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I can prove that $\lim_{x\to0}\frac{\cos(x)-1}x=0$ since

$$\sin^2(x)=1-\cos^2(x)$$

$$\implies2\sin'(x)\sin(x)=-2\cos'(x)\cos(x)$$

$$\sin'(x)\sin(x)=-\cos'(x)\cos(x)$$

at $x=0$, we have

$$0=-\cos'(0)$$

Thus, $\cos'(0)=\lim_{x\to0}\frac{\cos(x)-1}x=0$.

Can one produce the same result for the famous $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ by manipulating derivatives?

Particularly, can we calculate $\sin'(0)$ without first showing that $\sin'(x)=\cos(x)$?

Edit:

As has been shown, we need more than just trig identities to prove this, since trig identities work regardless of the radian/degrees while the limit does not. So consider the following information:

$$0\le\frac{\sin(x+t)-\sin(x)}t\le\cos(x)\ \forall\ x\in(0,\frac\pi2),\ t\in\left(0,\frac\pi2-x\right)$$

The last inequality proven geometrically in this answer.

Thus, we get

$$0\le\sin'(0)\le\cos(0)$$

As of yet, I'm unsure what other information should be required, mainly how to deal with the units issue.

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Simply Beautiful Art
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  • Do you mean $\lim_{x\to 0}\frac{\sin(x)-\sin(0)}{x}=\sin'(0)=\cos(0)=1$? – A.Γ. Dec 30 '16 at 00:21
  • @A.G. Yes, but you may not use the knowledge that $\sin'(x)=\cos(x)$, just as I have not used $\cos'(x)=-\sin(x)$. – Simply Beautiful Art Dec 30 '16 at 00:22
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    How do you know that $\sin^{\prime}=\cos$ ? Usually this is derived as a consequence of the limit in your title. – Rene Schipperus Dec 30 '16 at 00:27
  • @SimpleArt How do you propose we prove that $\sin'(0)$ is finite, then? Or at least that $\lim_{x\to0}\sin'(x)\sin(x)=0$? – Arthur Dec 30 '16 at 00:29
  • @ReneSchipperus Well, suppose I don't know that $\lim\frac{\sin x}x=1$, since it is what we are trying to find and consequently I don't know that $\sin'=\cos$. So in short, I don't know that fact. – Simply Beautiful Art Dec 30 '16 at 00:29
  • @Arthur By looking at a graph, it stands to good reason that one can assume $\sin'(x)$ is a bounded function. – Simply Beautiful Art Dec 30 '16 at 00:30
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    @SimpleArt How would you know from the graph that nothing funny happens around $x=10^{-\text{Graham's number}}$? – Arthur Dec 30 '16 at 00:33
  • Its not clear what you are taking as your starting point, you need to begin with some limits as given. Usually the limit in the title is proved by the squeeze and is the cornerstone for many further limits. – Rene Schipperus Dec 30 '16 at 00:34
  • @Arthur You could look at this answer to get an approximate understanding of $\sin'$. – Simply Beautiful Art Dec 30 '16 at 00:35
  • @ReneSchipperus I take as a starting point all derivative rules and trig identities and I want to prove this by "solving" for $\sin'(0)$. – Simply Beautiful Art Dec 30 '16 at 00:36
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    @ReneSchipperus From looking at the question, I'd say it's pretty clear he's assumed that 1) $\sin'$ exists and is bounded near $0$, and that 2) normal differentiation rules, specifically the chain rule, applies. – Arthur Dec 30 '16 at 00:36
  • @Arthur If you take a good look at that image, you will notice that for $0<x<\pi/2,0<t<\pi/2-x$, we have the following bounds:$$0\le\frac{\sin(x+t)-\sin(x)}t\le\cos(x)$$ – Simply Beautiful Art Dec 30 '16 at 00:39
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    If all you are given is that $s(x)$ and $c(x)$ are functions which satisfy $s(x)^2+c(x)^2 = 1$ for all real $x$ and $s(0) = 0$ and $c(0) = 1$, then both [$s(x) = \sin(x)$, $c(x) = \cos(x)$] and [$s(x) = -\sin(x)$, $c(x) = \cos(x)$] are solutions. So it won't be possible to decide whether $s'(0) = 1$ or $s'(0) = -1$. – JimmyK4542 Dec 30 '16 at 01:07
  • @JimmyK4542 I do not think that is a big problem, please see the comment above yours. – Simply Beautiful Art Dec 30 '16 at 01:09
  • I don't know if you can find the limit by solving $\sin'(0)$, but you could find it if you can prove the following inequality : $\sin x \leq x \leq \tan x$, $x \in (0, \frac{\pi}{2})$, which might be something you would be okay with. – Desura Dec 30 '16 at 01:10
  • @Desura That is clearly one of the normal ways of doing the problem, and I want to tackle this with derivative specifically. If you are interested in such "regular" proofs, see here. – Simply Beautiful Art Dec 30 '16 at 01:12
  • What can we assume about sin(x)? In calculus the sine-function is defined as the inverse of an integral which has been shown not to be reducables to simple powers. This is Int(1/sqrt(1-t^2),(t,0,x)) =(def)= arcsin(x). For small x we can replace the sqrt by 1 and obtain int ~= x. Taking now the sine of the approximate equation we get sin(x) ~= sin(arcsin(x))== x which proves the relation in question. – Dr. Wolfgang Hintze Dec 30 '16 at 01:12
  • @Dr.WolfgangHintze I would probably define it via the unit circle, which is where I get the inequality $0\le\frac{\sin(x+t)-\sin(x)}t\le\cos(x)$ in the comments above and all the trig identities (including the one I used in my question) – Simply Beautiful Art Dec 30 '16 at 01:13
  • @Simple Art: I understand. I deliberately wanted to take a purely arithmetic - i.e. non-geometric - approach as an alternative. Of course the arcsin is - as the word says - the arc ON the unit circle from the point (1,0) to the point (cos(t), sin(t)). – Dr. Wolfgang Hintze Dec 30 '16 at 01:20
  • @Dr.WolfgangHintze Personally though, I do not believe that to be the standard definition of $\sin$. But whatever floats your boat I guess. – Simply Beautiful Art Dec 30 '16 at 01:23
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    @SimpleArt: The point of my comment was that using only the information $\sin^2(x)+\cos^2(x) = 1$, $\sin(0) = 0$, $\cos(0) = 1$ and manipulating derivatives will not be enough to determine $\sin'(0)$. If you have other assumptions about $\sin(x)$ and $\cos(x)$, then you might be able to prove $\sin'(0) = 1$. – JimmyK4542 Dec 30 '16 at 01:25
  • @JimmyK4542 Yes, I can see that now. Thank you for your interesting contributions :D – Simply Beautiful Art Dec 30 '16 at 01:26
  • @Simple Art: At least that's how I learned it ages ago in my calculus course where a hierarchy of functions was built in this manner: step 1 all powers can be intergrated easily, except 1/x. This integral is then defined as the logarithm. It inverse is the exponential function. Next came algebraic functions like the one sqrt above etc. Of course you can define sin(x) otherwise, e.g. via the infinite product. Then your relation is trivial. – Dr. Wolfgang Hintze Dec 30 '16 at 01:34

