2

My TA told me in problem section one day that every trig identity could be derived from just 2: the Pythagorean identity and the double-angle identity (or he might have said the half-angle identity). I'm a bit dubious that every trig identity could be derived from just these two. What would you say the minimum number of identities one must know to say derive every identity on the wikipedia page?

If it is possible to derive all of the identities from just 2, can anyone recommend a source that takes that approach?

user165538
  • 21
  • A rather unhelpful answer is that you only need Euler's identity $e^{i\phi}=\cos\phi +i \sin \phi$ and the interpretation of this as a point on the complex unit circle. But unless you know complex algebra, that probably doesn't help! – Semiclassical Jul 21 '14 at 14:38
  • Do you count $\cos 0=1$ and $\cos \pi/2=0$ as identities? If not, just using the cosine angle addition formula can get you all the rest. – Hayden Jul 21 '14 at 14:40
  • On a more concrete level: I think the cosine addition formula should be enough, but I don't see how the double-angle formula would be enough. – Semiclassical Jul 21 '14 at 14:41
  • @Hayden I don't count those as identities. But how would you say derive the power reduction formulas from that? – user165538 Jul 21 '14 at 14:42
  • @Semiclassical Does that imply that you can do ALL of trigonometry with only complex algebra? – user165538 Jul 21 '14 at 14:45
  • @user165538 Repeated use of Double-angle formulas and angle-addition formulas, half-angle formulas, etc, which are all definitely provable from the cosine angle addition formula. However, I should not that not every formula on that page can be proven just using those; infinite-products and related results likely won't be doable. – Hayden Jul 21 '14 at 14:46
  • @user165538: I would say so, yes. For example, see the discussion on Wikipedia of De Moivre's formula. – Semiclassical Jul 21 '14 at 14:47
  • OK, thanks. :-) – user165538 Jul 21 '14 at 14:48
  • @Hayden: Depends on what means by 'trigonometric identity' in part. I should think something like the identity $\prod_k \cos\theta_k$ could be done using angle-addition alone as a substitute for using Euler's formula. (Not that it would probably look nice!) On the other hand, the infinite product expansion of trig functions in terms of their zeroes I'd say is more properly considered as a complex analysis identity. – Semiclassical Jul 21 '14 at 14:51
  • @Semiclassical Agreed, I believe the complex analytic approach would be much more fruitful because one could encode the very definition of cosine and sine into Euler's identity. – Hayden Jul 21 '14 at 14:56

1 Answers1

2

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.

John Hughes
  • 93,729
  • It sounded like the question was more geared towards which trig identities followed from Pythagorean theorem + angle-addition formula, proceeding entirely from algebra. – Semiclassical Jul 21 '14 at 15:43
  • Perhaps so. If that's what OP meant, s/he presumably won't accept my answer. :( Without continuity, though, I have grave doubts about uniqueness of sine and cosine. Without that, it would be surprising if all the formulas held. In particular, the calculus ones certainly won't, since once you know the first and second derivative of sine and cosine, they're uniquely determined by uniqueness of solutions to differential equations... (Indeed, without continuity, it'd be hard to find derivatives. :) ) – John Hughes Jul 21 '14 at 15:52
  • well, presumably the proofs would be involving naive assumptions about continuity which are implicit in us using sine/cosine to do Euclidean geometry – Semiclassical Jul 21 '14 at 15:55
  • Yeah...but those "assumptions," it seems to me, are exactly what the OP was trying to get away from, trimming them down to a bare minimum. Whether you believe that directly assuming continuity or doing so indirectly via some Euclidean reasoning is more "minimal" is, I guess, a matter of taste. – John Hughes Jul 21 '14 at 17:22
  • I read it more as "could I really crunch out all the stuff on that page algebraically just from two identities". – Semiclassical Jul 21 '14 at 17:29