Defining the Cosine Function from First Principles, intuitively

Question

Throughout most of my mathematics education, the cosine function has been defined formally either using its series expansion, as $\mathfrak{Re}(\exp i\theta)$, or as the unique solution to $y+y''=0$ with $y(0)=1$ and $y'(0) = 0$.

Although these definitions make the mathematics more convenient, I've always been interested to see what it would be like to try to formalise it directly with the geometric intuition from the unit circle, assuming only the notion of distance (i.e. that we have a complete metric space with metric $d\colon\mathbb R^2 \times \mathbb R^2 \to \mathbb R$).

Since, by geometric intuition, we have that the cosine function is the $x$-coordinate obtained by walking along the unit circle, my idea was to take a line segment from the point $(1,0)$ to another point on the circle such that the distance is $\theta$, and split it into two line segments whose sum of lengths is also $\theta$. We then split into 3 line segments, and so on, in a limiting manner, in such a way that the sum of lengths is always $\theta$, as illustrated below with $\theta = 2\pi/5$.

[Desmos link]

When this procedure converges, the $x$- and $y$-coordinates of the limiting point should be the cosine and sine of $\theta$ respectively. The problem is I can't find an easy way to express the $n$th coordinate of this procedure (assuming we are splitting the line into $n-1$ segments) so that I can take the limit.

I was going to try and do things the other way round, that is, define $\cos^{-1}\colon [-1,1] \to [0,\pi]$ by approaching the circle from below and finding the length. I did manage to do this, and got that \begin{align*} \cos^{-1}(x) &= \sum_{k=1}^\infty d\left[\left(\frac{(k-1)x}{n}, \sqrt{1-\left(\frac{(k-1)x}{n}\right)^2}\right), \left(\frac{kx}{n}, \sqrt{1-\left(\frac{kx}{n}\right)^2}\right)\right]\\[4pt] &= \sum_{k=1}^\infty \sqrt{2 \left(1-\sqrt{1-\frac{(k-1)^2 x^2}{n^2}} \sqrt{1-\frac{k^2 x^2}{n^2}}\right)-\frac{2 k(k-1) x^2}{n^2}}, \end{align*}

and I suppose one can extrapolate from here to obtain the $\cos$ and $\sin$ functions and extend them appropriately.

But it feels like the other way, that is, finding that limiting point, should be possible. I appreciate any assistance with this.

Answer 1

Let $(x_1, y_1)$ be the first point in your construction. Using the similar right triangles $(0, 0)$–$\left(\frac{x_1 + 1}{2}, \frac{y_1}{2}\right)$–$(1, 0)$ and $\left(\frac{x_1 + 1}{2}, \frac{y_1}{2}\right)$–$\left(\frac{x_1 + 1}{2}, 0\right)$–$(1, 0)$, we compute

$$(x_1, y_1) = \left(1 - \frac{(\theta/n)^2}{2}, \frac\theta n\sqrt{1 - \frac{(\theta/n)^2}{4}}\right).$$

How do we go from $(x_k, y_k)$ to $(x_{k + 1}, y_{k + 1})$? This is just a rotation, which is a job easily done by complex multiplication by the constant $x_1 + iy_1$:

$$x_{k + 1} + iy_{k + 1} = (x_k + iy_k)(x_1 + iy_1).$$

Iterating this $n$ times gives

$$x_n + iy_n = (x_1 + iy_1)^n.$$

At this point, some calculus would show that this approaches $e^{i\theta}$, but we don’t want to use calculus, so instead we’ll expand this using the binomial theorem:

$$x_n + iy_n = \sum_{k = 0}^n \binom{n}{k} x_1^{n - k}i^ky_1^k.$$

The real part is

$$\begin{split} x_n &= \sum_{k = 0}^{\lfloor n/2\rfloor} \binom{n}{2k} (-1)^kx_1^{n - 2k}y_1^{2k} \\ &= \sum_{k = 0}^{\lfloor n/2\rfloor} \binom{n}{2k} (-1)^k\left(1 - \frac{(\theta/n)^2}{2}\right)^{n - 2k}\left(\frac{\theta}{n}\right)^{2k}\left(1 - \frac{(\theta/n)^2}{4}\right)^k, \end{split}$$

which I suppose is an expression.

Miraculously, in the limit as $n \to \infty$, we have

$$\binom{n}{2k} \frac{1}{n^{2k}} \to \frac{1}{(2k)!}, \quad \left(1 - \frac{(\theta/n)^2}{2}\right)^{n - 2k} \to 1, \quad \left(1 - \frac{(\theta/n)^2}{4}\right)^k \to 1,$$

so the $k$th term converges to simply

$$\frac{(-1)^k}{(2k)!}\theta^{2k}.$$

After checking some details about exchanging the sum with the limit, we get back the power series we know and love.

$$\cos \theta = \sum_{k = 1}^\infty \frac{(-1)^k}{(2k)!}\theta^{2k}.$$

We can get $\sin \theta$ from the imaginary part in a similar way.

Answer 2

Let $\alpha_n$ be the central angle of the first trangle in the $n$-th iteration.

It is an isosceles triangle with base length $\frac{\theta}n$ and leg length $1$. Hence $\alpha_n$ is given by the equation $$\sin\frac{\alpha_n}2 = \frac{\theta}{2n}$$ or $\alpha_n = 2\arcsin\left(\frac{\theta}{2n}\right)$.

