I came across this question: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

From the comments, Joren said:

L'Hopital Rule is easiest: $\displaystyle\lim_{x\to 0}\sin x = 0$ and $\displaystyle\lim_{x\to 0} = 0$, so $\displaystyle\lim_{x\to 0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{\cos x}{1} = 1$.

Which Ilya readly answered:

I'm extremely curious how will you prove then that $[\sin x]' = \cos x$

My question: is there a way of proving that $[\sin x]' = \cos x$ without using the limit $\displaystyle\lim_{x\to 0}\frac{\sin x}{x} = 1$. Also, without using anything else $E$ such that, the proof of $E$ uses the limit or $[\sin x]' = \cos x$.

All I want is to be able to use L'Hopital in $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$. And for this, $[\sin x]'$ has to be evaluated first.

Alright... the definition that some requested.

Def of sine and cosine: Have a unit circumference in the center of cartesian coordinates. Take a dot that belongs to the circumference. Your dot is $(x, y)$. It relates to the angle this way: $(\cos\theta, \sin\theta)$, such that if $\theta = 0$ then your dot is $(1, 0)$.

Basically, its a geometrical one. Feel free to use trigonometric identities as you want. They are all provable from geometry.

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    $\begingroup$ Define sine and cosine first. $\endgroup$ – user296602 Jul 27 '16 at 21:21
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    $\begingroup$ @T.Bongers What? Er... Well.. The usual sine and cosine. Definition from the trigonometric cycle (the unit circumference). =). $\endgroup$ – Physicist137 Jul 27 '16 at 21:22
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    $\begingroup$ The usual backup is the geometric proof. $\endgroup$ – user137731 Jul 27 '16 at 21:23
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    $\begingroup$ One usually untangles this in the other end. Use the limit to find the derivative, find the limit without l'Hopital. $\endgroup$ – Arthur Jul 27 '16 at 21:31
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    $\begingroup$ The derivative is a limit by definition. You can't prove something about the derivative without using limits. Geometric proof still relies on the limit in the title of the OP $\endgroup$ – Yuriy S Jul 27 '16 at 21:50

[This answer has significant overlap with my answer to a related but different question, Intuitive understanding of the derivatives of $\sin x$ and $\cos x$. ]

Disclaimer: I am not sure how to make the following totally rigorous, but it uses the unit circle definition and avoids using the limit in question, while seeming pretty convincing, hopefully enough to be worthy of an answer.

The curve $\gamma(t)=(\cos(t),\sin(t))$ has everywhere unit length and unit speed, as follows readily from the definition. $\|\gamma(t)\|^2=\cos^2(t)+\sin^2(t)=1$ everywhere by definition, because $(\cos(t),\sin(t))$ is on the unit circle. $\|\gamma'(t)\|=1$ everywhere because it is rate of change of the distance traveled around the circle with respect to $t$, and $t$ is the distance traveled by definition of radian (and we are using radians here, or else the result would be false).

From $\|\gamma(t)\|^2=\gamma(t)\cdot \gamma(t)\equiv 1$ we conclude from the product rule that $\gamma'(t)\cdot \gamma(t)\equiv 0$, so that $\gamma'(t)$ is a unit vector perpendicular to $(\cos(t),\sin(t))$. There are only two such vectors in the plane, namely $(-\sin(t),\cos(t))$ and $(\sin(t),-\cos(t))$, but it is easy to rule out the latter, for instance by noticing where $\sin$ is increasing and decreasing. Hence $\gamma'(t)=(-\sin(t),\cos(t))$ for all $t$, and in particular the derivative of $\sin$ is $\cos$.

Remark: As far as I know the most significant gap in the above is the rigorous justification that $\gamma$ is even differentiable (which is equivalent to knowing that $\cos$ and $\sin$ are differentiable).

Added: It can be shown that $\gamma$ is differentiable without further assumptions about trig functions. There is probably a nicer way to do so than what follows, but the following is something. Note that we are dealing with the unit speed parametrization of the curve $x^2+y^2=1$. Near $(1,0)$ for example, we have $x=\sqrt{1-y^2}$, so we can parametrize the curve smoothly as $\alpha(y) = (\sqrt{1-y^2}, y)$. Note that $\alpha$ is smooth on $-1<y<1$ and is tracing the right half of the circle. We can then reparametrize with respect to arclength as outlined here. The arclength along the circle from $(1,0)$ to $\alpha(y)$ with $y>0$ is the radian measure $\theta$ of the angle (by definition of radian), so $\theta(y)=\int_0^y\|\alpha'(t)\|\,dt=\int_0^y\frac{1}{\sqrt{1-t^2}}\,dt$ (which is $\arcsin(y)$). Note that $\theta(y)$ is smooth, hence so is the inverse function $y(\theta)=\sin(\theta)$ by the inverse function theorem.

