[Algebra]
Problem: Compute $\sin ^{2} 4^{\circ}+\sin ^{2} 8^{\circ}+\sin ^{2} 12^{\circ}+\cdots+\sin ^{2} 176^{\circ}$.

Solution: Because $\cos 2 x=1-2 \sin ^{2} x, \sin ^{2} x=\frac{1-\cos 2 x}{2}$. Thus the desired sum can be rewritten as

$$
\frac{1-\cos 8^{\circ}}{2}+\frac{1-\cos 16^{\circ}}{2}+\cdots+\frac{1-\cos 352^{\circ}}{2}=\frac{44}{2}-\frac{1}{2}\left(\cos 8^{\circ}+\cos 16^{\circ}+\cdots+\cos 352^{\circ}\right) .
$$

If $\alpha=\cos 8^{\circ}+i \sin 8^{\circ}$, then $\alpha$ is a primitive $45^{\text {th }}$ root of unity, and $1+\alpha+\alpha^{2}+\alpha^{3}+\cdots+\alpha^{44}=0$. Hence $\alpha+\alpha^{2}+\cdots+\alpha^{44}=-1$, and because the real part of $\alpha^{n}$ is simply $\cos 8 n^{\circ}$,

$$
\cos 8^{\circ}+\cos 16^{\circ}+\cdots+\cos 352^{\circ}=-1 .
$$

Thus the desired sum is $22-(1 / 2)(-1)=\mathbf{4 5} / \mathbf{2}$.
The problem asks to simplify the sum

$$
\sin ^{2} a+\sin ^{2} 2 a+\sin ^{2} 3 a+\cdots+\sin ^{2} n a
$$



where $a=4^{\circ}$ and $n=44$. Because $\cos 2 x=1-2 \sin ^{2} x, \sin ^{2} x=\frac{1-\cos 2 x}{2}$. Thus the desired sum can be rewritten as

$$
\frac{1-\cos 2 a}{2}+\frac{1-\cos 4 a}{2}+\cdots+\frac{1-\cos 2 n a}{2}=\frac{n}{2}-\frac{1}{2}(\cos 2 a+\cos 4 a+\cdots+\cos 2 n a) .
$$

Let $Q=\cos 2 a+\cos 4 a+\cdots+\cos 2 n a$. By the sum-to-product identity,

$$
\begin{aligned}
\sin 3 a-\sin a & =2 \cos 2 a \sin a \\
\sin 5 a-\sin 3 a & =2 \cos 4 a \sin a \\
& \vdots \\
\sin (2 n+1) a-\sin (2 n-1) a & =2 \cos 2 n a \sin a
\end{aligned}
$$

Thus

$$
\begin{aligned}
Q \cdot 2 \sin a & =(\sin 3 a-\sin a)+(\sin 5 a-\sin 3 a)+\cdots+(\sin (2 n+1) a-\sin (2 n-1) a) \\
& =\sin (2 n+1) a-\sin a
\end{aligned}
$$

With $a=4^{\circ}$ and $n=44$, the difference on the right side becomes $\sin 356^{\circ}-\sin 4^{\circ}$; note that the terms in this difference are opposites, because of the symmetry of the unit circle. Hence

$$
\begin{aligned}
Q \cdot 2 \sin 4^{\circ} & =-2 \sin 4^{\circ}, \text { and } \\
Q & =-1 .
\end{aligned}
$$

Thus the original sum becomes $44 / 2-(1 / 2)(-1)=\mathbf{4 5} / \mathbf{2}$.

Final answer: $\frac{45}{2}$