expand RHS and equate real parts,

$\cos (2\mathrm{nx})=\sum _{k=0}^n(-1{)}^{k}C(2n,2k){\cos }^{2n-2k}(x){\sin }^{2k}(x)$

The $x$'s in question are roots of the equation,

$\cos (2\mathrm{nx})=0$

hence,

$0=\sum _{k=0}^n(-1{)}^{k}C(2n,2k){\cos }^{2n-2k}(x){\sin }^{2k}(x)$

divide by ${\cos }^{2n}(x)$ on both sides and let $u={\tan }^2(x)$,

$0=\sum _{k=0}^n(-1{)}^{k}C(2n,2k)u^k$

If $z=e^{i\theta }$ then

${\tan }^2\theta =1-4/(z^2+1{)}^2-4/(z^2+1)$

So construct a poly that $z$ satisfies (we can have the angles go through the full 360 degrees rather than 90 degrees - this just quadruples the sum you require).

If you substitute $u=1/(z^2+1)$ the resulting poly in $u$ should allow you to evaluate the equation for ${\tan }^2\theta$ above, summed over all required angles.

Hopefully this makes some sense - if not I'll be forced to work through the gory details!

${\tan }^2\theta =-1-4/(z^2+1{)}^2+4/(z^2+1)$.

Consider (for simplicity) the case when there are just two terms. We want $S$ where

$S={\tan }^{2}15+{\tan }^{2}45={\tan }^2(\pi /8)+{\tan }^2(3\pi /8)$

$4S={\tan }^2(\pi /8)+{\tan }^2(3\pi /8)+{\tan }^2(5\pi /8)+{\tan }^2(7\pi /8)+{\tan }^2(9\pi /8)+{\tan }^2(11\pi /8)+{\tan }^2(13\pi /8)+{\tan }^2(15\pi /8)$

Now if $z=e^{i\theta }=\cos \theta +i\sin \theta$ then

${\tan }^2\theta =-(z-1/z{)}^2/(z+1/z{)}^2=-1-4/(z^2+1{)}^2+4/(z^2+1)$

Furthermore, if

$z^8+1=0$ (A)

then $\theta =\pi /8$, $3\pi /8$, ..., $15\pi /8$

So actually,

$4S=\sum -1-4/(z^2+1{)}^2+4/(z^2+1)$

where the sum is over the roots of the poly above.

If we write $u=1/(z^2+1)$ then

$(1/u-1{)}^4+1=0\Rightarrow (u-1{)}^4+u^4=0$ (B)

Now each root of (B) corresponds to 2 roots of (A) (a root and the same but the opposite sign).

Therefore

$4S=2\sum -1-4.u^2+4.u$

$2S=\sum -1-4.u^2+4.u$

Where the sum is over the roots of(B), of which there are 4:

$S=-2+\sum -2.u^2+2.u$

It is at this point that the construction of the polynomial becomes significant, becausethe coefficients tell us about the sums of the roots taken 1, 2, etc. at a time. In particular, the coefficient of $u^3$ in (B) is $-4$, therefore

$\sum u=4/2$

and therefore

$S=-2+4-2\sum u^2$

Finally, $\sum u^2=$ (sum of roots taken1 at a time) ${}^2-$ 2(sum of roots taken2 at a time) $=(4/2{)}^2-2.(6/2)$

Hence $S=-2+4-2[16/4-6]=6$

Having $n$ points in the quadrant rather than just 2 should change the computations very little.

$4S=\sum _z-1-4/(z^2+1{)}^2+4/(z^2+1)$

where the sum is over the $4n$ roots of the polynomial

$z^{4n}+1=0$

Therefore

$2S=\sum _u-1-4u^2+4u$

where the sum is over the $2n$ roots of

$(u-1{)}^{2n}+u^{2n}=0$

Therefore

$2S=-(2n)+4(2n/2)-4[(2n/2{)}^2-2.2n(2n-1)/2.2]$

$S=2n^2-n$