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Comment by mysteria | Hacker News Reader

So what's the correct answer?

9 degrees. arcsin(arccos(arctan(tan(cos(sin(9)))))) basically makes a set of sin-cos-tan layers that arctan-arccos-arcsin unwrap one-by-one, which should result in nothing having changed, unless the functions used weren't accurate.

That's incorrect, you have to choose the proper inverse branch if you want the answer to be 9.

There is no choice here - each inverse is uniquely determined. That's similar to how 3 and -3 are both square roots of 9 (i.e., solutions to x^2=9), but sqrt(9)=3 as it denotes the principal square root, which by convention is always the non-negative value. Of course, in a different context we might design functions to have multi-valued properties, like atan2(x,y) != atan(y/x) in general (atan2 takes quadrant in account and returns full range [-pi, pi], atan only returns principal values in [-pi/2, pi/2]) as practical applications benefit from preserving quadrant beyond just the principal inverse (or not failing when x=0!)

The inverse branches are not unique, you might think there is no choice being made but picking the standard branch is a choice b/c I can always shift the result by 2π by picking a different branch of the inverse. The answer is not unique & the assumption is that the calculators are using the standard branch.

Of course, but the choice is standard and thus the answer is 9. I can define a non-standard sqrt(x) which sometimes gives the positive root and sometimes the negative one, and then sqrt(sqrt(16)) could be -2 or undefined (if I defined sqrt(16)=-4) but that's just silly - the only reasonable interpretation for what the calculator should show for sqrt(sqrt(16)) is simply 2.

Yes, that's what those functions do.

I was with you until I remembered the default unit for angles in calculators is degrees, not radians.

The page also specifies it's degrees mode.

You can assume that sin(9) is within the range of all the functions that are post-composed w/ it so what you end up w/ in the end is arcsin(sin(9)). Naively you might think that's 9 but you have to be careful b/c the standard inverse branch of sin is defined to be [-1, 1] → [-π/2, π/2].

Edit: The assumption is that the calculators are using specific branches of the inverse functions but that's still a choice being made b/c the functions are periodic there are no unique choices of inverse functions. You have to pick a branch that is within the domain/range of periodicity.

arcsin(arccos(arctan(tan(cos(sin(9)))))) = 9 (in degrees mode - when regular trig functions output pure numbers, those numbers get interpreted as degrees for the next function and similar for inverses - calculator style), because each intermediate lands in the principal-value domain of the next inverse (e.g., arctan(tan(x)) = x when x \in (-90°, 90°) and the intermediates happen to be in those ranges). Specifically, sin(9°) ≈ 0.156434, cos(0.156434°) ≈ 0.999996, arctan(tan(0.999996°)) = 0.999996°, arccos(0.999996)≈0.156434°, arcsin(0.156434)≈9°.

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