I have the equation, $$\tag{1} K_{11}\sum_i \cos^4\alpha_i+K_{12}\sum_i 2\cos^3\alpha_i\sin\alpha_i+K_{22}\sum_i\sin^2\alpha_i\cos^2\alpha_i=\sum_i\rho_i\cos^2\alpha_i$$
In the paper it states, "For equal intervals in 180 degree segments the sums of odd powers are zero; therefore,"
$$\tag{2} K_{11}\sum_i \cos^4\alpha_i+K_{22}\sum_i\sin^2\alpha_i\cos^2\alpha_i=\sum_i\rho_i\cos^2\alpha_i$$
The way I interpret what they mean is that, if I sum the values of $\cos^3\alpha_i$ for 0 degree and 180 degree I get a value of 0. Similarly, if I sum the values of $\cos^3\alpha_i$ for 1 degrees and 181 degree I get a value of 0. And so on and so forth all the way up to 179 degree and 359 degree, the sum of the values of $\cos^3\alpha_i$ will be 0. Therefore, we can cancel out the $K_{12}\sum_i 2\cos^3\alpha_i\sin\alpha_i$ term of Eqn(1). However, if I were to make the same sums with the term $\sum_i 2\cos^3\alpha_i\sin\alpha_i$ I see that I do not get values of zero. So what am I missing?
Note that $$\sum_{\alpha=x}^{180^\circ+x} \cos(\alpha)=\sum_{\alpha-x=0}^{180^\circ}\cos(\alpha-x)=\sum_{y=0}^{180^\circ}\cos y$$
and this last sum telescopes to $0$ by the identity $\cos(180^\circ-y)=-\cos y$.
The same happens with $\sum \cos^3\alpha\sin\alpha$ since the sign stays the same because of the identity $\sin(180^\circ-y)=\sin y$.