Drexel dragonThe Math ForumDonate to the Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Math Forum » Discussions » Math Topics » geometry.puzzles.independent

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
F. Alexander Norman

Posts: 141
Registered: 12/3/04
Re: sine series
Posted: Nov 7, 2003 3:58 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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

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.


Date Subject Author
Read sine series
Read Re: sine series
F. Alexander Norman

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum 1994-2015. All Rights Reserved.