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Topic: sine series
Replies: 1   Last Post: Nov 7, 2003 3:58 PM

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 F. Alexander Norman Posts: 141 Registered: 12/3/04
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 well-known, 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

Date Subject Author
11/7/03 Ryan
11/7/03 F. Alexander Norman