Talk:Derivative of Sine Function

Why oh why?

Can somebody explain to me why this is true?:

$\displaystyle \lim_{h \to 0}\frac{\sin(h)}{h} = 1\mbox{ and }\lim_{h \to 0}\frac{\cos(h)-1}{h} = 0$

I can't really see that.

Thanks a lot! --Hannes 20:09, 16 October 2008 (UTC)

Good question. They still need to be proved as theorems separately. It will happen. --Matt Westwood 20:48, 16 October 2008 (UTC)

It comes from the Taylor Series Representations of sin and cos, see here: [1]
From this it is easy to see that $\frac{\sin x}{x}=\cos x$. Since $\cos0=1$, then the limit is 1. I think the other proof is similar. We'll get to Taylor series expansions eventually... --cynic 01:06, 17 October 2008 (UTC)

$\dfrac{\sin x}{x}=\cos x$ what?!?!? --Matt Westwood 05:16, 17 October 2008 (UTC)

Yep, that would be strange. Should it be $\displaystyle \lim_{x \to 0}\frac{\sin x}{x}=\cos x$? --Hannes 13:52, 17 October 2008 (UTC)

um, my bad. did this way too late last night. forgot to look at the denominators, which don't fit. Also, forgot to think about it at all. --cynic 20:15, 17 October 2008 (UTC)

Does anyone fancy proving $\displaystyle \lim_{h \to 0}\frac{\sin(h)}{h} = 1\mbox{ and }\lim_{h \to 0}\frac{\cos(h)-1}{h} = 0$ without using (like I did) the assumption that $D_x \left({\sin x}\right) = \cos x$ and $D_x \left({\cos x}\right) = -\sin x$? I tell a lie - I haven't done the latter result, but the former is up on Limit of Sine of X over X. --Matt Westwood 20:40, 27 January 2009 (UTC)

Alright, the first one is done, I'll get to the second one later. I suspect it follows the same pattern. --Cynic (talk) 22:04, 27 January 2009 (UTC)