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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.

Zoey
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  • See, for instance, the CORDIC algorithm – HallaSurvivor Dec 27 '21 at 02:27
  • Sorry, but I explicitly asked for how to do it without a computer, so giving me the complicated method computers use doesn't help. – Zoey Dec 27 '21 at 02:30
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    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. – HallaSurvivor Dec 27 '21 at 02:31
  • 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, i, etc. – Zoey Dec 27 '21 at 02:35
  • 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. – HallaSurvivor Dec 27 '21 at 02:39
  • The problem is that for large inputs, we need to use many terms of this sum (most calculus classes will show you how to figure out how many terms of the sum to use in order to be within a certain error tolerance for whatever input you like). The power of the CORDIC algorithm is that we can still get good approximations, but the number of terms we use doesn't change as our input grows! – HallaSurvivor Dec 27 '21 at 02:41
  • I mean, they expect 10th graders to quickly and easily find it out by themselves without being taught, it can't be as complicated as the CORDIC method if they expect people who have never been taught matrices or matrix multiplication, or Capital Pi notation, to find it out. – Zoey Dec 27 '21 at 02:42
  • I am still very confused on the way to do it without any infinite sequences or matrices, or anything. The way they expect it to be done in school. Which wouldn't use any of those due to us having been expected to calculate them without a calculator before we were even taught that such things existed. It bothered me then, and it is bothering me now. I just want to have some idea of how we were expected to do it without any of the methods you all are talking about. Edit: Another thing, how the heck are were supposed to do infinite series without a calculator. We don't have time to spend hours. – Zoey Dec 27 '21 at 02:55

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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.

Eric
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