# How To Calculate A Square Root By A Method You Will Never Actually Use

Sunday’s comics post got me thinking about ways to calculate square roots besides using the square root function on a calculator. I wondered if I could find my own little approach. Maybe something that isn’t iterative. Iterative methods are great in that they tend to forgive numerical errors. All numerical calculations carry errors with them. But they can involve a lot of calculation and, in principle, never finish. You just give up when you think the answer is good enough. A non-iterative method carries the promise that things will, someday, end.

And I found one! It’s a neat little way to find the square root of a number between 0 and 1. Call the number ‘S’, as in square. I’ll give you the square root from it. Here’s how.

First, take S. Multiply S by two. Then subtract 1 from this.

Next. Find the angle — I shall call it 2A — whose cosine is this number 2S – 1.

You have 2A? Great. Divide that in two, so that you get the angle A.

Now take the cosine of A. This will be the (positive) square root of S. (You can find the negative square root by taking minus this.)

Let me show it in action. Let’s say you want the square root of 0.25. So let S = 0.25. And then 2S – 1 is two times 0.25 (which is 0.50) minus 1. That’s -0.50. What angle has cosine of -0.50? Well, that’s an angle of 2 π / 3 radians. Mathematicians think in radians. People think in degrees. And you can do that too. This is 120 degrees. Divide this by two. That’s an angle of π / 3 radians, or 60 degrees. The cosine of π / 3 is 0.5. And, indeed, 0.5 is the square root of 0.25.

I hear you protesting already: what if we want the square root of something larger than 1? Like, how is this any good in finding the square root of 81? Well, if we add a little step before and after this work, we’re in good shape. Here’s what.

So we start with some number larger than 1. Say, 81. Fine. Divide it by 100. If it’s still larger than 100, divide it again, and again, until you get a number smaller than 1. Keep track of how many times you did this. In this case, 81 just has to be divided by 100 the one time. That gives us 0.81, a number which is smaller than 1.

Twice 0.81 minus 1 is equal to 0.62. The angle which has 0.81 as cosine is roughly 0.90205. Half this angle is about 0.45103. And the cosine of 0.45103 is 0.9. This is looking good, but obviously 0.9 is no square root of 81.

Ah, but? We divided 81 by 100 to get it smaller than 1. So we balance that by multiplying 0.9 by 10 to get it back larger than 1. If we had divided by 100 twice to start with, we’d multiply by 10 twice to finish. If we had divided by 100 six times to start with, we’d multiply by 10 six times to finish. Yes, 10 is the square root of 100. You see what’s going on here.

(And if you want the square root of a tiny number, something smaller than 0.01, it’s not a bad idea to multiply it by 100, maybe several times over. Then calculate the square root, and divide the result by 10 a matching number of times. It’s hard to calculate with very big or with very small numbers. If you must calculate, do it on very medium numbers. This is one of those little things you learn in numerical mathematics.)

So maybe now you’re convinced this works. You may not be convinced of why this works. What I’m using here is a trigonometric identity, one of the angle-doubling formulas. Its heart is this identity. It’s familiar to students whose Intro to Trigonometry class is making them finally, irrecoverably hate mathematics:

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

Here, I let ‘S’ be the squared number, $\cos^2\left(\theta\right)$. So then anything I do to find $\cos\left(\theta\right)$ gets me the square root. The algebra here is straightforward. Since ‘S’ is that cosine-squared thing, all I have to do is double it, subtract one, and then find what angle 2θ has that number as cosine. Then the cosine of θ has to be the square root.

Oh, yeah, all right. There’s an extra little objection. In what world is it easier to take an arc-cosine (to figure out what 2θ is) and then later to take a cosine? … And the answer is, well, any world where you’ve already got a table printed out of cosines of angles and don’t have a calculator on hand. This would be a common condition through to about 1975. And not all that ridiculous through to about 1990.

This is an example of a prosthaphaeretic rule. These are calculation tools. They’re used to convert multiplication or division problems into addition and subtraction. The idea is exactly like that of logarithms and exponents. Using trig functions predates logarithms. People knew about sines and cosines long before they knew about logarithms and exponentials. But the impulse is the same. And you might, if you squint, see in my little method here an echo of what you’d do more easily with a logarithm table. If you had a log table, you’d calculate $\exp\left(\frac{1}{2}\log\left(S\right)\right)$ instead. But if you don’t have a log table, and only have a table of cosines, you can calculate $\cos\left(\frac{1}{2}\arccos\left(2 S - 1 \right)\right)$ at least.

Is this easier than normal methods of finding square roots? … If you have a table of cosines, yes. Definitely. You have to scale the number into range (divide by 100 some) do an easy multiplication (S times 2), an easy subtraction (minus 1), a table lookup (arccosine), an easy division (divide by 2), another table lookup (cosine), and scale the number up again (multiply by 10 some). That’s all. Seven steps, and two of them are reading. Two of the rest are multiplying or dividing by 10’s. Using logarithm tables has it beat, yes, at five steps (two that are scaling, two that are reading, one that’s dividing by 2). But if you can’t find your table of logarithms, and do have a table of cosines, you’re set.

This may not be practical, since who has a table of cosines anymore? Who hasn’t also got a calculator that does square roots faster? But it delighted me to work this scheme out. Give me a while and maybe I’ll think about cube roots.

## Author: Joseph Nebus

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

## 9 thoughts on “How To Calculate A Square Root By A Method You Will Never Actually Use”

1. Neat trick! Alas your accuracy would only be about as good as the precision of your cosines. If I had to calculate square roots without any other resources I’d go with Heron’s formula, even though the fractions start getting really nasty after 4 or 5 iterations.

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1. Thank you. And yes, one drawback of a scheme like this is that you’re limited by the accuracy of your table. Essentially, by the work that you’ve done (or had the book-compiler do) ahead of time, in anticipation of need. You gain in speed, at the cost of precision.

Most non-specialist trig tables would carry four digits of precision, which for the examples I gave here — square roots of 0.5 and of 81 — would give perfect answers. For something a little more realistic, like the square root of 0.8660 (itself chosen as approximately half the square root of 3), this would give an approximate square root of 0.9305. The square of that is 0.8658. Which isn’t perfect, sure, but is not bad for how little work is demanded.

A slightly more specialist table, one that goes to six digits of precision, would imply a square root of 0.8660 as 0.930591, the square of which is slightly more than 0.8659996. An eight-digit table would get a square root of 0.9305912100, the square of which is 0.8660000001. There are many purposes for which that’s accurate enough.

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