Deriving the power series for sine and cosine from geometry

Question

A few years ago, I saw an article that showed that it was possible to derive the power series for sine and cosine using a tail-to-tip chord approximation in the unit circle.

Basically, one first create a chord of length $0 \leqslant \theta \leqslant 2$ starting from the origin point $(1,0)$ to a point $P_1$ on the unit circle.

Now, if you divide the length $\theta$ by a given constant $n$ and create a sequence of multiple tail-to-tip chords, then you have $n$ points on the unit circle from $P_1$ to $P_n$ and the total length [of all chords] remains $\theta$.

You want to find the couple $(x_n,y_n)$ by solving the tail-to-tip recurrence system given by

$$ \begin{cases} (x_k - x_{k-1})^2 + (y_k - y_{k-1})^2 = \left(\frac\theta n\right)^2 \\ (x_k)^2 + (x_k)^2 = 1 \end{cases} $$

Trying to first the first couple $(x_1,y_1)$ we have

$$ \begin{cases} (x_1 - x_0)^2 + (y_1 - y_0)^2 = \left(\frac\theta n\right)^2 \\ (x_1)^2 + (y_1)^2 = 1 \end{cases} $$

And it's not hard to show that

$$ x_0 \neq 0, y_0 = 0 \implies x_1 = \frac{(x_0)^2 - (\theta/n)^2 + 1}{2x_0}, y_1 = \sqrt{1 - (x_1)^2} $$

We know that $P_0 = (1,0)$ so we can easily compute $P_1$ from the formula above. Further terms can be computed using iterative substitution, aka the "plug and chug" method. Indeed, we can show that

$$ x_0 \neq 0, y_0 = 0 \implies x_k = \frac{(x_{k-1})^2 - (\theta/n)^2 + 1}{2x_{k-1}}, y_k = \sqrt{1 - (x_k)^2} $$

Now, expanding this whole expression becomes very cumbersome for $k > 2$ and we want to find an explicit formula instead, that is, one which doesn't involve $x_{k-1}$ in the expression.

I tried to do the substitution up to $k = 3$ but I failed to identify a recurring pattern. From the geometric intuition, though, we should have

$$ x_n \sim \cos(\theta) \quad \text{and} \quad y_n \sim \sin(\theta) \quad \text{as} \ n \to \infty $$

A lot of steps have been skipped in the original article, and somehow, the final result is as follows

$$ x_n = \sum_{k=0}^{\lfloor n/2 \rfloor}(-1)^k \bigg({n \atop 2k}\bigg) \bigg(1 - \frac{(\theta/n)^2}{2}\bigg)^{n-2k} \bigg(\frac\theta n\bigg)^{2k} \bigg(1 - \frac{(\theta/n)^2}{4}\bigg)^k $$

and

$$ y_n = \sum_{k=0}^{\lfloor n/2 \rfloor}(-1)^k \bigg({n \atop 2k+1}\bigg) \bigg(1 - \frac{(\theta/n)^2}{2}\bigg)^{n-2k-1} \bigg(\frac\theta n\bigg)^{2k+1} \bigg(1 - \frac{(\theta/n)^2}{4}\bigg)^{k+(1/2)} $$

In the limit as $n \to \infty$ and glossing over some convergence details, it can be shown that the two above expressions simplify to the power series of cosine and sine, respectively.

To me, it seems not so trivial to derive those two expressions. I like the concept, and I would like some help or guidance on what steps could have been used to get those two results.

Answer 1

I don't write the direct solution, but present the method.

In general, the whole procedure seems unnecessarily complicated. Basically, the angle in radians is the arc length parameter on the unit circle, starting at (x,y)=(1,0). The chord approximates the arc length s, i.e. $\theta \to s$ as $n \to \infty$.

On mathematical grounds, I'd also question the method, because cos and sin are nowadays DEFINED by the series. So, it could rather be a method for motivating the series definition of sin and cos.

But as for the method: For the case with n chords, denote the points by $P_{n,k}\, $, $0 \leq k \leq n$, and $P_{n,0} = (1,0)^t$. Your formula for P1 holds in any case for $P_{n,1}$ with $\theta$ replaced by $\theta / n$:

$P_{n,1} = \begin{bmatrix} 1 - (\frac \theta n)^2 / 2 \\ \frac\theta n \cdot \sqrt{1-(\frac \theta n)^2 / 4} \, \end{bmatrix}$

You know, that this is ($\cos(\phi) , \sin(\phi)$) for some angle $phi$ related to $\theta$.

Whith these entries, build the matrix which provides rotation about the origin through that angle $\phi$, but take $\frac\theta n$ as the argument:

$ R(\frac \theta n) = \begin{bmatrix}1-(\frac \theta n)^2/2 & -\frac\theta n \cdot \sqrt{1-(\frac \theta n)^2 / 4} \\ \frac\theta n \cdot \sqrt{1-(\frac \theta n)^2 / 4} & 1-(\frac \theta n)^2/2 \end{bmatrix}$

The first column is the coordinates of the point $P_{n,1}$ which is $(1,0)$ rotated through $\phi$. The second column is $(0,1)$ rotated through $\phi$, so it makes a right angle at the origin with $P_{n,1}$.

By using the rotation matrix, the coordinates of the points are obtained as

$P_{n,k} = R(\frac \theta n)^k \begin{bmatrix}1\\0\end{bmatrix}$

Now, compute $R(\frac \theta n)^n$ and prove the formula by induction over n. (Your formula has misprints, I guess.)

EDIT: I've just found a full calculation using the binomial theorem here: Defining the Cosine Function from First Principles, intuitively