# How to prove that $d \sin(x)/dx = \cos(x)$ without circular logic such as L'Hôpital's rule?

How do I prove that the derivative of $\sin$ is $\cos$ without resorting to L'Hôpital's rule (circular logic)?

This part is easy:

\begin{align*} \sin'(x) &= \lim_{\Delta x \to 0} \frac{\sin(x + \Delta x) - \sin(x)}{\Delta x} \\ \sin'(x) &= \lim_{\Delta x \to 0} \frac{\cos(x)\sin(\Delta x) + \sin(x) \cos(\Delta x) - \sin(x)}{\Delta x} \\ \sin'(x) &= \lim_{\Delta x \to 0} \left(\cos(x)\frac{\sin(\Delta x)}{\Delta x} + \sin(x)\frac{\cos(\Delta x) - 1}{\Delta x}\right) \\ \end{align*}

but where do I go from here?

• I'm tempted to suggest Taylor series... but there should be some way without using it... Commented Mar 11, 2014 at 1:20
• @2012ssohn: How do you obtain the Taylor series? Commented Mar 11, 2014 at 1:21
• @2012ssohn Yeah...
– user122283
Commented Mar 11, 2014 at 1:21
• @Mehrdad - here you go. Commented Mar 11, 2014 at 1:23
• Depends of your definition of sine. You can define the sine function as the function $\sin: \mathbb{R}\to \mathbb{R}$ by the formula $\sin(x):=\sum_{n=0}^\infty (-1)^n x^{2n+1}/(2n+1)!$. This is not circular. But depends of how do you define sine Commented Mar 11, 2014 at 1:26

Starting off where you finished and credits to this pdf:

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

Part I: prove that $\displaystyle \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} = 1$.

Assuming that $OA = 1$ (i.e. it's a unit circle), we have $\sin \theta = BC$. We then note that

[Area of triangle $AOD$] > [Area of sector $AOC$] > [Area of triangle $AOC$]

which means

$$\frac{1}{2} \frac{\sin \theta}{\cos \theta} > \frac{1}{2} \theta > \frac{1}{2} \sin \theta$$

From this we get $$\cos \theta < \frac{\sin \theta}{\theta} < 1$$

and since $\displaystyle \lim_{\theta \to 0} \cos \theta = 1$, we have that $\displaystyle \lim_{\theta \to 0} = 1$.

Part II: prove that $\displaystyle \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = 0$.

$$\lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = \lim_{\Delta x \to 0} \frac{(1 - \cos \Delta x) (1 + \cos \Delta x)}{(1 + \cos \Delta x)\Delta x} = \lim_{\Delta x \to 0} \frac{1 - \cos^2 \Delta x}{(1 + \cos \Delta x)\Delta x}$$

$$= \lim_{\Delta x \to 0} \frac{\sin^2 \Delta x}{(1 + \cos \Delta x)\Delta x} = \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} \frac{\sin \Delta x}{1 + \cos \Delta x}$$

We have proven in part I that $\displaystyle \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} = 1$, and we note that $\displaystyle \frac{\sin \Delta x}{1 + \cos \Delta x} = \frac{0}{1+1} = 0$.

We finally put all this together to get that $$\boxed{\displaystyle \sin'(x) = \cos x \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} - \sin x \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = \cos x \cdot 1 - \sin x \cdot 0 = \cos x}$$

• I've fixed a couple of typos; feel free to revert. Commented Mar 11, 2014 at 1:54
• Thanks @ShreevatsaR! I got lost here and there 'cause it was such a long response x_x Commented Mar 11, 2014 at 2:00

Since you are unsatisfied and do not understand the proof, I will further explain. As we learned in precalc, $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim_{x\rightarrow 0} \frac{1- \cos x}{x} = 0$. Please search a proof of these identities in another question if still confused, as that is a much simpler problem that can be solved using the squeeze theorum. Make these substitutions and you will have your answer.

• -1 for not understanding the question. Commented Mar 11, 2014 at 1:27
• Is this not what 2012ssohn posted? Exact same thinking. Commented Mar 11, 2014 at 1:39
• "Exact same thinking"? Where in the site you linked to does it mention the squeeze theorem that 2012ssohn posted? Commented Mar 11, 2014 at 1:46
• I apologize for failing to mention the squeeze theorum; I made the incorrect assumption that you had already learned about the squeeze theorum and limits of the two functions mentioned above. By the time I was taking derivatives (calc), I had already learned those two identities in trig/precalc. My fault for not recognizing that you may have learned the information in a different order, or your school may have taught it in a different way. Commented Mar 11, 2014 at 1:52

In my introduction to limits, we used a lemma

$$\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1.$$

This was proved by drawing a picture and using the definition of the sine function and the definition of radians, along with the geometric "axiom" "the shortest distance between two points is a straight line segment".

That lemma can then be used to calculate $\lim_{x\rightarrow 0} \frac{1-\cos x}{x}$ and $\sin'(x)$.

• Pictures aren't proofs... Commented Mar 11, 2014 at 1:38
• No, they are aids to proofs. Commented Mar 11, 2014 at 1:40
• What I'm saying is that all you did was to suggest I draw a picture, which is completely useless since the whole difficulty of the question relied on proving that rigorously. Commented Mar 11, 2014 at 1:41
• @Mehrdad On the contrary, some of the most elegant proofs are pictures! Commented Mar 11, 2014 at 1:42
• I think the next steps depend a lot on exactly how you define a radian. Commented Mar 11, 2014 at 1:58