A non-circular argument that uses the Maclaurin series expansions of $\sin x$ and $\cos x$ to show that $\frac{d}{dx}\sin x = \cos x$ I have often heard it said that it is circular to use the Maclaurin series expansions of sine and cosine to show that $\frac{d}{dx}\sin x = \cos x$ because the series expansions themselves use the fact that $\frac{d}{dx}\sin x = \cos x$. However, I find this claim objectionable because the expansions only require you to know the derivatives  of sine and cosine when $x=0$. I was wondering if the following argument that $\frac{d}{dx}\sin x = \cos x$ is rigorous:

*

*Define $\sin \theta$ as the $y$-coordinate when you move $\theta$ units counterclockwise around the unit circle, starting from the point $(1,0)$.

*Define $\cos \theta$ as the $x$-coordinate when you move $\theta$ units counterclockwise around the unit circle, starting from the point $(1,0)$.

*From this, it follows that $\cos 0=1$, and that $\sin 0=0$, which will be useful later on.

*Prove that $\lim_{\theta\to0} \frac{\sin \theta}{\theta}=1$ using a geometric argument, such as the one Robjohn provided in this post.

*We can use this limit to find $\frac{d}{dx}\sin x|_{x=0}$:

\begin{align}
\frac{d}{dx}\sin x|_{x=0} &= \lim_{\Delta x \to 0}\frac{\sin (0+\Delta x)-\sin 0}{\Delta x} \\
&= \lim_{\Delta x \to 0}\frac{\sin (\Delta x)}{\Delta x} \\
&= 1 \\
&= \cos 0
\end{align}

*

*Here is where I am a little unsure about my argument. We know that $\frac{d}{dx}\sin x|_{x=0}=\cos0$, but without knowing that in general that $\frac{d}{dx}\sin x = \cos x$, this might well be just a coincidence; it seems unjustified to find the derivative of $\cos x$ at $x=0$ in order to find the second derivative of $\sin x$ at $x=0$. However, I think a way to get around this is to use the second symmetric derivative. This still has the problem that the second symmetric derivative is only equal to the second derivative if the second derivative exists, and it seems hard to show that the second derivative of $\sin x $ exists when you use my approach. However, assuming that it is valid to use the second symmetric derivative, we have:

\begin{align}
\frac{d^2}{dx^2}(\sin x)|_{x=0}&=\lim_{\Delta x \to 0} \frac{\sin(0+\Delta x)-2\sin 0+\sin(-\Delta x)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{\sin(\Delta x)+\sin(-\Delta x)}{\Delta x} \\
&= 0
\end{align}

*

*And though it may be ugly, presumably we can take the third  and fourth symmetric derivatives to derive the Maclaurin series expansion of $\sin x$. If it possible to do the same for $\cos x$, then there seems to be no trouble in proving that $\frac{d}{dx}\sin x = \cos x$ just from the series expansions. This argument turned out to be a lot more cumbersome than I envisaged, but I still wonder if this kind of argument is valid, and whether there is a simpler alternative.

 A: I like to think about it as follows: using the definitions you gave for sine and cosine, it's possible (although maybe a bit ugly) to prove the 'angle addition formulas':
$$ \sin(a+b) = \sin(a)\cos(b) + \sin(b) \cos(a)$$
$$ \cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$$
using only geometric considerations. From here on, one can easily see that
$$\frac{d}{dx} \sin(x) = \lim_{h\rightarrow 0}(\frac{\sin(x+h)-\sin(x)}{h}) = \lim_{h\rightarrow 0} \big( \sin(x)\frac{\cos(h)-1}{h} + \cos(x) \frac{\sin(h)}{h} \big) $$
Now, since $\frac{\cos(h)-1}{h} = - \frac{\sin^2(h)}{h(1+\cos(h))}$, we can see that because $\lim_{h\rightarrow 0} \frac{\sin(h)}{h} = 1$ the limit on the right hand side only gets a contribution from the second term, which equals $\cos(x)$. Using the other angle addition formula, you can also prove that $\frac{d}{dx} \cos(x) = -\sin(x)$. This automatically implies that both functions are smooth, so you should be allowed to use Taylor's theorem to deduce their expansion (which turns out to converge for all $x$).
The ugly part would be the proof of the 'angle addition formulas' which need some case distinctions depending on in which quadrant you are looking. However, I think you can make some shortcuts. For example, it's kind of obvious from the definitions that $\cos(\frac{\pi}{2} - x) = \sin(x)$ and vice versa, so you only need to prove the first one. Moreover, $\sin(\pi+x) = -\sin(x)$ is also clear, so you can assume $a+b \leq \pi$.
