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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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    @Gerry It did baffle me when someone wrote "Circular definition [for the sine]..." and explained how we can defined the sine in terms of the unit circle. I thought: "Why would you say it is circular?" Then I went "Ohh... **circle**"2012-03-07

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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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    What is your definition for $B(x)$ , it does not seem to be defined earlier. It suddenly appears. Also the proof becomes harder to follow after this point.2018-10-20