In addition to parametrizing rational points on the unit circle, the matrix method, and one-fold origami, here’s a way to get the Pythagorean triples using the hyperbolic sine, cosine, and tangent.

It begins with the fundamental identity

,

which looks like the Pythagorean identity if you divide by and rearrange terms: .

That is, given a Pythagorean triple (where ), if we define by the condition , then we will also necessarily have . Conversely, if and are both rational for , then we can get a Pythagorean triple by scaling .

But, how do we find such where both and are rational numbers?

We exploit the fact that is a rational number if* and only if* and are rational numbers too. This fact follows from a few hyperbolic function identities:

- (aka “half-angle formula”)

The first identity shows that is a rational number if and are, and the other identities, combined with the fundamental identity (disguised in the forms and ) show the converse.

This means that every Pythagorean triple arises by scaling , where is a positive rational number. Suppose , where *a* and *b* are positive integers. (Note that *a* < *b*, because .) Using the identities above, we find that

And this quickly leads us to rediscovering the Pythagorean triple parametrization , where, to be primitive, we require that *a* and *b* be relatively prime integers not both odd.

If you’re wanting to brush up on your trig identities, find the analogous derivation that uses sine, cosine, and tangent.

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