More generally, I think something along these lines, with the implicit and inverse function theorems, can be used to show that algebraic curves in the plane have smooth unit speed parametrizations away from singular points, but my attempts to search for a better statement and reference for such claims have been unfruitful.

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    $\begingroup$ Seems nice, however "has everywhere unit length and unit speed" - I don't understand this part $\endgroup$ – Yuriy S Jul 27 '16 at 22:01
  • $\begingroup$ @You'reInMyEye: For a point $(a,b)$ in the plane being thought of as a vector, its length is $\|(a,b)\|=\sqrt{a^2+b^2}$, the distance from the origin to the point. "Unit length" means the length is one, and this is true for points on the unit circle by definition of unit circle. For a curve, which is a function from (an interval in) $\mathbb R$ to the plane, the speed is the magnitude of the derivative, which measures the rate of change of the distance. In this case it is always speed $1$, because the variable $t$, the radian measure, is the distance. "Unit speed" means $\|\gamma'(t)\|=1$. $\endgroup$ – Jonas Meyer Jul 27 '16 at 22:04
  • $\begingroup$ Doesn't the product rule use the definition of a derivative which in itself employs a limit? $\endgroup$ – GeorgSaliba Jul 27 '16 at 22:12
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    $\begingroup$ @GeorgSaliba: The only problem with that, remarked at the end of the answer, is assuming that $\gamma'$ exists. If it exists, then the product rule holds as always. But I do not have proof that $\gamma'$ exists, hence I have a gap. Does this raise more questions than it answers? $\endgroup$ – Jonas Meyer Jul 27 '16 at 22:14
  • $\begingroup$ I guess that as @You'reInMyEye said, there might be a problem in the question. $\endgroup$ – GeorgSaliba Jul 27 '16 at 22:16

There is a way, based on the definition of circular functions via power series (not for high school!):

One shows the power series $\;\displaystyle\sum_n(-1)^n\frac{x^{2n+1}}{(2n+1)!}\;$ and $\;\displaystyle\sum_n(-1)^n\frac{x^{2n}}{(2n)!}\;$ are convergent for all $x$. Their sums are denoted $\sin x$ and $\cos x$ respectively.

As all sums of power series, these sums are differentiable, and their derivatives are the sums of their series ‘derived term by term’, i.e.

  • $\sin' x=\displaystyle\sum_{n\ge 0}(-1)^n\frac{(2n+1)x^{2n}}{(2n+1)!}=\sum_{n\ge 0}(-1)^n\frac{x^{2n}}{(2n)!}=\cos x$.
  • $\cos' x=\displaystyle\sum_{n\ge 0}(-1)^n\frac{2n\,x^{2n-1}}{(2n)!}=\sum_{n\ge 1}(-1)^n\frac{x^{2n-1}}{(2n-1)!}=\sum_{n\ge 0}(-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!}=-\sin x.$

Note: Actually one defines this the sine and cosine of any complex number, and one can show they're holomorphic functions;

  • $\begingroup$ My high school definitely covered series expansions, through the AP Calculus BC program $\endgroup$ – Brevan Ellefsen Jul 27 '16 at 22:38
  • $\begingroup$ @Brevan Ellefsen: including derivability, radius of convergence and so on? You're ready for complex analysis… $\endgroup$ – Bernard Jul 27 '16 at 22:47
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    $\begingroup$ indeed these were all covered. It was of course given through basic examples such as these and some conditions were not given formally. However, I would fully expect any BC Calculus student to follow this proof $\endgroup$ – Brevan Ellefsen Jul 27 '16 at 22:51
  • $\begingroup$ One more typo in my collection! Fixed. Thanks for pointing it! $\endgroup$ – Bernard Jul 27 '16 at 22:57
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    $\begingroup$ I'm not using Taylor's expansion. Re-read what I wrote: the rigourous definition of sine and cosine is through power series. Of course it is heavily inspired by what is obtained from the naive definition of these functions. But it only requires results on power series. $\endgroup$ – Bernard Jul 27 '16 at 23:21

$$\frac{\sin(x+h)-\sin x}h=\frac{\sin x\cos h+\cos x\sin h-\sin x}h=\sin x\frac{\cos h-1}h+\cos x\frac{\sin h}h.$$

