DevilSolution Posted November 12, 2013 Posted November 12, 2013 (edited) I appolagise for my ignorance, my original post was incorrect, its not sinx * o/h, its (h * sinx) OR (o / sinx) depending on the side you require, the sine is o/h. Because the number will always be below 1 relative to the graph i drew its obvious that by dividing hypotenuse by it will give a number bigger than H and H is always the longest so o/sinx = h, h*sinx=o. I have done some research into how to compute this and have found the Taylor series seems to relate, although while trying to sieve through the wiki page it looks quite complex. Here's a link if anyone can verify it to be of some computational use in finding the sin, cos, tan. http://en.wikipedia.org/wiki/Taylor_series I would certainly like to wrap my head around this so any pointers would be greatly appreciated. Regards. Edited November 12, 2013 by DevilSolution
studiot Posted November 13, 2013 Posted November 13, 2013 (edited) For programming, you need Clenshaw's recurrence method, on the trigonometric recurrence formulae: cos(nt) = 2cos(t)cos({n-1}t) - cos({n-2}t) sin(nt) = 2cos(t)sin({n-1}t) - sin({n-2}t) I have already supplied the refernce, but here is another one Fike CT 1968 Computer Evaluation of mathematical functions. incidentally my first reference explains why the series for sin(t) [math]\sin (t) = \sum {\frac{{{{\left( { - 1} \right)}^k}{t^{\left( {2k + 1} \right)}}}}{{\left( {2k + 1} \right)!}}} [/math] is unsuitable for computer programming. This series does not begin to converge until k>> |t| So for large t ( which a computer can be expected to respond to ) this can mean many, many terms. Edited November 13, 2013 by studiot
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