Someone back on this post requested proof that the derivative of cos(x) is -sin(x). I may not have passed Calc III, this time around, but Calc I is eminently doable. To avoid spaminess here, find the proof here. (Now with Maple goodness.)

# Math I Can Do

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This page contains a single entry by Erica published on *December 14, 2005 7:27 PM*.

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How do you justify lim h-> 0 [ sin(h) / h] ?

That part I lazily let Maple do, but it's also doable by L'Hopital's rule since it's the limit of one function that's 0 over another that's 0. The derivative with respect to h of sin(h) is cos(h), which is 1 for h=0; the derivative of h with respect to h is 1, so it's 1/1 = 1.

but, I mean, isn't using l'hopital a little circular? using the derivative of cos to prove the derivative of sin?

Hmm. Most of the similar proofs I've seen online use

lim h-> 0 [sin(h)/h]

as a given, and proceed to use that fact to prove the opposite - that d/dx [sin(x)] is cos(x). I suppose I don't find it any more suspect than using the cos(x+h) identity - I mean, how was that identity discovered in the first place?