Re: [PD] arccos / inverse cos in pd ?

Jerome Etienne

10 Dec 2003 10 Dec '03

10:51 p.m.

thanks :)

i attached the patch just in case somebody find it usefull.

On Wed, Dec 10, 2003 at 04:49:42PM -0500, Mathieu Bouchard wrote:

arccos(x) = arctan(tan(arccos(x))) = arctan(sqrt(1-x^2)/x) and pd has [atan].

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