So what's the correct answer?

gs17 2 days ago | parent | next [-]

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.

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measurablefunc 2 days ago | parent [-]

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

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madars 2 days ago | parent | next [-]

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!)

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measurablefunc 2 days ago | parent [-]

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.

	▲	madars 2 days ago \| parent [-]
		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.

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deruta 2 days ago | parent | prev | next [-]

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

	▲	gs17 2 days ago \| parent [-]
		The page also specifies it's degrees mode.

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saagarjha 2 days ago | parent | prev [-]

Yes, that's what those functions do.

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measurablefunc 2 days ago | parent | prev | next [-]

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.

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madars 2 days ago | parent | prev | next [-]

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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2 days ago | parent | prev [-]

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