cos = adj/hyp sin = opp/hyp

This is probably the way you first met sin and cos. From a few more trig identities, and a bit of geometry, we can 'prove' the derivatives of cos and sin:

$d/\mathrm{dx}(\sin x)=\cos x$

$d/\mathrm{dx}(\cos x)=-\sin x$

Then we can use Maclaurin's theorem. This says that if a function $f$ is infinitely differentiable at 0, then

$f(x)=f(0)+f^{(1)}(0)x+[f^{(2)}(0)/2!]x^2+\dots +[f^{(r)}(0)/r!]x^r+\dots$

where this series converges. It so happens that for cos and sin we get series that converge everywhere, and these series are:

$\cos x=1-x^2/2!+x^4/4!+\dots +(-1{)}^r{x}^{2r}/(2r)!+\dots$

$\sin x=x-x^3/3!+\dots +(-1{)}^{r+1}{x}^{2r+1}/(2r+1)!+\dots$

Also $e^x=1+x^2/2!+x^3/3!+\dots$

And the formula for $e^{i\theta }$ can be obtained by comparing these series.

However, in higher analytical work, it is more usual to start by defining sin and cos as the power series, and then deducing the ratios. We do this using the formula for $e^{i\theta }$.

The set of points $z=e^{i\theta }$ $0\le \theta <2\pi$ form a unit circle in the complex plane.

For this next bit, a diagram would help, but I can't manage to draw a good one, so you might want to try sketching one yourself.

If we choose a point $z$ in the first quadrant of this circle (ie real and imaginary parts $>0$) given by $z=e^{i\theta }$ and draw a line connecting $z$ to the origin, then the formula for $e^{i\theta }$ tells us that the angle between this line and the positive real axis is $\theta$.

To see this, drop a line down from $z$ to the real axis such that this line is perpendicular to the real axis. You now have a right-angled triangle, where the hypotenuse has length 1, the adjacent side has length $\cos \theta$ (the real part of $z$) and the opposite side has length $\sin \theta$ (the imaginary part of $z$).