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?

One way to fill the gap is by using the theory of differential equations: First the derivatives of sin and cos can be geometrically established. For a very rigorous proof, one should start from a very advanced presentation of elementary geometry, define an angle in a very precise way, define a function which measures the angles,(all this is done in this book) then go on to establish the addition formulae and the derivatives. For details you can consult any good book on calculus such as the one by Apostol. Hence it can be verified that $\sin$ satisfies the differential equation $y''+y=0, y(0)=0, y'(0)=1$. 
Now by the method of power series of solving differential equations we find that the solution of the above differential equation is an infinite series $\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$. By the existence and uniqueness theorem of differential equations there should only be one solution so $\sin (x) =\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ Furthermore since power series are unique so this is also the Taylor series.
