This is a sketch not a proof, but if it's not enough let me know:

The objective is to construct a poly whose roots are

cot²(pk/(2m+1))

If the poly is monic, then the second highest coefficient will be (to within a minus sign) the sum of the roots, i.e. your expression.

How do we construct the polynomial? Consider (for odd p)

(cosf+isinf)^p=(cospf+isinpf)

=ápower series in products of , ñ

Take the imaginary part (I'll leave you to figure out the coefficients a, b, ...)

sinpf = asin^p f+bsin^p-2fcos²f+...

Now take out a factor of sin^pf from the RHS (I'll leave you to figure out the m, n...)

sinpf = sin^p f(mcot^p-1f+ncot^p-3f+...)

Now we know (from the LHS) the p¢ for which the RHS is zero. So letting z=cot²f the RHS is a poly in z, the roots of which are cot² p¢.

So the sum of the roots is -n/m. You now need to figure out what to choose for p, f and what m, n are.

Let me know if this is not enough.

I don't know whether you managed to solve the problem, so let me go through a particular example (m=2) using the approach outlined above.

Consider

(cos(theta)+isin(q))⁵=cos(5q)+isin(5q)

=1(cos(q))⁵(isin(q))⁰+5(cos(q))⁴(isin( q))¹+10(cos(q))³(isin(q))²+10(cos(q))² (isin(q))³+5(cos(q))¹(isin(q))⁴+1 (cos(q))⁰(isin(q))⁵

Take imaginary part:

sin(5q)=5(cos(q))⁴(sin(q))¹-10(cos(theta)) ²(sin(q))³+1(cos(q))⁰(sin(q))⁵

Taking out sin⁵(q) from the RHS:

sin(5q)=sin⁵(q)[5cot⁴(q)-10cot²(q) +1]

The LHS is zero when q = 0, p/5, 2p/5, 3p/5, 4p/5, p, etc. For the RHS, sin(q) is zero when q = 0, p etc. Therefore

5cot⁴(q)-10cot²(q)+1

is zero when q = p/5, 2p/5, 3p/5, 4p/5, 6p/5 etc.

Write z=cot²(q), the equation

5x²-10z+1=0 has distinct roots

z=cot²(p/5), cot²(2p/5)

But the sum of the roots of this quadratic are10/5=2, hence cot²(p/5)+cot²(2p/5)=2

I'll leave you to generalise it for any m, if you haven't already done so.