[PD] arccos / inverse cos in pd ?
Mathieu Bouchard
matju at sympatico.ca
Wed Dec 10 22:49:42 CET 2003
On Wed, 10 Dec 2003, Jerome Etienne wrote:
> im looking for a way to compute a arccos in pd ? i fail to find an
> object directly computing it, did i miss it ? is there a way to
> compute it in a more indirect way (e.g. with a "limited developpment"
> really free translation of the french "developpement limite") ?
sin(t)^2 + cos(t)^2 = 1
so
sin(arccos(x)) = sqrt(1-x^2)
so
tan(arccos(x)) = sin(arccos(x))/cos(arccos(x)) = sqrt(1-x^2)/x
so
arccos(x) = arctan(tan(arccos(x))) = arctan(sqrt(1-x^2)/x)
and pd has [atan].
Using this trigonometric identity is easier and faster than computing
partial taylor series.
________________________________________________________________
Mathieu Bouchard http://artengine.ca/matju
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