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Re: sine series
Posted:
Nov 7, 2003 3:58 PM


On Fri, 7 Nov 2003, Cameron Purchase wrote: ... >... I can calculate > the sine of any angle by the well known series: > > X  X^3/3! + X^5/5!...... > > but how would I go about doing this in reverse so to speak? That is > to say if I had the ratios of the right angled triangle and no > recourse to a calculator with trigometrical functions or tables how > would I work out the angle? How would I convert the .5 side of the > triangle to the .523 of its radian measure? > ...
You could use the less wellknown, but analogous, series for the inverse sine
arcsin(x) = x + s^3/6 + 3x^5/40 + ...
just using the first three terms we get
arcsin(.5) = .523177...
which is pretty close to the actual value .523598...
The series above for arcsin came from the same place that you got the series for sine  the Taylor series expansion, which you can find in any calculus book.
Sandy



