Defining the Cosine Function from First Principles, intuitively

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.