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In an informal exam tonight, my professor asked me to demonstrate that for $f(x)=\sin(x),\, f'(x)=\cos(x)$ using the definition of the derivative, $f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$.

Well, plugging things in, we get $$\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin(x)}{h}$$ $$=\lim_{h\rightarrow0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}$$$$=\lim_{h\rightarrow0}\frac{\sin(x)\left(\cos(h)-1\right)+\cos(x)\sin(h)}{h}$$

And here I managed to stump him. In order to prove that this equals $\cos(x)$, we need to demonstrate that $\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}=0$ and that $\lim_{h\rightarrow0}\frac{\sin(h)}{h}=1$. You can't simply plug in $h=0$ because that would lead to an indeterminate form. L'Hôpital's Rule does indeed get you the required limits, but that requires the very derivative we're trying to prove.

Is there a way, using the definition of the derivative, to actually prove that the derivative of sine is cosine, without a proof by construction or a proof by contradiction?

DonielF
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  • In case anyone's wondering, I managed to satisfy my professor by using Taylor Series, glossing over the fact that those equalities can only be proven to hold using the derivatives we're trying to find... – DonielF Jan 28 '20 at 02:54
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    These two identities are often proved using a geometric argument and the squeeze theorem. – Michael Burr Jan 28 '20 at 02:55
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    The $sinh/h=1$ can be shown through inscribed and circumscribed triangles as well as an accompanied circle sector all spanned by angle $h$. This is a non calculus approach and I am sure it can be found on this site somewhere. I am glad though that you realize that both Taylor series and Hospital Rule is NOT the correct way to prove these limits – imranfat Jan 28 '20 at 02:56
  • I mean, if you don't want to use Taylor series, then what definition of $\sin(x)$ and $\cos(x)$ are you using. I personally treat those as the definition but there are other equivalent definitions – QC_QAOA Jan 28 '20 at 02:56
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    The mentioned proof using a geometric argument and the squeeze theorem is here. – dxdydz Jan 28 '20 at 02:57
  • @dxdydz Exactly, thanks! – imranfat Jan 28 '20 at 02:58
  • @Michael and others: Do you want to write that up as an answer? – DonielF Jan 28 '20 at 04:16
  • @QC_QAOA If you can show how to derive the Taylor Series for these functions without relying on prior knowledge of their derivatives, go right ahead and post that as an answer. – DonielF Jan 28 '20 at 04:17
  • @DonielF But thats just it, how do you define the function $\sin(x)$. I personally define it by the infinite series $x-x^3/6+\cdots$ but there are other equivalent ways to define it. Some other ways are $(e^{ix}-e^{-ix})/(2i)$ or the function satisfying $y''(x)=-y(x); y(0)=0;y'(0)=1$. These all give the same function but are different ways to define $\sin(x)$. – QC_QAOA Jan 28 '20 at 04:23
  • @QC_QAOA What's wrong with using the unit circle definition? – DonielF Jan 28 '20 at 04:27
  • @QC_QAOA Have you never heard of the geometric definition of the sine function? It is quite literally the first definition anyone is ever taught, before calculus is even encountered. And I am not talking about trigonometric ratios: I am talking about the proper parametrization of the unit circle centered at the origin, which uniquely defines the sine and cosine functions if done carefully. – Angel Nov 08 '21 at 20:43
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    @Angel Yes, I have heard and understand that that is a definition. My line of questioning for OP was for them to state what they where using as a definition; which (up until their final comment) they neglected to do. – QC_QAOA Nov 08 '21 at 20:54

2 Answers2

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Without a proper definition, you can't prove anything. For instance, the limit

$$\lim_{t\to0}\frac{\sin t}t$$ has no meaning as long as the function $\sin t$ has not been defined, or at least some properties given. And it is not so easy to have a definition that avoids a circular argument.

The analytical definition relies on the complex exponential, via Euler's formula $$\sin t=\Im e^{it}$$ where the exponential is the entire function $$e^z=\sum_{n=0}^\infty\dfrac{z^n}{n!}$$ or equivalently a solution of the ODE $$f'(z)=f(z).$$

This obviously implies the derivative of the sine "by definition".

A slightly more geometric approach is by analytical geometry, from the equation of the unit circle, $$x^2+y^2=1,$$ giving by differentiation, $$y+x\frac{dx}{dy}=0.$$

Now if we accept the formula for the element of arc, $$ds^2=dx^2+dy^2,$$ we have

$$s=\int_0^s ds=\int_0^y\frac{dy}{\sqrt{1-y^2}}=f(y)$$ which defines a functional relation between $s$ and $y$. And by the derivative of the inverse function, let $g(s):=f^{(-1)}(s)$, we obtain

$$g'(s)=\sqrt{1-g^2(s)}.$$

A purely geometric argument is more difficult, as it requires to prove facts about lengths/areas of triangles and circular segments, and relate the axioms of geometry to calculus.

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We define $\sin$ and $\cos$ to be functions $\mathbb{R}\rightarrow\mathbb{R}$ such that $\sin(t+2\pi)=\sin(t)$ and $\cos(t+2\pi)=\cos(t)$ for every $t\in\mathbb{R}$, $\sin(0)=0$ and $\cos(0)=1$, and $\cos(t)^2+\sin(t)^2=1$ for every $t\in\mathbb{R}$. This uniquely defines the functions in question, and it does using nothing but analytic geometry, without defining its derivatives a priori, or even continuity a priori.

The equation $$\lim_{h\to0}\frac{1-\cos(h)}{h}=0$$ can actually be proven from $$\lim_{h\to0}\frac{\sin(h)}{h}=1$$ directly, since $$\lim_{h\to0}\frac{1-\cos(h)}{h}=\lim_{h\to0}\frac{1-\cos(h)}{h}\frac{1+\cos(h)}{1+\cos(h)}=\lim_{h\to0}\frac{1-\cos(h)^2}{h[1+\cos(h)]}=\lim_{h\to0}\frac{\sin(h)^2}{h[1+\cos(h)]}=\lim_{h\to0}\frac{\sin(h)}{h}\frac{\sin(h)}{1+\cos(h)}=\lim_{h\to0}\frac{\sin(h)}{h}\lim_{h\to0}\frac{\sin(h)}{1+\cos(h)}=\lim_{h\to0}\frac{\sin(h)}{1+\cos(h)}=0.$$

As such, one merely needs to prove $$\lim_{h\to0}\frac{\sin(h)}{h}=1.$$ Here is the standard proof. For $h\in(0,\pi)$, $$\sin(h)\leq{h}\leq\frac{\sin(h)}{\cos(h)},$$ while for $h\in(-\pi,0)$, $$\frac{\sin(h)}{\cos(h)}\leq{h}\leq\sin(h).$$ For both cases, it follows that $$1\leq\frac{h}{\sin(h)}\leq\frac{1}{\cos(h)}$$ which implies $$\cos(h)\leq\frac{\sin(h)}{h}\leq1.$$ Since $$\lim_{h\to0}\cos(h)=1$$ and $$\lim_{h\to0}1=1,$$ it follows by the squeeze theorem that $$\lim_{h\to0}\frac{\sin(h)}{h}=1.$$

Angel
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