Thanks to an answer from Jack D'Aurizio, we can compute the following integral through Riemann sums:

$$\begin{align} \int_a^b\cos(x)\ dx & =\lim_{n\to +\infty}\frac{b-a}{n}\sum_{k=1}^{n}\cos\left(a+\frac{(b-a)k}{n}\right) \\ & = \lim_{n\to +\infty}\frac{(b-a)}{n}\,\cos\left(\frac{(b-a)+(a+b) n}{2 n}\right)\frac{\sin\left(\frac{b-a}{2}\right)}{\sin\left(\frac{b-a}{2 n}\right)} \\ & =\sin(b)-\sin(a) \\ \end{align}$$

And so we have

$$\int_a^b\cos(x)\ dx=\sin(b)-\sin(a)$$

without the use of the FTOC.

Now, if we include the knowledge of the FTOC, does this imply that


And thus we have effectively calculated the derivative of a trig function without that silly $\lim_{h\to0}\frac{\sin(h)}h$? Or is the implication not so direct? If so, how would we make the connection?

And just out of interest, are there any derivatives that can be calculated through Riemann sums and reversed FTOC?

  • $\begingroup$ That is an incredibly complicated way to show $\sin' x = \cos x$. It depends on identities that are more involved than the simple trigonometric addition formula and estimate for $\sin x$. Why do you call the usual way of finding $\sin' x$, which depends on $\lim_{h \rightarrow 0} (\sin h)/h$, silly? $\endgroup$ – KCd Sep 19 '16 at 13:01
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    $\begingroup$ My guess is that the fact that the telescoping sum converges to $\sin(b)-\sin(a)$ is equivalent to $\lim_{h\to 0} \frac{\sin(h)}{h} =1$. $\endgroup$ – Matthew Leingang Sep 19 '16 at 13:02
  • $\begingroup$ @KCd Just that the limit is not algaebraically doable in any easy sense. I felt this method was more algebraic. But sure, it is incredibly complicated. $\endgroup$ – Simply Beautiful Art Sep 19 '16 at 13:06
  • $\begingroup$ But like the usual method of finding $\sin' x$, this method you write about depends on the addition formula for a trig function and that is definitely not algebraic. $\endgroup$ – KCd Sep 19 '16 at 13:15

Like user Matthew Leingang commented, it is not possible to ignore the limit $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{1}$$ when you are dealing with any analytic (non-algebraic) properties of $\sin x$. The limit of Riemann sum in your question evaluates to $\sin b - \sin a$ precisely because of the limit $(1)$ (check the factor $$\frac{b - a}{n}\cdot\dfrac{1}{\sin\dfrac{b - a}{2n}}$$ in your Riemann sum).

It should be obvious that the limit formula $(1)$ can't be derived using any sort of algebraic tricks precisely because function $\sin x$ is transcendental. An analogy is that the proof of irrationality of $\sqrt{2}$ involves only basic algebra (and a little bit of number theory), but the proof of irrationality of $\pi$ is simply not possible through algebra.

Regarding computing derivatives via the complicated route of "evaluating limit of Riemann sum and then using FTC" I should say the process is similar is establishing the formula $$n^{2} - (n - 1)^{2} = 2n - 1\tag{2}$$ from the complicated formula $$1 + 3 + \cdots + (2n - 1) = n^{2}\tag{3}$$ The simpler and preferable route is to prove $(3)$ using $(2)$ and not the other way round.


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