sine function above, it works for all $\inline x\in \mathbb{R}$ .

We can easily check that these both satisfy our initial assumption that f'' = -f. And we can also verify another basic relationship between sine and cosine from calculus by taking the first derivative of each, which is that

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$

and also

$\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-\sin(x)$

Now that we've developed functions for sine and cosine, we can go on to derive all the other trigonometric functions -- tangent, cosecent, and so on -- from these.

For more on the "real analysis" view of sine and cosine, see this famous primer, especially Chapter 7, Section 4.

Posted by Andrew on Thursday June 18, 2009 | Feedback?

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