User talk:Qzekrom/Cosine


 * The Taylor series for sine is only available once you know how to differentiate sine indefinitely. To do this you need to know that the derivative of sine is cosine and the derivative of cosine is minus sine. Your "proof" that the derivative of sine is cosine already assumes that the derivative of sine is cosine. There's a very simple geometrical argument involving circles and triangles that prove that the derivative of sine is cosine and that the derivative of cosine is minus sine. Moreover, how do you know that the final series gives cosine? You need to be able to compute the Taylor series of cosine, i.e. be able to differentiate cosine indefinitely. — Fly by Night  ( talk )  01:05, 8 January 2012 (UTC)

I do know how to differentiate sine indefinitely. I noticed the factorial in the denominator and I knew that the derivative of axn is naxn - 1. So I extended it to infinity! Cheers, The Doctahedron, 23:47, 12 January 2012 (UTC)


 * An article on this topic will of course need to cover many different approaches to defining it (which will have to be done with motivation and without circularity), its properties, its applicability and uses, etc. As such, other than that the current content seems to be WP:OR rather than sourced, there is as yet little to comment on. Happy writing! — Quondum☏✎ 07:16, 25 January 2012 (UTC)