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If $\frac{d(\sin{x})}{dx}= \cos{x}$ is proven using the limit $\lim \limits_{x \to 0}\frac{\sin{x}}{x}=1$ (as it is in most textbooks), would it be circular to then use $\frac{d(\sin{x})}{dx}= {\cos{x}}$ and L'Hospital to prove the limit $\lim \limits_{x \to 0}\frac{\sin{x}}{x}=1$ (because limit $\lim \limits_{x \to 0}\frac{\sin{x}}{x}= \frac{\cos{0}}{1})$ later in the same textbook?

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Yes indeed it would be circular. – André Nicolas Mar 6 '12 at 16:05
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@johnthemathteacher: And how are you going to establish that the Taylor series of $\sin(0+h)$ is $\sim h$ without knowing what the derivative of $\sin(x)$ is, or without computing $\lim_{h\to 0}\frac{\sin(0+h)}{h}$? – Arturo Magidin Mar 6 '12 at 16:23
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What you can do is: once you have established that $\frac{d}{dx}\sin x = \cos x$, if you are having trouble remembering what $\lim\limits_{h\to 0}\frac{\sin h}{h}$ is, then you can use L'Hopital's rule to remind you of the value, though not to prove its value ex nihilo. – Arturo Magidin Mar 6 '12 at 16:25
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@johnthemathteacher: Since the vast majority of the modern developments do not start from the power series, for the vast majority of developments using LHopital's rule would be circular; in order to justify that the LHopital's rule argument is not circular you would likely have to specify that you are defining $\sin x$ as a power series, and deriving its properties from that starting point. Otherwise, for most people, the argument will be circular. I would not be satisfied with an unspoken "and it is not necessarily circular because there is other roundabout way of proving this". – Arturo Magidin Mar 6 '12 at 17:19
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Why is it that no one has pointed out the pun involved in using "circular" to describe a problem about $\sin x$? – Gerry Myerson Mar 7 '12 at 2:33
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1 Answer

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You can establish that the derivative of $\sin x$ is $\cos x$ without the fact that $\sin x = x +o(x)$.

Let $ x$ be in radians and define $\cos x$ and $\sin x$ as the coordinates of the point $P(\cos x ,\sin x)$ in the unit circle that spans a circular sector of area $x/2$.

We can define the area of this sector in terms of $x$, as

$$A(x) = \frac{x\sqrt{1-x^2}}{2}+\int\limits_x^1 \sqrt{1-t^2}dt$$

Then we define $\cos x$ as the unique real number in $[-1,1]$ such that

$$A(\cos x) = \frac{x}{2}$$

and $\sin x = \sqrt{1-\cos^2 x}$

We can see that for $-1\leq x\leq1$, $A(x)$ is differentiable, and we get

$$A'(x) = -\frac{1}{2\sqrt{1-x^2}}$$

But since $B(\cos x) = 2A(\cos x) = x$, we have that $B$ is the inverse of $\cos x$, so we have that

$$(\cos x)' = B^{-1}(x)'$$

$$(\cos x)' = \frac{1}{B'(B^{-1}(x))}$$

$$(\cos x)' = \frac{1}{\frac{-1}{\sqrt{1-B^{-1}(x)^2}}}$$

$$(\cos x)' = -\sqrt{1-\cos^2 x}$$ $$(\cos x)' = -\sin x$$

Since $\sin x = \sqrt{1-\cos^2 x}$, we have that

$$(\sin x)' = \frac{-2 \cos x (-\sin x)}{2\sqrt{1-\cos^2 x}}=\frac{\cos x \sin x}{\sin x} = \cos x$$

So, from my point of view, it isn't.

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