I am interested in evaluating the sum of:
$$S_{N}=\sum_{k=1}^{N}\sin^2\left({x\over 2k}\right)\tag1$$
Expanded $(1)$
$$\sin^2\left({x\over 2}\right)+\sin^2\left({x\over 4}\right)+\sin^2\left({x\over 6}\right)+\cdots+\sin^2\left({x\over 2N}\right)\tag2$$
Using $$\sin^2(x)+\cos^2(x)=1$$
$$S_{N}=N-\sum_{k=1}^{N}\cos^2\left({x\over 2k}\right)\tag3$$
Using $$\cos^2(x)-\sin^2(x)=\cos(2x)$$
$(1)$+$(3)\implies$
$$2S_{N}=N-\sum_{k=1}^{N}\cos\left({x\over k}\right)\tag4$$
Recall $${1\over 2}+\sum_{k=1}^{N}\cos(kx)={\sin(N+1/2)x\over 2\sin(x/2)}\tag5$$
$$\cos(A)-\cos(B)=2\sin\left({A+B\over 2}\right)\sin\left({A-B\over 2}\right)\tag6$$
$$\cos(kx)-\cos\left({x\over k}\right)=2\sin\left[x\left({k^2+1\over 2k}\right)\right]\sin\left[x\left({k^2-1\over 2k}\right)\right]\tag7$$
$(4)$+$(5)\implies$
$$2S_{N}+{\sin(N+1/2)x\over 2\sin(x/2)}=N+{1\over 2}+2\sum_{k=1}^{N}\sin\left[x\left({k^2+1\over 2k}\right)\right]\sin\left[x\left({k^2-1\over 2k}\right)\right]\tag8$$
How can we evaluate $(1)?$