Evaluating the sum $ \sum_{n = 1}^{44} {\sin^{2}}(n^{\circ}) ~ {\cos^{2}}(n^{\circ}) $. I want to find the sum
$$
{\sin^{2}}(1^{\circ})  ~ {\cos^{2}}(1^{\circ}) +
{\sin^{2}}(2^{\circ})  ~ {\cos^{2}}(2^{\circ}) +
{\sin^{2}}(3^{\circ})  ~ {\cos^{2}}(3^{\circ}) + \cdots +
{\sin^{2}}(44^{\circ}) ~ {\cos^{2}}(44^{\circ}).
$$
I thought of using the identity $ \sin(2 x) = 2 \sin(x) \cos(x) $, so
$$
  [2 \sin(1^{\circ}) \cos(1^{\circ})]^{2} +
  [2 \sin(2^{\circ}) \cos(2^{\circ})]^{2} + \cdots +
  [2 \sin(44^{\circ}) \cos(44^{\circ})]^{2}
= 4 y.
$$
By the identity above, I get
$$
{\sin^{2}}(2^{\circ}) + {sin^{2}}(4^{\circ}) + \cdots + {\sin^{2}}(88^{\circ}) = 4 y,
$$
but then I don’t know how to simplify this or if it could be done in other simpler ways.
Thanks.

Note that $\sin^2 \theta=\frac{1-\cos 2\theta}{2}$. As there are $44$ terms in the sum, this means that
$$
4y=\frac{44}{2}-\frac{1}{2}\left(\cos 4^\circ + \cos 8^\circ + \dots + \cos 176^\circ \right) \, .
$$
Now, since $\cos(180^\circ -\theta)=-\cos \theta$, the terms in parentheses will cancel in pairs:
\begin{align}
\cos 4^\circ + \cos 176^\circ&= 0 \\
\cos 8^\circ + \cos 172^\circ&= 0 \\
\cdots \\
\cos 88^\circ + \cos 92^\circ&= 0
\end{align}
So the parenthesized sum vanishes, meaning that $4y=\frac{44}{2}=22$, and thus $y=\frac{11}{2}$.
