Prove that the derivative of sine is cosine

Question

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?

Answer 1

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 $$\lim_{h\to0}\frac{\sin(h)}{h}=1.$$

Answer 2

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.