2 Answers2

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I don't think it is possible to derive that $\sin'(0)=1$ using only "trigonometric identities", due to the fact that trigonometric identities (at least what I call trigonometric identities - see (*)) are blind with respect to the essential fact that makes $\frac{\sin(x)}{x} \to 1$, which is, informally speaking, the measurement by "radians".

What we can do using only trigonometric identities is derive the fact that $\sin'(x)=\sin'(0) \cos(x)$ and $\cos'(x)=-\sin'(0)\sin(x)$. For the first, consider the identity $$\sin(x+y)=\sin(x)\cos(y) +\sin(y)\cos(x).$$ Differentiating with respect to $y$, we get $$\sin'(x+y)=\sin(x)\cos'(y)+\sin'(y)\cos(x).$$ Evaluating at $y=0$, $$\sin'(x)=\sin(x)\cos'(0)+\sin'(0)\cos(x)$$ $$\therefore \sin'(x)=\sin'(0)\cos(x),$$ since you concluded that $\cos'(0)=0$. Analogously, using the identity for $\cos(x+y)$, one reaches the other formula.

(*) The question is quite unclear. Using only "trigonometry", we are left with a fair amount of freedom on the functions $\sin$, $\cos$ as real functions (essentially, changing $\sin(x)$ to $\sin(kx)$ for some constant $k \neq 0$ does not change trigonometry, which is what we perceive in practice as a "change of units" on the angles. And this amounts to changing $\sin'(0)$ as well, by the same factor). More explicitly with respect to the question, I consider a "trigonometric identity" to entail (not iff) that it is invariant under changing the functions $\sin(x)$, $\cos(x)$ by $\sin(k x)$, $\cos(kx)$. As such, it is thus impossible to prove that $\sin'(0)=1$ using only "trigonometric identities", because there is always a factor on the derivative which can come from the constant $k$.

Aloizio Macedo
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    Hm, you make very good points at the end. It comes to be that regardless of units, I can derive $\cos'(0)=0$, but not for $\sin'(0)$. I'll think about that if I can figure out any ways to improve the question. – Simply Beautiful Art Dec 30 '16 at 01:35
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We can easily see that the trigonometric relations

$$\sin (0)=0, \cos (0)=1, \sin ^2(x)+\cos ^2(x)=1$$

are not sufficient to prove

$$\lim_{x\to0}{\sin(x)\over x} = 1$$

Indeed, define two functions

$$\text{sen}(x)=\sin (a x), \text{ces}(x)=\cos (a x)$$

These satisfy the conditions above but

$$\lim_{x\to 0} \, \frac{\text{sen}(x)}{x}=a$$

which obviously need not be unity.

Dr. Wolfgang Hintze
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