Inverse cosine

In the previous two posts, we looked at why Mathematica and SymPy did not simplify sinh(arccosh(x)) to (x^2 - 1) as one might expect. After understanding why sinh(arccosh(x)) doesn't simplify nicely, it's natural to ask why sin(arccos(x)) does simplify nicely.

In this post I sketched a proof of several identities including

sin(arccos(x)) = (1 - x^2)

saying the identities could be proved geometrically. Let x be between 0 and 1. Construct right triangle with hypotenuse of length 1 and one side of length x. Call the acute angle formed by these two sides . Then cos = x, and so arccos(x) = , and sin = (1 - x^2). This proves the identity above, but only for 0 <x < 1.

If we make branch cuts along (-, -1] and [1, ) we can extend arccos(z) uniquely by analytic continuation. We can extend the definition to the branch cuts by continuity, but from one direction. We either have to choose the extension to be continuous from above the branch cuts or from below; we have to choose one or the other because the two limits are not equal. As far as I know, everyone chooses continuity from above, i.e. continuity with quadrant II, by convention.

In any case, we can define arccos(z) for any complex number z, and the result is a number whose cosine is z. Therefore the square of its cosine is z^2, and the square of its sine is 1 - z^2. So we have

sin^2(arccos(z)) = 1 - z^2.

But does that mean

sin(arccos(z)) = (1 - z^2)?

Certainly it does if we're working with real numbers, but does it for complex numbers?

Recall what square root means for complex numbers: it is the analytic function with branch cut (-, 0] that agrees with the real square root function along the real line, and is defined along the branch cut to be continuous with quadrant II. Since

sin(arccos(z)) = (1 - z^2)

for real -1 <z < 1, the two sides of the equation are equal on a set with a limit point, and so as analytical functions they are equal on their common domain.

The only remaining detail is whether the two functions are equal along the branch cut (-, -1] where we've extended the function by continuity. As

Since we've defined both the arccos and square root functions by continuous extension from the second quadrant, equality on the branch cut also follows by continuity.

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