Infinite series expansion of $\sin (x)$ Are there any other ways to demonstrate that $$\sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^kx^{1+2k}}{(1+2k)!}$$
without using the definition of Taylor series of complex exponentials, and similarly for $\cos(x)$?
 A: This is from Simmons' Calculus. It's in an exercise. 
$$ \cos x \leq 1$$
$$ \int_0^x\!\cos t \,\mathrm{d}t\leq \int_0^x\! \,\mathrm{d}t$$
$$ \sin x \leq x$$
$$ \int_0^x\!\sin t \,\mathrm{d}t\leq \int_0^x\! t \,\mathrm{d}t$$
$$ \left.-\cos t\right|_0^x\leq \frac{ x^2}{2}$$
$$ 1-\cos x\leq \frac{ x^2}{2}$$
$$ \cos x\geq 1-\frac{ x^2}{2}$$
Continuing, you see that $\sin x$ is less than its expansion when truncated after progressively higher odd numbers of terms and, in alternation, that $\cos x$ is greater than its expansion truncated after progressively higher even numbers of terms.
I don't have the book in front of me. I think this was intended more to suggest the expansion than to rigorously prove it, but my theoretical understanding isn't quite up to identifying what's lacking or to correcting anything. Still, I thought it was interesting when I saw it and I hope it's relevant.
A: Here is a mosquito-nuking solution: one can use Lagrangian inversion:
$$f^{(-1)}(x)=\sum_{k=0}^\infty \frac{x^{k+1}}{(k+1)!} \left(\left.\frac{\mathrm d^k}{\mathrm dt^k}\left(\frac{t}{f(t)}\right)^{k+1}\right|_{t=0}\right)$$
and let $f(t)=\arcsin\,t$; probably the only deal-breaker here is that the expressions for the derivatives get progressively unwieldy. However, if one takes limits as $t\to 0$ for these derivatives, one recovers the familiar sequence $1,0,-1,0,1,\dots$.

There is a version of Lagrange inversion that uses the coefficients of the original power series instead of the function itself. Mathematica natively supports this operation through the InverseSeries[] construction, but here is an implementation of one of the simpler algorithms for series reversion, due to Henry Thacher:
a = Rest[CoefficientList[Series[ArcSin[x], {x, 0, 20}], x]];
n = Length[a];
Do[
    Do[
      c[i, j + 1] = Sum[c[k, 1]c[i - k, j], {k, 1, i - j}];
      , {j, i - 1, 1, -1}];
    c[i, 1] = Boole[i == 1] - Sum[a[[j]] c[i, j], {j, 2, i}]
    , {i, n}];
Table[c[i, 1], {i, n}]

and then compare with the output of Rest[CoefficientList[Series[Sin[x], {x, 0, 20}], x]].
Other methods, including a modification of Newton's method for series, have been presented, but I won't get into them here.
A: We can start with the basic definition of $e$:
$$
e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n
$$
then raise $e$ to a real power $x$:
$$
\begin{align}
e^x&=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}\\
   &=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n
\end{align}
$$
Next we can extend this to imaginary exponents:
$$
e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}{n}\right)^n\tag{1}
$$
One way to look at $(1)$ is using the Binomial Theorem to get a series for $e^{ix}$.
$$
\begin{align}
e^{ix}
&=\lim_{n\to\infty}\left(1+\frac{ix}{n}\right)^n\\
&=\lim_{n\to\infty}\sum_{k=0}^n\binom{n}{k}\left(\frac{ix}{n}\right)^k\\
&=\lim_{n\to\infty}\sum_{k=0}^\infty\frac{P(n,k)}{n^k}\frac{(ix)^k}{k!}\\
&=\sum_{k=0}^\infty\frac{(ix)^k}{k!}\tag{2}
\end{align}
$$
Passing the limit inside the sum is legal because $\frac{P(n,k)}{n^k}\to 1$ monotonically, and because the final sum converges absolutely.
Another way to look at $(1)$ is using the geometry of complex numbers.
Recall that for a complex number, $z$, we have
$$
\begin{align}
\left|z^n\right|&=|z|^n\tag{3a}\\
\arg\left(z^n\right)&=n\arg(z)\tag{3b}
\end{align}
$$
Furthermore, recall that
$$
\begin{align}
\textstyle\left|1+\frac{ix}{n}\right|&=\textstyle\sqrt{1+\left(\frac{x}{n}\right)^2}\tag{4a}\\
\textstyle\arg\left(1+\frac{ix}{n}\right)&=\textstyle\tan^{-1}\left(\frac{x}{n}\right)\tag{4b}
\end{align}
$$
Using $\mathrm{(3a)}$ and $\mathrm{(4a)}$, we get
$$
\begin{align}
\left|e^{ix}\right|
&=\left|\lim_{n\to\infty}\textstyle\left(1+\frac{ix}{n}\right)^n\right|\\
&=\lim_{n\to\infty}\textstyle\left(1+\left(\frac{x}{n}\right)^2\right)^\frac{n}{2}\\
&=\lim_{n\to\infty}\textstyle\left(1+\left(\frac{x}{n}\right)^2\right)^{n^2\frac{1}{2n}}\\
&=\lim_{n\to\infty}\textstyle\left(e^{x^2}\right)^\frac{1}{2n}\\
&=1\tag{5}
\end{align}
$$
Using $(3\mathrm{b})$ and $(4\mathrm{b})$, we get
$$
\begin{align}
\arg(e^{ix})
&=\arg\left(\lim_{n\to\infty}\textstyle\left(1+\frac{ix}{n}\right)^n\right)\\
&=\lim_{n\to\infty}\textstyle n\;\tan^{-1}\left(\frac{x}{n}\right)\\
&=x\;\lim_{n\to\infty}\textstyle\tan^{-1}\left(\frac{x}{n}\right)\left/\frac{x}{n}\right.\\
&=x\tag{6}
\end{align}
$$
Using $(5)$ and $(6)$, we see that $e^{ix}$ has length $1$ and argument $x$.  Converting $e^{ix}$ to rectangular coordinates, we get
$$
e^{ix}=\cos(x)+i\sin(x)\tag{7}
$$
Comparing the real and imaginary parts of $(2)$ and $(7)$, we get the series for $\sin(x)$ and $\cos(x)$.
A: There's the way Euler did it.  First recall that
$$
\sin(\theta_1+\theta_2+\theta_3+\cdots) = \sum_{\text{odd }k \ge 1} (-1)^{(k-1)/2} \sum_{|A| = k}\  \prod_{i\in A} \sin\theta_i\prod_{i\not\in A} \cos\theta_i.
$$
Then let $n$ be an infinitely large integer (that's how Euler phrased it, if I'm not mistaken) and let
$$
x= \frac{\theta}{n} + \cdots + \frac{\theta}{n}
$$
and apply the formula to find $\sin x$.  Finally, recall that (as Euler would put it), since $\theta/n$ is infinitely small, $\sin(\theta/n) = \theta/n$ and $\cos(\theta/n) = 1$.  Then do a bit of algebra and the series drops out.
The algebra will include things like saying that
$$
\frac{n(n-1)(n-2)\cdots(n-k+1)}{n^k} = 1
$$
if $n$ is an infinite integer and $k$ is a finite integer.
