GoKawiilI’ve written several posts about applying trig functions to inverse trig functions. I intended to write two posts, one about the three basic trig functions and one about their hyperbolic counterparts. But there’s more to explore here than I thought at first. For example, the mistakes that I made in the first post lead to a couple more posts discussing error detection and proofs.
I was curious about how Mathematica would handle these identities. Sometimes it doesn’t simplify expressions the way you expect, and for interesting reasons. It handled the circular functions as you might expect.
So, for example, if you enter Sin[ArcCos[x]] it returns √(1 − x²) as in the table above. Then I added an h on the end of all the function names to see whether it would reproduce the table of hyperbolic compositions.
For the most part it did, but not entirely. The results were as expected except when applying sinh or cosh to arccosh. But Sinh[ArcCosh[x]] returns
and Tanh[ArcCosh[x]] returns
Why doesn’t Mathematica simplify as expected?
Why didn’t Sinh[ ArcCosh[x] ] just return √(x² − 1)? The expression it returned is equivalent to this: just square the (x + 1) term, bring it inside the radical, and simplify. That line of reasoning is correct for some values of x but not for others. For example, Sinh[ArcCosh[2]] returns −√3 but √(2² − 1) = √3. The expression Mathematica returns for Sinh[ArcCosh[x]] correctly evaluates to −√3.
Defining ArcCosh
To understand what’s going on, we have to look closer at what arccosh(x) means. You might say it is a function that returns the number whose hyperbolic cosine equals x. But cosh is an even function: cosh(−x) = cosh(x), so we can’t say the value. OK, so we define arccosh(x) to be the positive number whose hyperbolic cosine equals x. That works for real values of x that are at least 1. But what do we mean by, for example, arccosh(1/2)? There is no real number y such that cosh(y) = 1/2.
To rigorously define inverse hyperbolic cosine, we need to make a branch cut. We cannot define arccosh as an analytic function over the entire complex plane. But if we remove (−∞, 1], we can. We define arccosh(x) for real x > 1 to be the positive real number y such that cosh(y) = x, and define it for the rest of the complex plane (with our branch cut (−∞, 1] removed) by analytic continuation.
... continue reading