What trig identities must one know to derive the others?

Solution 1:

I think that Apostol's Calculus book might contain this idea in detail:

If you are willing to believe that sine and cosine are continuous, and have proved that a continuous function on a dense subset of an interval has a unique continuous extension to the interval, then once you know

(1) $\sin(0) = 0$, $\cos(0) = 1$, $\sin(\pi/2) = 0$; $\cos(x) > 0$ for $x \in [0, \pi/2)$,

(2) $\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)$ for all $a, b$, and

(3) $\sin(-x) = -\sin(x)$ for all $x$.

you can derive $1 = \cos^2(a) + \sin^2(a)$ by setting $b = a$ in the second formula. You then can find that cosine is even (set $a = 0$).

Assumption 3 may not be necessary, but I confess, I forget how to show that the sine is odd without it.

Then you can set $b = -a$ to get

$$cos(2a) = \cos^2(a) - \sin^2(a) = 2\cos^2(a) - 1.$$

Applying this to $a = x/2$, you get

$$ \cos(x) = 2\cos^2(x/2) - 1 \\ \cos(x/2) = \sqrt{\frac{\cos(x) + 1}{2}} $$ from this, you can determine cosine of all numbers of the form $\frac{\pi}{2} \frac{1}{2^n}$; using the addition formula, you can determine cosine at all points of the form

$$\frac{\pi}{2} \frac{k}{2^n}$$ where $k$ is an integer. These form a dense subset of the interval $[0, \pi/2]$.

You can then also show that for $x$ small, $\sin x$ is also small, so that (using the addition/subtraction formulas) cosine is continuous on this dense subset; it therefore has a unique continuous extension to $[0, \pi/2]$. The same goes for sine, and you're on your way.

@Semiclassical suggested, in comments above, that an addition formula, together with $\sin^2 + \cos^2 = 1$, might suffice, but that a half-angle formula might not. The conjecture about half-angle formulas is correct, as the following shows:

Let \begin{align} Sin(x) = \begin{cases} \sin(x) & x = \frac{p}{2^k} \pi & \text{for $p, k, \in \mathbb Z$} \\ 0 & \text{otherwise} \end{cases}\\ Cos(x) = \begin{cases} \cos(x) & x = \frac{p}{2^k} \pi & \text{for $p, k, \in \mathbb Z$} \\ 0 & \text{otherwise} \end{cases} \end{align}

Then $Sin$ and $Cos$ satisfy $Sin^2 + Cos^2 = 1$ and the half-angle formulas, but they are not the same functions as $\sin$ and $\cos$, and hence need not satisfy the other formulas. In particular, they fail to satisfy the addition formula.