How to prove: $$2\cos \alpha \sin(\alpha N)=\sin(\alpha(N-1)) \quad\text{where}\quad \alpha=\frac{\pi k}{N+1}$$ with $k$ an integer.
Some common trig identities don't work.
The problem arises from the following exercise. In particular, showing that the equality in the last component of the vector. (In this exercise, $\Delta x = 1/(N+1)$.)
Using $$2\sin p \cos q = \sin(p+q) + \sin(p-q) \tag{1}$$ with $p = \alpha N$ and $q = \alpha$ gives $$2\sin(\alpha N)\cos\alpha = \sin(\alpha N + \alpha) + \sin(\alpha N - \alpha) \tag{2}$$ But $$\alpha N + \alpha = \alpha (N+1) = \frac{\pi k}{N+1}(N+1) = \pi k \tag{3}$$ Since $k$ is an integer, $\sin \pi k = 0$; thus, the first term on the right-hand side of (2) vanishes, leaving the desired relation. $\square$