Rigorous proof of the Taylor expansions of sin $x$ and cos $x$ We learn trigonometric functions in high school, but their treatment is not rigorous.
Then we learn that they can be defined by power series in a college.
I think there is a gap between the two.
I'd like to fill in the gap in the following way.
Consider the upper right quarter of the unit circle $C$ = {$(x, y) \in \mathbb{R}^2$; $x^2 + y^2 = 1$, $0 \leq x \leq 1$, $y \geq 0$}.
Let $\theta$ be the arc length of $C$ from $x = 0$ to $x$, where $0 \leq x \leq 1$.
By the arc length formula, $\theta$ can be expressed by a definite integral of a simple algebraic function from 0 to $x$.
Clearly $\sin \theta = x$ and $\cos\theta = \sqrt{1 - \sin^2 \theta}$.
Then how do we prove that the Taylor expansions of $\sin\theta$ and $\cos\theta$ are the usual ones?

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.
