Given that $\int_0^b \sin x \, dx$ exists, evaluate it using the identity $$ \cos(j-1) \theta - \cos(j+1)\theta = 2\sin\theta \sin j \theta$$ to calculate certain Riemann sums.
Given a partition over $[a,b] => [0,b]$ subdivided uniformly into $n$ equal subintervals, let's evaluate the Riemann upper sum of $f$.
$$ S(P)=\sum_{j=0}^n f(d_j) (x_j -x_{j-1})$$ $$S(P)=\frac{b-0}n \sum_{j=0}^n \sin \left(0 + \frac{(b-0)j}{n} \right) $$ $$ S(P) =\frac{b}{n} \sum_{j=0}^n \sin \left(b\cdot\frac{j}{n} \right) $$
Placing in context the formula given at the top ($\theta \rightarrow b$, $j \rightarrow \frac{j}{n}$):
$$ \cos\left[\left(\frac{j}{n}-1\right) b\right] - \cos\left[\left(\frac{j}{n}+1\right)b\right] = 2\sin\left(b\right) \sin\left( \frac{j}{n} \cdot b\right)$$ $$\sin\left( \frac{j}{n} \cdot b\right) = \frac{1}{2\sin\left(b\right)} \cdot \left[ \cos\left[\left(\frac{j}{n}-1\right) b\right] - \cos\left[\left(\frac j n + 1 \right)b\right] \right] $$ Continuing with $S(P)$:
$$S(P)= \frac{b}{n} \cdot \frac{1}{2\sin\left(b\right)} \sum_{j=0}^n \left[ \cos\left[\left(\frac{j}{n}-1\right) b\right] - \cos\left[\left(\frac{j}{n}+1\right)b\right] \right]$$
I have seen other models but not exacting the same interval and exacting the same formula as the one at the top (using a similar formula (but not the same) . I am trying to apply this formula given at the top to see how it works. I remain uncertain about two elements.
.
the 1st element is the premise that the upper sum is $ S(P) =\frac{b}{n} \sum_{j=0}^n \sin \left(b\cdot\frac{j}{n} \right) $ as this does not seem to take into the values when $b =[k\pi,2kpi]$ where sines' values are negative. How can the sum be written to take this into account?
.
The 2nd element concern the last line. How to determine the telescoping rule in this case? I have seen a model on mathexchange Riemann sum of $\sin(x)$ stating that the sum is in the form $\sum_{j=1}^n (g(j)-g(j+1))= g(1) - g(n+1)$. But,first, I did not find any reference on internet confirming this; second, isn't the sum actually in the form $\sum_{j=1}^n (g(j-1)-g(j+1))$?
Any help, input is much appreciated.
I would have left this as a comment but there is not enough space. You need to take $\theta=b/n$ if you want to use those identities $$ S_n=\frac{b}{n}\sum_{j=1}^n \sin (jb/n)=\theta\sum_{j=1}^n\sin j\theta=\frac{\theta}{2\sin\theta}\sum_{j=1}^n [\cos(j-1)\theta - \cos(j+1)\theta] $$ Note that $$ \sum_{j=1}^n\cos(j-1)\theta=\sum_{j=0}^{n-1}\cos j\:\theta, \qquad \sum_{j=1}^n\cos(j+1)\theta=\sum_{j=2}^{n+1}\cos j\:\theta $$ So subtracting, the sum become $\cos 0+\cos \theta -\cos n\theta-\cos (n+1)\theta$, which is $$ 1+\cos\theta - \cos b-\cos\theta \cos b+\sin\theta\sin b $$ Taking the limit $\theta \to 0$ now, noting $\lim_{\theta\to 0}\sin\theta/\theta =1$, and $\lim_{\theta}\cos \theta=1$, you find the value of the integral to be $1-\cos b$.