Rigorous proof of the Taylor expansions of sin $x$ and cos $x$

This is what I consider the 'standard proof' for $\sin (x)$, and the proof that I would give tomorrow if suddenly asked of me. ($\cos(x)$ is more or less identical)

1. Recall Taylor's Theorem. If we want to pay a bit more attention to the basis of the proof, then Taylor's Theorem can be proven from the mean value theorem.
2. Recall that the derivative of $\sin (x)$ is $\cos (x)$, and that the derivative of $\cos(x)$ is $-\sin (x)$. These are often done geometrically.
3. Note that successive derivatives of $\sin$ look like $\cos(x), -\sin(x), -\cos(x), \sin(x), ...$ cyclically.
4. Use Taylor's Theorem to expand $\sin(x)$ around $0$.
5. We have to check to see that the expansion we've attained actually converges to $\sin(x)$. To do this, we'll use the Lagrange form of the remainder, which is obtained during the standard proofs of Taylor's Theorem. The trick here is that the $k$th remainder term is always bounded in absolute value by $\dfrac{x^{k+1}}{(k+1)!}$ because the successive derivatives are always just a $\sin$ or $\cos$ (up to sign). It's easy to see that for any $x$, as $k \to \infty$, this error term goes to $0$. Thus the expansion does converge to $\sin(x)$.
6. Repeat for $\cos(x)$ if desired.

Since $x^2 + y^2 = 1$, $y = \sqrt{1 - x^2}$,

$y' = \frac{-x}{\sqrt{1 - x^2}}$

By the arc length formula,

$\theta = \int_{0}^{x} \sqrt{1 + y'^2} dx = \int_{0}^{x} \frac{1}{\sqrt{1 - x^2}} dx$

We consider this integral on the interval [$-1, 1$] instead of [$0, 1$]. Then $\theta$ is a monotone strictly increasing function of $x$ on [$-1, 1$]. Hence $\theta$ has the inverse function defined on [$\frac{-\pi}{2}, \frac{\pi}{2}$]. We denote this function also by $\sin\theta$. We redefine $\cos\theta = \sqrt{1 - \sin^2 \theta}$ on [$\frac{-\pi}{2}, \frac{\pi}{2}$].

Since $\frac{d\theta}{dx} = \frac{1}{\sqrt{1 - x^2}}$,

(sin $\theta)' = \frac{dx}{d\theta} = \sqrt{1 - x^2} =$ cos $\theta$.

On the other hand, $(\cos\theta)' = \frac{d\sqrt{1 - x^2}}{d\theta} = \frac{d\sqrt{1 - x^2}}{dx} \frac{dx}{d\theta} = \frac{-x}{\sqrt{1 - x^2}} \sqrt{1 - x^2} = -x = -\sin\theta$

Hence

$(\sin\theta)'' = (\cos\theta)' = -\sin\theta$

$(\cos\theta)'' = -(\sin\theta)' = -\cos\theta$

Hence by the induction on $n$,

$(\sin\theta)^{(2n)} = (-1)^n\sin\theta$

$(\sin\theta)^{(2n+1)} = (-1)^n\cos\theta$

$(\cos\theta)^{(2n)} = (-1)^n\cos\theta$

$(\cos\theta)^{(2n+1)} = (-1)^{n+1}\sin\theta$

Since $\sin 0 = 0, \cos 0 = 1$,

$(\sin\theta)^{(2n)}(0) = 0$

$(\sin\theta)^{(2n+1)}(0) = (-1)^n$

$(\cos\theta)^{(2n)}(0) = (-1)^n$

$(\cos\theta)^{(2n+1)}(0) = 0$

Note that $|\sin\theta| \le 1$, $|\cos\theta| \le 1$.

Hence, by Taylor's theorem,

$\sin\theta = \sum_{n=0}^{\infty} (-1)^n \frac{\theta^{2n+1}}{(2n+1)!}$

$\cos\theta = \sum_{n=0}^{\infty} (-1)^n \frac{\theta^{2n}}{(2n)!}$

QED

Remark:

When you consider the arc length of the lemniscate instead of the circle, you will encounter $\int_{0}^{x} \frac{1}{\sqrt{1 - x^4}} dx$. You may find interesting functions like we did with $\int_{0}^{x} \frac{1}{\sqrt{1 - x^2}} dx$. This was young Gauss's approach and he found elliptic functions.