The best approach I've seen uses the fact that $\exp z=\lim_{n\to\infty} (1+\frac{z}{n})^n$ for all complex $z$. Then, for real $x$, we show that, for large $n$, $1+\frac{ix}{n}$, is "close enough" to $\cos \frac{x}{n} + i\sin\frac{x}{n}$, so that $(1+\frac{ix}{n})^n$ is "close enough" to $\cos x + i\sin x$. (Here, $\cos$ and $\sin$ are the geometric definitions of the trig functions.)
So, first define $\exp z$ via the normal power series, and then show that $\exp z = \lim_{n\to\infty} (1+\frac{z}{n})^n$.
Then, define $\operatorname{cis} x =\cos x + i\sin x$, and prove, using (geometric) properties of the trig functions, that $\operatorname{cis} x\operatorname{cis} y =\operatorname{cis} (x+y)$. Then note that by induction, $(\operatorname{cis} x)^n = \operatorname{cis} nx$.
Next, we are going to compute $\exp(ix)$ by writing $1+\frac{ix}n$ in polar coordiates, $r_n \operatorname{cis}(\theta_n)$.
As mentioned above, the goal is to show that, for large enough $n$, $1+\frac{ix}{n}$ is "close enough" to $\operatorname{cis} \frac{x}n$, so that the limit of $(1+\frac{ix}n)^n$ is the same as the limit of $(\operatorname{cis}\frac x n)^n = \operatorname{cis} x$, and therefore $\exp(ix)=\operatorname{cis} x$.
First, note that $r_n=\sqrt{1+\frac{x^2}{n^2}}$, and show that $r_n^n\to 1$ as $n\to \infty$. (This is relatively easy, noting that $(r_n)^{2n^2}\to \exp (x^2)$.)
So this shows that $\exp(ix)$ ends up on the unit circle for real $x$, and is equal to $\lim_{n\to\infty} \operatorname{cis}(n\theta_n)$.
What we know about $\theta_n$ is that $\sin \theta_n = \frac{x}{nr_n}$.
Next you need to show is that $\lim_{n\to\infty} n\theta_n = x$. Then you've shown that $\exp(ix)=\operatorname{cis} x$.
The key to proving this last limit is to show that, for small $\theta$, $|\theta-\sin(\theta)|\leq C|\sin \theta|^2$ for some constant $C$. Note, though, that proving this is not easy since we are assuming we do not know the power series expansion for $\sin$.
If you can show that, then you can show, for large $n$, $|\theta_n - \frac{x}{nr_n}|\leq C|\frac{x}{r_nn}|^2<C|\frac{x}{n}|^2$. So $|n\theta_n - \frac{x}{r_n}|<C\frac{|x^2|}{n}$.
So, since $\frac{x}{r_n}\to x$, $n\theta_n\to x$, and therefore, $\lim \operatorname{cis} n\theta_n = \operatorname{cis} x$
So all that's left to prove is that for small enough $\theta$, $|\theta - \sin \theta| \leq C|\sin \theta|^2$. (You could actually replace $2$ with $1+\epsilon$ for fixed $\epsilon>0$.)
The geometric reason for this last one is as follows. Let $P=(1,0)$ and $Q=(\cos \theta,\sin \theta)$. Take the lines tangent to the circle at $P$ and $Q$ and let their intersection be $R$. Claims (for small $\theta>0$:)
$$|PR|=|QR|=\tan\frac{\theta}2$$
$$\sin\theta < \theta < |PR|+|QR|=2\tan\frac{\theta}2$$
The first half of the inequality is because $\sin \theta$ is the shortest distant from $Q$ to the $x$-axis, and $\theta$ is the length of the path from $Q$ to the $x$-axis along the circle. The second inequality is a little less intuitive - basically, this is due to the rule that the shortest path from $P$ to $Q$ that does not go inside the circle is along the circle.
(For small $\theta<0$, we have to reverse all the signs, but the results are the same.)
Thus we see:
$$|\theta-\sin \theta| < |2\tan\frac\theta 2 - \sin\theta|$$
But $2\tan\frac{\theta}2 = \frac{2\sin \frac{\theta}2}{\cos\frac\theta 2}$. Mutliply numerator and denominator by $\cos \frac{\theta}2$ and we see that:
$$2\tan\frac{\theta}2 = \frac{\sin \theta}{\cos^2\frac{\theta}2}$$
So:
$$|2\tan\frac\theta 2 - \sin\theta| = |\sin \theta \frac{\sin^2\frac \theta 2}{\cos^2\frac\theta 2}|$$
Multiplying the numerator and denominator by $4\cos^2\frac \theta 2$, we get:
$$|2\tan\frac\theta 2 - \sin\theta| = |\sin\theta \frac{\sin^2\theta}{4\cos^4\frac{\theta}2}|$$
So with $\theta$ small enough that $\cos\frac\theta 2>\frac{1}{2}$, we have
$$|\theta-\sin \theta| < |2\tan\frac\theta 2 - \sin\theta| < 4|\sin^3\theta|<4|\theta|^3$$
So we now know that $\exp ix = \operatorname{cis} x$. That means, in turn, that $$\sin x = \frac{\exp(ix)-\exp(-ix)}{2i}$$ and $$\cos x = \frac{\exp(ix)+\exp(-ix)}{2}$$
From there we can derive the power series for $\sin$ and $\cos$ using the power series for $\exp$.