If the limits for $h\to 0$ exist, we have

$$\sin'x=a\sin x+b\cos x$$ for two constants $a, b$ and for all $x$.


$$\cos'x=a\cos x-b\sin x.$$

Then, deriving $\cos^2x+\sin^2x$,

$$\cos x\cos'x+\sin x\sin'x=a\cos^2x-b\sin x\cos x+a\sin^2x+b\sin x\cos x=0,$$ so that $a=0$.

Seems harder to prove that $b=1$, as this depends on the angular unit, which is not expressed in the above relations.

  • $\begingroup$ I haven't understood..: How so $b$ depends on angular unit? $\endgroup$ – Physicist137 Jul 28 '16 at 17:26
  • $\begingroup$ In the equations that I used, it is said nowhere that $x$ is in radians, hence $b$ cannot be determined. $\endgroup$ – Yves Daoust Jul 28 '16 at 17:44
  • $\begingroup$ But why $x$ has to be radians so $b=1$? Have $x$ in degrees. Then $h$ must be in degrees since $x+h$ is in the first equation. We know: $b = \lim\frac{\sin h}{h}$. If we convert to radians by factor $g$ we get the same result $\lim\frac{\sin hg}{hg} = \lim\frac{\sin u}{u} = \lim\frac{\sin h}{h} = b$. Thus it shouldn't depend on angular unit. $\endgroup$ – Physicist137 Jul 28 '16 at 18:13
  • $\begingroup$ When $h$ is in degrees, $b=0.01745329252\cdots$. Try it with a calculator. $\endgroup$ – Yves Daoust Jul 28 '16 at 18:54
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    $\begingroup$ @Physicist137: maybe, but the numerator is not in degrees. $\endgroup$ – Yves Daoust Jul 28 '16 at 19:26

The geometric definition of circular functions which you have mentioned in your question is the most intuitive/popular/accessible definition. However rarely do textbooks try to justify this definition in a rigorous manner. The justification of this definition is based on two notions (and any one of them can be used):

  1. Any arc of a circle has a well defined length. This follows by a rigorous definition of arc-length and for the case of a circle the existence of length of an arc follows by the monotone nature of $f(x) = \sqrt{1 - x^{2}}$ in $[0, 1]$.
  2. Any sector of a circle has a well defined area. This follows by a rigorous definition of area of plane regions and for the case of a circle the existence of area of a sector follows from the continuity of $f(x) = \sqrt{1 - x^{2}}$ in $[0, 1]$.

As I said any of the two justifications can be used for defining circular functions and it is possible to establish that definition based on these concepts of area and arc-length are equivalent.

Having chosen one of these justifications we now have two competing routes to take in order to find derivatives of circular functions:

  1. Show that $\sin x < x < \tan x$ for $x \in (0, \pi/2)$ and then deduce $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ This is the traditional route and combined with addition formulas for circular functions this easily gives the derivatives of circular functions.
  2. Use the justification of length of an arc of a circle (or the area of a sector of a circle) in a concrete manner (via the use of integrals) and arrive at equations like $$\theta = \int_{\cos \theta}^{1}\frac{dx}{\sqrt{1 - x^{2}}} = \int_{0}^{\sin \theta}\frac{dx}{\sqrt{1 - x^{2}}}\tag{1}$$ for $\theta \in [0, \pi/2]$ (the equation above is based on the notion of length of arc of a circle). By using inverse function theorem and rule for derivatives of inverse functions it directly gives us $(\sin \theta)' = \cos \theta$ for $\theta \in [0, \pi/2]$.

I prefer the conventional route based on $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$ because it is simpler to present. The second approach mentioned above essentially defines the $\arcsin$ function as an integral and due to the improper integral involved the approach is technically somewhat inconvenient. If we wish to go via this route then it is much simpler instead to use the definition $$\arctan x = \int_{0}^{x}\frac{dt}{1 + t^{2}}\tag{2}$$ which can be easily deduced from the geometric definition of $\sin \theta, \cos \theta$ (BTW deriving addition formulas is somewhat difficult when we use the approach based on integrals).

