My Calculus 1 teacher gave an assignment for the class that I'm having trouble answering it completely. The assignment is to prove that:
$$
\int_a^b \cos(x) dx = \sin (x)|_a^b
$$
I managed to find a video from Flammable Maths titled "A RIDICULOUS LIMIT! The Integral of $\cos(x)$ in its Riemann Sum form!", which helped me a lot. The problem is that he uses the lower limit as $0$ and upper limit as $\pi/2$. I tried my best to generalize it to my case, but I'm stuck with this:
$$
\Delta b=b-a
$$
$$
\lim_{n \to \infty} \frac{\Bigl(\frac{1}{2}\Bigr)\biggl(1+\Bigl(\frac{\cos\Delta b}{1-\cos(\frac{\Delta b}{n})}\Bigr)-\cos\Bigl(\frac{\Delta b}{n}+\Delta b\Bigr)\biggr)} {1-\cos\frac{\Delta b}{n}}
$$
I already tried applicating de l'Hopital on that, as seen on the video as the next step, but it didn't really help since it gave me a sign function according to Wolfram.
Any help on further steps is greatly appreciated.
Best Answer
The problem boils down to using the formula for a sum of cosines when the arguments are in arithmetic progression: $$\sum_{k=0}^{n-1}\cos(a+kd)=\frac{\sin(nd/2)}{\sin(d/2)}\cos\left(a+\frac{(n-1)d}{2}\right)$$ You can find a trigonometric proof on page 2 here. You may also use complex numbers to prove it.
The Riemann sum (with left endpoints) is: $$\frac{b-a}{n}\sum_{i=0}^{n-1}\cos\left(a+i\frac{b-a}{n}\right) \Leftrightarrow\\ \frac{\frac{b-a}{n}}{\sin\frac{b-a}{2n}}\sin\frac{b-a}{2}\cos\left(a+\frac{n-1}{n}\frac{b-a}{2}\right)$$ by applying the identity above and making a few simplifications. Now you just have to take limits and we can use $\lim_{t\to 0}\frac{\sin t}{t}=1$. We get $$2\sin\frac{b-a}{2}\cos\frac{a+b}{2}=\sin b-\sin a $$