Revisiting Pythagoras goes linear

Let’s say for some reason that you know \sin \theta and \cos \theta, for some angle \theta, and you need to figure out what \theta is. Let’s say, furthermore, that you live in some benighted age which doesn’t have calculators or even trigonometric tables.

There are a few approaches to this. One is what I did in my “Pythagoras goes linear” post. We can fit a linear model to the points

(x_i, y_i) = (\cos \theta_i \sin \theta_i)

where the $\theta_i$ range over an arithmetic sequence with endpoints 0 and $\latex \pi/2$, namely

0, {\pi \over 2n}, {2\pi \over 2n}, \cdots, {n\pi \over 2n}.

The model you get out is quite simple:

\theta \approx \pi/4 - 0.7520 \sin \theta + 0.7520 \cos \theta

This is actually just a few lines of R.

 n = 10^6
 theta = (c(0:n))*pi/(2*n);
 x = cos(theta); y = sin(theta);

I’ll save you the output, but be impressed: r2 = 0.9992$, and the mean square error is 0.0126 radians or 0.72 degrees. So we can write \theta \approx \pi/4 + 0.7520 (-\cos \theta + \sin \theta), in radians. In degrees this is \theta \approx 45^o + 43.08^o(-\cos \theta + \sin \theta).

For example, say you have an angle with \sin \theta = 5/13, \cos \theta = 12/13 — the smaller angle of a 5-12-13 right triangle. Then we get \theta \approx 45^o + 43.08^o (-7/13) \approx 21.8^o. In fact \theta = \tan^{-1} 5/12 \approx 22.6^o — we’re not off by much! And the error is never more than a couple degrees, as you can see in the plot below.

arctan error

This was inspired by a post by Jordan Ellenberg that I came across recently: How to compute arctangent if you live in the 18th century, which refers back to my Pythagoras goes linear. A better approximation, although nonlinear, is

\theta \approx (3y)/(2 + x)

where x = \sin \theta, y = \cos \theta. This is essentially a simplification of the rule that Ellenberg’s source (Hugh Worthington’s 1780 textbook, The Resolution of Triangles) gives, which can be translated into our notation as

\theta = {86*pi \over 180} {y \over x/2 + 1}. Applying this to our test angle with x = 12/13, y = 5/13, we get \theta \approx 15/38 = 0.39474 radians, while in truth \tan^{-1} 5/12 = 0.39479.

This formula \theta \approx (3y)/(2+x) is so nice that I can’t help but suspect there’s a simple derivation. Any takers?


4 thoughts on “Revisiting Pythagoras goes linear

  1. I assume you mean x = \cos\theta and y = \sin\theta.

    Here’s one way you could invent that approximation: you recall that
    \cos x \le \frac{\sin x}{x} \le 1
    and you’d like to know more precisely where in that range the middle function falls. Maybe you make the guess that it’s close to a convex combination of the two functions on either side:
    \frac{\sin x}{x} \approx (1-\lambda)\cos x + \lambda\cdot 1
    The choice \lambda = \frac{2}{3} makes the x^2 terms equal, and then rearranging yields
    x \approx \frac{3\sin x}{2+\cos x}
    as desired.

    (Sorry for any latex errors; there’s no preview, alas.)

  2. This looks like it’s related to multivariable Padé approximants. Let me take quick stab at it, apologies for no LaTeX.

    Assume that there is some approximating function of the form

    t ~= (a0+a1*x+a2*y)/(1+b1*x+b2*y).

    Multiply across and rewrite in terms of theta as:

    t ~= a0 + (a1-b1*t) sin t + (a2-b2*t) cos t

    Now, replace the trigonometric function with their series representations:

    t ~= a0 + (a1-b1*t)[t – t^3/6 +O(t^5)] + (a2-b2*t)[1-t^2/2+t^4/24+O(t^6)]

    Collecting terms up to t^4, we have

    t ~= (a0+a2)t^0 + (a1 – b2) t + (-a2/2 – b1) t^2 +(-a1/6 + b2/2) t^3 + (a2/24 + b1/6)t^4

    Now just match the powers and boom, five equations in five variables. Mathematica gives me a1 -> 3/2, b2 -> 1/2, and all others 0. Therefore, the final function form would be:

    t ~= (3y/2)/(1+x/2) ~= 3y/(2+x)

    and we’re done. Padé approximants are quite nifty and have cool applications. My personal favorite is using exp(x) ~= (2+x)/(2-x) to quickly calculate interest rates in my head.


  3. Apologies, the x and y were switched in my last formula. As the first commenter pointed out, it should be x=cos t and y = sin t. I stand by the rest of the derivation, though.

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