Convergence and limit of $\sum_{j=1}^n\sin\left(\frac{j-1}n\right)\left\{\cos\left(\frac jn\right)-\cos\left(\frac{j-1}n\right)\right\}$ The title says it all - I'm trying to find a way of proving the convergence and evaluating the limit of $a_n=\sum_{j=1}^n\sin\left(\frac{j-1}n\right)\left\{\cos\left(\frac jn\right)-\cos\left(\frac{j-1}n\right)\right\}$. This exercise is from the field of integration theory, so probably Riemann sums are to be used on this one but I can't find any way that makes sense. I'd appreciate any hints.

We have:
$$ \lim_{n\to +\infty} \sum_{j=1}^{n}\sin\left(\frac{j-1}{n}\right)\left[\cos\left(\frac{j}{n}\right)-\cos\left(\frac{j-1}{n}\right)\right]=\color{red}{\frac{\sin(2)-2}{4}}.$$
Sketch of proof: we have:
$$ \cos\left(\frac{j}{n}\right)-\cos\left(\frac{j-1}{n}\right) = \int_{\frac{j-1}{n}}^{\frac{j}{n}}(-\sin x)\,dx=-\frac{1}{n}\,\sin\left(\frac{j-1}{n}\right)+O\left(\frac{1}{n^2}\right), $$
hence our limit is the same as:
$$ -\lim_{n\to +\infty}\sum_{j=1}^{n}\frac{1}{n}\sin^2\left(\frac{j-1}{n}\right) = -\int_{0}^{1}\sin^2(x)\,dx. $$
