How do we prove that the slope of $f(x) = \cos(x)$ is $0$ at $x = 0$ without using the fact that $\frac{d}{dx} \cos(x) = -\sin(x)$?

Question

In the proof of $\frac{d}{dx} \sin(x) = \cos(x)$ using the first principle of derivatives:

$\lim_{\Delta{x} \to 0}\frac{\sin(x+\Delta{x})-\sin(x)}{\Delta{x}} = \lim_{\Delta{x} \to 0}\frac{\sin(x)\cos(\Delta{x}) + \cos(x)\sin(\Delta{x}) - \sin(x)}{\Delta{x}} = \lim_{\Delta{x} \to 0}\frac{\cos(x)\sin(\Delta{x})}{\Delta{x}} + \lim_{\Delta{x} \to 0}\frac{\sin(x)(\cos(\Delta{x}) - 1)}{\Delta{x}}$

we must show that:

$\lim_{\Delta{x} \to 0}\frac{\sin(\Delta{x})}{\Delta{x}} = 1$ (1) and
$\lim_{\Delta{x} \to 0}\frac{(\cos(\Delta{x}) - 1)}{\Delta{x}} = 0$ (2).

The proof of equation (1) can be done using the Sandwich Theorem, showing $\cos(\Delta{x}) < \frac{\sin(\Delta{x})}{\Delta{x}} < 1$. However, for the proof of equation (2), is there a more rigorous way to show that this is true other then identifying it as the slope of $f(x) = \cos(x)$ at $x=0$ and proving through graph that the slope at that point happens to be $0$?

I hope the way that I have worded my question is understandable. If not, please let me know.

Thank you!

Pretty sure one can show equation 2 with the Sandwich Theorem as well. Good Calculus books introduce equations 1 and 2 alongside each other, although the first one is far more prominent. — Torsten Schoeneberg, Aug 27 '23 at 15:47
In fact, one can reduce (2) to (1). See e.g. this quasi-duplicate: https://math.stackexchange.com/q/707469/96384 — Torsten Schoeneberg, Aug 27 '23 at 15:49
Does this answer your question? How to prove that $d \sin(x)/dx = \cos(x)$ without circular logic such as L'Hôpital's rule? — Torsten Schoeneberg, Aug 27 '23 at 15:50
cosine is even and differentiable, so the derivative must be zero, QED — Calvin Khor, Aug 27 '23 at 15:53
Hint: writing $h=\Delta x$,$$\frac{\cos h-1}{h}=\frac{-2\sin^2(h/2)}{h}=-\frac12h\left(\frac{\sin(h/2)}{h/2}\right)^2.$$ — J.G., Aug 27 '23 at 15:55
@CalvinKhor: That cosine is differentiable is what one wants to prove. — Torsten Schoeneberg, Aug 27 '23 at 16:28
Since $\cos$ is even, the slope at $x=0$ must be zero. Oops, late to the game, see that Calvin already got that. — copper.hat, Aug 27 '23 at 18:47
@copper.hat: And I already pointed out that only works if we know the slope exists. — Torsten Schoeneberg, Aug 27 '23 at 21:37

score 1 · Answer 1 · answered Aug 27 '23 at 22:51

If you are willing to accept 1. in the question, then there is a straightforward estimate.

We have $\cos x = \sqrt{1-\sin^2 x}$, and for $|x| \le {1 \over 2} $ the mean value theorem gives $|\sqrt{1-x^2} - 1 | \le K x^2$ for some constant $K$.

Hence $| \cos x -1 | \le K \sin^2 x$ for sufficiently small $x$ and so $| { \cos x -1 \over x } | \le K x ({ \sin x \over x} )^2 $ from which the desired result follows.

How do we prove that the slope of $f(x) = \cos(x)$ is $0$ at $x = 0$ without using the fact that $\frac{d}{dx} \cos(x) = -\sin(x)$?

1 Answers1