Mathematics Asked on November 24, 2021
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:
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}
begin{align}
frac{d^2}{dx^2}(sin x)|_{x=0}&=lim_{Delta x to 0} frac{sin(0+Delta x)-2sin 0+sin(-Delta x)}{Delta x} \
&= lim_{Delta x to 0} frac{sin(Delta x)+sin(-Delta x)}{Delta x} \
&= 0
end{align}
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 $$frac{d}{dx} sin(x) = lim_{hrightarrow 0}(frac{sin(x+h)-sin(x)}{h}) = lim_{hrightarrow 0} big( sin(x)frac{cos(h)-1}{h} + cos(x) frac{sin(h)}{h} big) $$ Now, since $frac{cos(h)-1}{h} = - frac{sin^2(h)}{h(1+cos(h))}$, we can see that because $lim_{hrightarrow 0} frac{sin(h)}{h} = 1$ 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 $a+b leq pi$.
Answered by Thomas Bakx on November 24, 2021
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