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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? –  mixedmath Aug 22 '12 at 7:08
    
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. –  William Aug 22 '12 at 7:10
    
@mixedmath Any correct proof will do. –  Makoto Kato Aug 22 '12 at 7:10
    
As I describe in this answer (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 (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. –  Blue Aug 22 '12 at 7:29
    
Anybody would like to post an answer using complex analysis? –  Makoto Kato Aug 22 '12 at 21:32

3 Answers 3

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.
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You don't repeat for cosine, you just use Taylor's theorem and differentiate the series term by term. But good answer ; +1 –  Patrick Da Silva Aug 22 '12 at 9:33
    
Could you elaborate on the step 2? –  Makoto Kato Aug 23 '12 at 22:49
    
@Makoto: Step 2 consists of proving that the derivative of $\sin x$ is $\cos x$, and that the derivative of $\cos x$ is $-\sin x$. If you don't know how to prove this, I'd recommend looking it up. One method would be to just do it, using the trig angle-sum identities and the limits of $\sin x / x$ and $(\cos x - 1)/x$ as $x \to 0$. –  mixedmath Aug 24 '12 at 4:17
    
@mixedmath Could you tell me where I can look it up? –  Makoto Kato Aug 24 '12 at 4:37
    
@Makoto: wikipedia is always a good place to start. –  mixedmath Aug 24 '12 at 4:45

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.

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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.

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