The coordinates of the last ($n$-th) point are then $$(\cos (n\alpha_n), \sin (n\alpha_n)) = \left(\cos\left(2n\arcsin\left(\frac{\theta}{2n}\right)\right), \sin\left(2n\arcsin\left(\frac{\theta}{2n}\right)\right)\right) \xrightarrow{n\to\infty} (\cos\theta, \sin\theta)$$

because of the limit $\lim_{t\to 0} \frac{\arcsin t}t = 1$ and the continuity of $\sin$ and $\cos$.

This shows that your last constructed point indeed converges to $(\cos\theta, \sin\theta)$. It doesn't give an algebraic expression for the coordinates, though.

Answer 3

There is an alternative, algebraicalky based method for setting cosine values based on defining the function in this matter. Except for the last of these four characteristics, the definitions should be evident from the usual geometric rendering on the unit circle:

• The cosine of $0$ is $1$.
• The cosine function is even and has a period of $2\pi$.
• The cosine function strictly, monotonically decreases as its argument increases from $0$ to $\pi$.
• The cosine function satisfies the multiple angle formulae we can prove via algebraic manipulation of the trigonometric functions; e.g. $\cos 2x = 2\cos^2 x -1$. This does not require foreknowledge of any specific function values.

From these properties we can identify $\cos(p\pi/q)$ as a root of a polynomial equation for any whole numbets $p$ and $q$, provided of course $q\ne 0$, thus getting a dense set of values over the real domain.

Start with $\cos(p\pi)$ ($q=1$). We know from the first two properties that the cosine will be $1$ when $p$ is even. For odd $p$ we apply the double angle formula to get

$2\ cos^2(p\pi)-1=1$

from which condition, $\cos(p\pi)=\pm 1$ for all odd $p$. But the third property above forces $\cos\pi<1$ meaning $\cos\pi$ can only be $-1$, and then the second property implies $\cos(p\pi)=-1$ for call odd $p$.

We keep going. Apply the double angle formula again, with $x=p\pi/2$ and $p$ odd, to get

$2\cos^2(p\pi)-1=-1$

for all odd $p$. We have thus gotten $\cos(p\pi/2)=0$ for all odd $p$. Even values of $p$ with $q=2$ are, of course, reducible to the $q=1$ case.

When $q$ isn't a power if $2$ we invoke higher multiple angle formulas. Say we want $\cos(p\pi/3$). We have the triple angle formula from the fourth property:

$\cos(3x)=4\cos^3 x-3\cos x$

Putting $x=\pi/3$ then gives

$4\cos^3(\pi/3)-3\cos(pi/3)+1=0$

which has two distinct roots, thus $\cos(pi/3)=-1$ or $\cos(pi/3)=1/2$. We already derived $\cos(\pi)=-1$, so our third property rejects $\cos(pi/3)=-1$. We must have $\cos(pi/3)=1/2$. The reader can similarly put $x=2\pi/3$ and get $\cos(2pi/3)=-1/2$. Property 2 takes care of other multiples of $\pi/3$.

We can keep going with higher denominators and, where necessary, higher multiple angle formulas, but apart from the more complicated algebra we are basically doing the same thing from here.

Note carefully the second property above. It is based on the symmetry and closed nature of the circle. Replace the circle with a hyperbola and we lose the closure. With it goes the periodic property. So while we can still define $\cosh 0=1, we can't gain any traction for rendering other "nice" values like we have for circular trigonometric functions.

Answer 4

Plot the unit circle on a Cartesian plane. For purposes of exposition, let’s use the usual $x,y$ coordinates and measure angles by walking the circle counterclockwise from $(x,y)=(1,0)$ as in your figures. Then the point at arc distance $t$ from $(1,0)$ has coordinates $(\cos t, \sin t)$ by the unit-circle definition of the sine and cosine.

We can also view $\mathbf x(t) = (\cos t, \sin t)$ as a parameterized position vector of the point at arc distance $t$ around the unit circle. Since making a rigorous measurement of the length of the arc seems inevitably to involve some kind of limiting process, let's use the kind of limiting process that allows us to compute $\mathbf x'(t),$ the derivative of $\mathbf x(t)$ with respect to $t.$

Now, $\mathbf x'(t)$ will be in the direction of the tangent to the circle at $\mathbf x(t),$ and $\lvert\mathbf x'(t)\rvert = 1,$ so it's a unit vector at a right angle counterclockwise from $\mathbf x(t).$ In particular, since $\mathbf x(0) = (1,0),$ it follows that $\mathbf x'(0) = (0, 1).$

Differentiating a second time, $\mathbf x''(t)$ is a unit vector at a right angle counterclockwise from $\mathbf x'(t),$ exactly opposite to $\mathbf x(t).$

Therefore $\mathbf x(t) + \mathbf x''(t) = 0.$

Breaking this out into components, letting $\mathbf x(t) = (f_1(t), f_2(t)),$ we have $\cos t = f_1(t)$ where $f_1(t) + f_1''(t) = 0.$ We also have $f_1(0) = 1$ and $f_1'(0) = 0.$

Show that the solution to the differential equation with those boundary conditions is unique, then rename $f_1$ to $y,$ and you have shown that the cosine function is the unique solution to $y+y''=0$ with $y(0)=1$ and $y'(0)=0,$ using the geometric intuition of the unit circle.