Prove $[\sin x]' = \cos x$ without using $\lim\limits_{x\to 0}\frac{\sin x}{x} = 1$

Question

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 $[\sin x]' = \cos x$.

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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.

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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.

Answer 1

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

Answer 2

[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.

Answer 3

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;

Answer 4

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).

Answer 5

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 $$\phi:=\sup_PL_P\leq\tan\phi\ .$$

Answer 6

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

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 $x$.

Similarly,

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

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.