1
$\begingroup$

I'm looking for a relatively simple algorithm that can quickly be done by hand to refine an initial estimate for the sine of an angle in degrees.

I've memorized a few landmark values for sine and come up with some simple techniques such that I can rapidly estimate the sine of any angle (in degrees) to within 10% error. What I'd like to do now is be able to take that estimate and refine it, presumably through some simple iterative algorithm that I can perform on a white board, for instance.

  • 0
    @Ragib: Thanks, that will be a useful formula to learn; I was not familiar with it. This is quite a bit more accurate than what I currently have, and would probably be more than sufficient for the kind of back of the envelope calculations where this is useful. However, I still wonder if there's a simple iterative approach that could refine the estimate to any arbitrary level of accuracy.2012-09-27

2 Answers 2

2

You should know $\sin(x)$ and $\cos(x)$ for $x = 0$, $30$, $60$, $90, \ldots, 360$ degrees. Memorize $\sin$ and $\cos$ of $6$ degrees and $12$ degrees and you can use the addition formulas to calculate $\sin$ and $\cos$ for all multiples of $6$ degrees (e.g. for $18$ degrees, write $18 = 30 - 12$). Then memorize $\sin$ and $\cos$ of $1$, $2$ and $3$ degrees and you can use the addition formulas to calculate $\sin$ and $\cos$ for all multiples of $1$ degree.

  • 0
    Thanks, a very nice idea. I already have the sines for all the multiples of 10 degrees memorized, so I will probably just add the sines and cosines for 5 degrees, 3 degrees, and 1 degree. Ideally, I'd still prefer an algorithm where I can get progressively more accurate for any angle, even fractional angles. But if nobody else comes up with one, I will accept this answer, it is definitely a good one.2012-09-27
0

Interpolating $\cos$ at $0$, $\pm{\pi\over2}$, and $\pm{\pi\over3}$ gives, expressed as a function of degrees, $p(x):=1-{49\over324\,000} x^2+{1\over 291\,600\,000}x^4\ .$ Drawing the two plots over the interval $\bigl[{-{\pi\over2}},{\pi\over2}\bigr]$ you cannot distinguish them with a naked eye.

  • 0
    Thanks, but I don't really consider 49/324000 something I can easily do with pencil and paper =).2012-09-27