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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?

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    Does the usual proof suffice? Show that the derivatives of $\sin$ and $\cos$ are $\cos$ and $-\sin$, respectively, and thus are cyclic. Then carry the expansion around $0$, and show that the error term goes to zero. Are you looking for something else?2012-08-22
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    There are rigorous proofs (see Rudin's _Principle of Mathematical Analysis_) of Taylor Theorems about power series expansions involving derivatives as coefficients and how these approximate certain type of functions. There are rigorous proofs of the derivative of sine and cosine. Applying the two, I think you can prove the power series that the power series expansion is what it should be.2012-08-22
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    @mixedmath Any correct proof will do.2012-08-22
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    As I describe in this answer (http://math.stackexchange.com/a/1103/409), the power series of sine and cosine have nice geometric interpretations (and a nicely combinatorial proof). I link to my note (http://dlnds.com/mathdocs/Zig-Zag-Up-Down-Secant-Tangent.pdf) that adapts the argument for the secant and tangent series (and back-adapts it to the sine and cosine series in a closing remark), but if you just want the original, look for the article "A Problem" by Leo Gurin in the American Mathematical Monthly #103, 1996.2012-08-22
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    Anybody would like to post an answer using complex analysis?2012-08-22

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