Deriving power series for $\sin x$ without using Taylor's Theorem or $\exp z$ Starting with defining $(\cos t, \sin t)$ from the unit circle, is it possible to derive the power series for $\sin(t)$:
$$\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \dots$$
Note: I will be answering my question, I hope this doesn't offend anyone. If is any issue with the proof, I am grateful for any improvement.
 A: Another method is using the inequality
$$\sin(x)\leq x$$
Integrate this between $0$ and $t$
$$-\cos(t)+\cos(0)\leq t^2/2$$
$$1-t^2/2\leq \cos(t)$$
Integrate this between $0$ and $x$:
$$x-x^3/3!\leq \sin(x)$$
Repeat this and you will get a lower and upper bound, which are just partial sums of the Taylor series. Note that you directly get the taylor series for $\cos(x)$.
A: I think this works:
Starting with defining $(cos(t), sin(t))$ from the unit circle, it can be shown by elementary means that $sin(t)$ satisfies $f''=-f$. Also by observation $f(0)=0$.
Consider a power series $f(x) = a_0 + a_1 x + a_2 x^2 + ... $ with the same two constraints, we have $a_0=0$ and, by differentiating twice and comparing coefficients, $(k+2)(k+1)a_{k+2} = -a_k$.
It can be immediately seen that for even $k$, $a_k=0$.
Also that:
$a_3 = -\frac{1}{(3)(2)}a_1$ = -$\frac{a_1}{3!}$
$a_5 = -\frac{1}{(5)(4)}a_3 = +\frac{a_1}{5!}$
... etc.
So $f(x) = a_1 (x - 
\frac{x^3}{3!} + \frac{x^5}{5!} - ...)$ which (by the Ratio Test) converges.
$a_1$ must have value 1 if $f$ is to behave similarly to $sin$ in the neighbourhood of 0 (as $sin(x) \to x$ as $x \to 0$).
So we have $sin(x)$ and $f(x)$ both satisfying $f''=-f$ and having value 0 at $x=0$.
EDIT:
However, as pointed out in the comments, there might be more than one function satisfying this property.
If the function can be written as a power series, we have found it!  It must be $f$ with $a_1=1$.
But maybe there is some weird function that can't be written as a power series... so this proof is not quite complete. Can anyone put the last nail in?
A: One approach is to use the differential equation $$y'' + y = 0$$ satisfied by $y = \sin x$. The differential equation has unique solution $$y = y(0)\cos x + y'(0)\sin x$$ Now consider the power series $$f(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots$$ then $f(x)$ is defined for all $x$ (because the series is convergent for all $x$) and by differentiating the series twice we can see that $f''(x) + f(x) = 0$. Hence $f(x) = f(0)\cos x + f'(0)\sin x$ and noting that $f(0) = 0, f'(0) = 1$ we get $f(x) = \sin x$.
The above proof uses the derivatives of $\sin x, \cos x$ and differentiation of power series.
A: By definition of the unit circle and $\cos(t),\sin(t)$, we have the following lemma :

If $g : \mathbb{R} \to \mathbb{R}^2$, $g(t) = (x(t),y(t))$ is a smooth parametrized curve in $\mathbb{R}^2$ such that $\ \|g(t)\|= 1$ for every $t$, then $g(t)$ moves on the unit circle.
Moreover, if $\|g'(t)\|=1$ for every $t$ then $g(t) = (\cos(t+a),\pm\sin(t+a))$ for some $a$.

Let $$x(t) = \sum_{k=0}^\infty (-1)^k \frac{t^{2k}}{(2k)!}, \qquad\qquad y(t) = \sum_{k=0}^\infty (-1)^k \frac{t^{2k+1}}{(2k+1)!}$$
both series converge and define smooth functions of $t \in \mathbb{R}$.
Consider the function $g : \mathbb{R} \to \mathbb{R}^2$ $$g(t) = (x(t),y(t))$$
By differentiating the power series term by term, we get $x'(t) = -y(t)$, $y'(t) = x(t)$, so that $$\frac{d}{dt}\|g(t)\|^2 = \frac{d}{dt}[x(t)^2+y(t)^2] = 2 x'(t) x(t) + 2 y'(t)y(t) = -2 y(t) x(t)+2  x(t) y(t) = 0$$
thus $$\|g(t)\|^2 = \|g(0)\|^2 = 1$$
and the same for $\|g'(t) \|^2 = \|g(t)\|^2$.
Hence the lemma applies and $$g(t) = (\cos(t+a),\pm\sin(t+a))$$
obviously $a=0$ since $g(0) = (1,0)$, and $g(t) = (\cos(t),\sin(t))$ since $g'(0) = (0,1)$.
