How exactly does one calculate trigonometric functions without a calculator. I recognize the whole Sohcahtoa thing, that is not what I am talking about. I am referring to when we are given something like sin(a), how exactly would this actually be calculated. I can't find any answers, it is just expected that people use calculators for it.

I know you don't need the sides of a triangle to even do it, as shown by how you can just place sin(x) into a calculator. But I was only ever taught how to do things in reference to triangles, and to use the calculator for it. Except when we were expected to do it without a calculator but with never being once taught how to do it without one.

  • $\begingroup$ See, for instance, the CORDIC algorithm $\endgroup$ Dec 27, 2021 at 2:27
  • $\begingroup$ Sorry, but I explicitly asked for how to do it without a computer, so giving me the complicated method computers use doesn't help. $\endgroup$
    – Zoey
    Dec 27, 2021 at 2:30
  • 1
    $\begingroup$ This isn't a complicated method, all things considered. See here for a worked example. And there's nothing stopping you from doing this all by hand. This is an extremely efficient way of computing sine/cosine/etc in practice. $\endgroup$ Dec 27, 2021 at 2:31
  • $\begingroup$ But the Wikipedia page it is around 10+ steps with a ton of matrixes, matrix multipication, products of long series, imaginary numbers, a bunch of variables such as σ, n, x, 𝛾<sub>i</sub>, etc. $\endgroup$
    – Zoey
    Dec 27, 2021 at 2:35
  • $\begingroup$ I mean, these functions are fairly complicated. If you want an "elementary" answer, you can use the taylor approximations to these functions. For instance, $\sin(x) = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$. If you sum only to $k=20$, say, then we find $\sin(5) \approx -0.958933165196599$. Since $\sin(5) = -0.95892427466313...$ this is not so bad an approximation. $\endgroup$ Dec 27, 2021 at 2:39

1 Answer 1


You do it without a calculator in school by memorizing common values - all multiples of 15 degrees and 18 degrees have simple evaluations in terms of square roots. There are tricks you can use to simplify the memorization (i.e. that adding 180 negates sin and cos, $90-x$ switches them, trig addition formulas …) so you’d need to memorize at most a dozen values in order to be able to give an expression in square roots for any multiple of 3 degrees.

If you are allowed a table of known values, you can linearly interpolate between them - this is historically how mathematicians and engineers did calculations before there were calculators.

If you have neither, then you use some kind of converging algorithm like a Taylor series expression. This is how such tables of known values were constructed before computers. No one ever really does this by hand in practice.


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