We have the following identities:
$$ \beta_\nu = b_\nu -\frac{1}{2}(b_{\nu-1}+b_{\nu+1}),\ \ \ \ (\nu=2,3,4,\ldots)\\ \beta_1=b_1-1/2 b_2 $$
$$s_n(x)=\sum_{\nu=1}^n b_\nu \sin(\nu x) \\ \sigma_\nu(x)= \sum_{\nu=1}^n \beta_\nu \sin(\nu x)$$
It's argued that we have the following equality:
$$(1-\cos(x))s_n(x)= \sigma_n(x)-1/2 b_n \sin((n+1)x)+1/2 b_{n+1}\sin nx $$
If I expand both expressions I get:
$$ (1-\cos x) s_n(x) = \sum_{\nu=1}^n b_\nu \sin \nu x - \sum_{\nu=1}^n b_\nu \cos x \sin \nu x$$
and $$\sigma_n(x) = \sum_{\nu=1}^n \beta_\nu \sin \nu x = (b_1-1/2b_2)\sin x +\sum_{\nu=2}^n \bigg( b_\nu -1/2 (b_{\nu-1}+b_{\nu+1}) \bigg)\sin \nu x =\\ \sum_{\nu=1}^n b_\nu \sin \nu x -1/2 b_2 \sin x - 1/2 \sum_{\nu =1}^{n-1} b_\nu\sin((\nu+1)x) - 1/2 \sum_{\nu=1}^{n-1}b_{\nu+2}\sin (\nu+1)x$$
I don't see how to get the above equality, is it with the use of identities of exponanetials? or something else?
Thanks in advance, btw here's a scan of the book: http://web.student.chalmers.se/~robiand/home/files/0.resources/Hilbert-Methods_of_mathematical_physics.pdf
It is a bit of sum manipulations and trigonometric formula. You were on the good track, let's start where you ended:
$$\sigma_n(x) = \sum_{\nu=1}^n b_\nu \sin \nu x -1/2 b_2 \sin x - 1/2 \sum_{\nu =1}^{n-1} b_\nu\sin((\nu+1)x) - 1/2 \sum_{\nu=1}^{n-1}b_{\nu+2}\sin (\nu+1)x$$
Moving $b_2 \sin x $ in the third sum terms:
$$\sigma_n(x) = \sum_{\nu=1}^n b_\nu \sin \nu x - 1/2 \sum_{\nu =1}^{n-1} b_\nu\sin((\nu+1)x) - 1/2 \sum_{\nu=0}^{n-1}b_{\nu+2}\sin (\nu+1)x$$ Adding $-1/2 b_n \sin((\nu+1)x$ on each side: $$\sigma_n(x) -1/2 b_n \sin((\nu+1)x) = \sum_{\nu=1}^n b_\nu \sin \nu x - 1/2 \sum_{\nu =1}^{n} b_\nu\sin((\nu+1)x) - 1/2 \sum_{\nu=1}^{n}b_{\nu+1}\sin \nu x$$
Adding $1/2 b_{n+1}\sin \nu x$ on each side:
$$\sigma_n(x) -1/2 b_n \sin((\nu+1)x)+1/2 b_{n+1}\sin nx = \sum_{\nu=1}^n b_\nu \sin \nu x - 1/2 \sum_{\nu =1}^{n} b_\nu\sin((\nu+1)x) - 1/2 \sum_{\nu=1}^{n-1}b_{\nu+1}\sin \nu x$$
Let define $A = \sigma_n(x) -1/2 b_n \sin((\nu+1)x)+1/2 b_{n+1}\sin \nu x$. As $\sin 0 = 0$ we can add a term corresponding to $\nu = 0$ in the third sum.
$$A = \sum_{\nu=1}^n b_\nu \sin \nu x - 1/2 \sum_{\nu =1}^{n} b_\nu\sin((\nu+1)x) - 1/2 \sum_{\nu=0}^{n-1}b_{\nu+1}\sin \nu x$$
$$A = \sum_{\nu=1}^n b_\nu \sin \nu x - 1/2 \sum_{\nu =1}^{n} b_\nu\sin((\nu+1)x) +b_{\nu}\sin (\nu-1)x$$
then we use the trigonometric formula $\sin (a+b) + \sin (a-b) = 2 \sin a\cos b $
$$A = \sum_{\nu=1}^n b_\nu \sin \nu x -\sum_{\nu =1}^{n} b_\nu\sin(\nu x)\cos(x) $$
$$A = (1-\cos x)\sum_{\nu=1}^n b_\nu \sin \nu x =(1-\cos x)s_n(x)$$ I hope this is clear.