Payton Benson

2022-03-12

Why can I cancel this $\sum _{i}2{\mathrm{cos}}^{3}\left({\alpha }_{i}\right){\mathrm{sin}\alpha }_{i}$ term for pairs 180degrees apart?
I have the equation,
${K}_{11}\sum _{i}{\mathrm{cos}}^{4}{\alpha }_{i}+{K}_{12}\sum _{i}2{\mathrm{cos}}^{3}{\alpha }_{i}{\mathrm{sin}\alpha }_{i}+{K}_{22}\sum _{i}{\mathrm{sin}}^{2}{\alpha }_{i}{\mathrm{cos}}^{2}{\alpha }_{i}=\sum _{i}{\rho }_{i}{\mathrm{cos}}^{2}{\alpha }_{i}$
In the paper it states, "For equal intervals in 180 degree segments the sums of odd powers are zero; therefore,"
${K}_{11}\sum _{i}{\mathrm{cos}}^{4}{\alpha }_{i}+{K}_{22}\sum _{i}{\mathrm{sin}}^{2}{\alpha }_{i}{\mathrm{cos}}^{2}{\alpha }_{i}=\sum _{i}{\rho }_{i}{\mathrm{cos}}^{2}{\alpha }_{i}$
The way I interpret what they mean is that, if I sum the values of ${\mathrm{cos}}^{3}{\alpha }_{i}$ for 0 degree and 180 degree I get a value of 0. Similarly, if I sum the values of ${\mathrm{cos}}^{3}{\alpha }_{i}$ 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 ${\mathrm{cos}}^{3}{\alpha }_{i}$ will be 0. Therefore, we can cancel out the ${K}_{12}\sum _{i}2{\mathrm{cos}}^{3}{\alpha }_{i}{\mathrm{sin}\alpha }_{i}$ term of Eqn(1). However, if I were to make the same sums with the term $\sum _{i}2{\mathrm{cos}}^{3}{\alpha }_{i}{\mathrm{sin}\alpha }_{i}$ I see that I do not get values of zero.

Veronica Riddle

Note that
$\sum _{\alpha =x}^{{180}^{\circ }+x}\mathrm{cos}\left(\alpha \right)=\sum _{\alpha -x=0}^{{180}^{\circ }}\mathrm{cos}\left(\alpha -x\right)=\sum _{y=0}^{{180}^{\circ }}\mathrm{cos}y$
and this last sum telescopes to 0 by the identity $\mathrm{cos}\left({180}^{\circ }-y\right)=-\mathrm{cos}y$
The same happens with $\sum {\mathrm{cos}}^{3}\alpha \mathrm{sin}\alpha$ since the sign stays the same because of the identity $\mathrm{sin}\left({180}^{\circ }-y\right)=\mathrm{sin}y$

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