  • $\begingroup$ Looks like downvotes without comment have become too frequent and fashionable. Would the downvoter care to comment? $\endgroup$ – Paramanand Singh Jul 30 '16 at 15:22

What is required here is not a proof of $\sin'=\cos$ without using $$\lim_{\phi\to0}{\sin\phi\over\phi}=1\tag{1}\ ,$$ but a proof of the basic limit $(1)$ using the "geometric definition" of sine provided by the OP. To this end we shall prove "geometrically" that $$\sin\phi<\phi\leq\tan\phi\qquad\left(0<\phi<{\pi\over2}\right)\ .\tag{2}$$ The inequalities $(2)$ imply $$\cos\phi\leq{\sin\phi\over\phi}<1\qquad\left(0<\phi<{\pi\over2}\right)\ ,$$ so that $(1)$ follows from $\lim_{\phi\to0}\cos\phi=1$ and the squeeze theorem.

Comparing segment length to arc length immediately shows that $$\sin\phi=2\sin{\phi\over2}\cos{\phi\over2}\leq2\sin{\phi\over2}<\phi\ .$$ In order to prove that $\phi\leq\tan\phi$ we somehow have to use how the length of curves is defined. I'm referring to the following figure. If $0\leq\alpha<\beta<{\pi\over2}$ then $$s'=\tan\beta-\tan\alpha={\sin(\beta-\alpha)\over\cos\beta\cos\alpha}=2\sin{\beta-\alpha\over2}\>{\cos{\beta-\alpha\over2}\over\cos\beta}\>{1\over\cos\alpha}>2\sin{\beta-\alpha\over2}=s\ .$$ It follows that the length $L_P$ of any polygonal approximation $P$ to the circular arc $AB$ is $\ <\tan\phi$, and this implies $$\phi:=\sup_PL_P\leq\tan\phi\ .$$

enter image description here


I'm not sure if this is how you want the proof, correct me if I'm wrong: $$\frac{\sin(x+h)-\sin(x)}{h}=\frac{\sin(x)\cos(h)+\sin(h)\cos(x)-\sin(x)}{h}$$

Geometrically speaking, if the angle $h$ is small, then $\cos(h)$ is very nearly equal to the radius $1$, while $\sin(h)$ can be approximated by the arc of length $\text{radius}\times\text{angle}$ (two parallel sides of a triangle) which is $1\times h$, so: $$\frac{\sin(x+h)-\sin(x)}{h}=\frac{\sin(x)+h\cos(x)-\sin(x)}{h}=\cos(x)$$

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    $\begingroup$ GeorgSaliba, you are still using the limit in disguise, when you say "$\sin(h)$ can be approximated by the arc of length radius $\times$ angle" $\endgroup$ – Yuriy S Jul 27 '16 at 21:45
  • $\begingroup$ @You'reInMyEye but it's a limit from a geometrical perspective, not analytical... $\endgroup$ – GeorgSaliba Jul 27 '16 at 21:46
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    $\begingroup$ There is no such thing as a limit from a geometric perspective. I think this question can't really be answered. It doesn't make sense to talk rigorously about a derivative, but not use limits. Derivatives are limits by definition. So your attempt is good, it's just the question is not very good $\endgroup$ – Yuriy S Jul 27 '16 at 21:50
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    $\begingroup$ @You'reInMyEye What I meant was a geometric "visualization" of a limit. I was about to post a comment asking to "define" the derivative! Because I couldn't come up with another way of expressing the rate of change! $\endgroup$ – GeorgSaliba Jul 27 '16 at 21:51

There is a very nice Khan Academy video on it. I won't get into too much detail but basically you construct the following enter image description here

Then you take limits of the inequality $$1\geq|\frac{\sin\theta}{\theta}|\geq|\cos\theta|$$

to get $$\lim_{\theta->0}1\geq|\frac{\sin\theta}{\theta}|\geq|\cos\theta|$$ $$1\geq|\frac{\sin\theta}{\theta}|\geq|1$$

and and therefore by the squeeze theorem we have $\lim_{\theta->0}\frac{\sin\theta}{\theta}=1$.

Having proved that for yourself you can then use the definition of a derivative that is $$f'(x)=\lim_{h->0}\frac{f(x+h)-f(x)}{h}$$ for $f(x)$ beeing $\cos(x)$. Hope that helps.


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