I Don’t Have Any Good Ideas For Finding Cube Roots By Trigonometry


So I did a bit of thinking. There’s a prosthaphaeretic rule that lets you calculate square roots using nothing more than trigonometric functions. Is there one that lets you calculate cube roots?

And I don’t know. I don’t see where there is one. I may be overlooking an approach, though. Let me outline what I’ve thought out.

First is square roots. It’s possible to find the square root of a number between 0 and 1 using arc-cosine and cosine functions. This is done by using a trigonometric identity called the double-angle formula. This formula, normally, you use if you know the cosine of a particular angle named θ and want the cosine of double that angle:

\cos\left(2\theta\right) = 2 \cos^2\left(\theta\right) - 1

If we suppose the number whose square we want is \cos^2\left(\theta\right) then we can find \cos\left(\theta\right) . The calculation on the right-hand side of this is easy; double your number and subtract one. Then to the lookup table; find the angle whose cosine is that number. That angle is two times θ. So divide that angle in two. Cosine of that is, well, \cos\left(\theta\right) and most people would agree that’s a square root of \cos^2\left(\theta\right) without any further work.

Why can’t I do the same thing with a triple-angle formula? … Well, here’s my choices among the normal trig functions:

\cos\left(3\theta\right) = 4 \cos^3\left(\theta\right) - 3\cos\left(\theta\right)

\sin\left(3\theta\right) = 3 \sin\left(\theta\right) - 4\sin^3\left(\theta\right)

\tan\left(3\theta\right) = \frac{3 \tan\left(\theta\right) - \tan^3\left(\theta\right)}{1 - 3 \tan^2\left(\theta\right)}

Yes, I see you in the corner, hopping up and down and asking about the cosecant. It’s not any better. Trust me.

So you see the problem here. The number whose cube root I want has to be the \cos^3\left(\theta\right) . Or the cube of the sine of theta, or the cube of the tangent of theta. Whatever. The trouble is I don’t see a way to calculate cosine (sine, tangent) of 3θ, or 3 times the cosine (etc) of θ. Nor to get some other simple expression out of that. I can get mixtures of the cosine of 3θ plus the cosine of θ, sure. But that doesn’t help me figure out what θ is.

Can it be worked out? Oh, sure, yes. There’s absolutely approximation schemes that would let me find a value of θ which makes true, say,

4 \cos^3\left(\theta\right) - 3 \cos\left(\theta\right) = 0.5

But: is there a way takes less work than some ordinary method of calculating a cube root? Even if you allow some work to be done by someone else ahead of time, such as by computing a table of trig functions? … If there is, I don’t see it. So there’s another point in favor of logarithms. Finding a cube root using a logarithm table is no harder than finding a square root, or any other root.

If you’re using trig tables, you can find a square root, or a fourth root, or an eighth root. Cube roots, if I’m not missing something, are beyond us. So are, I imagine, fifth roots and sixth roots and seventh roots and so on. I could protest that I have never in my life cared what the seventh root of a thing is, but it would sound like a declaration of sour grapes. Too bad.

If I have missed something, it’s probably obvious. Please go ahead and tell me what it is.

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Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

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