The name comes from the fact that these functions transcend "normal" algebraic operations. That is, you can't express them as a finite set of additions, multiplications, roots, or their inverses. While there are an infinite number of such functions, several are of particular importance because they a) are solutions to ordinary differential equations and b) come up a lot in normal situations. (Devotees of classical physics would argue that (a) implies (b)). In fact, we're so familiar with them, it's easy to forget that they cannot be expressed in closed form. We can deal only in approximations.
Exponential:
From our convergence results, it's fairly easy to show that (exp x)(exp y) = exp(x + y). It should also be fairly clear that exp 0 = 1 (since the first term, 00/0! = 1 is the only non-zero term in the sequence). It takes a bit more finagling to get that exp 1 = e, from which we finally can derive that exp x = ex.
The really important result is that d/dx exp x = exp x. That's why it pops up in the solution of so many differential equations.
Natural log: log x = exp-1x.
Since exp x is strictly increasing, we can invert it, though the domain is now just the positive real numbers (since exp x > 0). As with the exponential, the most useful result is related to it's derivative: d/dx log x = 1/x.
Trig functions: We all learned about these in high school as ratios of the sides of a right triangle: SOHCAHTOA (sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent). While that's an easy way to remember them (even if a bit politically incorrect as kids of my generation were taught to say it like it was the name of an indian chief), the analytic versions are more interesting (at least, to me).
The fact that those relate to either right triangles or the unit circle is less than obvious. The way you get there is a case study in incremental proof.
Start by noting that, when x = 0, both sequences are degenerate and sin 0 = 0, cos 0 = 1. It should also be obvious that negating x does nothing to cos, but it flips the sign of all the terms on sin. So, cos -x = cos x and sin -x = sin x. It is only slightly harder to take the derivatives and note that sin'x = cos x and cos'x = -sin x.
Armed with those facts, we can now show that g(x) = (sin x)2 + (cos x)2 = 1. Take the derivative:
g'(x) = 2(sin x)(cos x) + 2(cos x)(-sin x) = 0.
If the derivative is zero everywhere, then g(x) is constant. Thus,
g(x) = g(0) = 02 + 12 = 1.
Thus, (cos x, sin x) describe a point on the unit circle where x is the angle in radians and the normal trig identities fall into place. As a bonus, formulating the the trig functions as infinite alternating series allows us to derive arbitrarily precise bounds on estimates of such (you're never off by more than the first term you drop), as well as giving us a means for estimating π (though, unless you're grinding a telescope lense or launching a satellite, 22/7 will do just fine for the latter).
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