Determine the value of the sum

$$
S=\sin^{2} 1^{\circ}+\sin^{2} 2^{\circ}+\sin^{2} 3^{\circ}+\ldots+\sin^{2} 89^{\circ}
$$

#
Solution

Since $\sin x = \cos (90^\circ - x)$ and $\sin^2 x + \cos^2 x = 1$, it follows that

$$
\sin^2 1^\circ + \sin^2 89^\circ = \sin^2 2^\circ + \sin^2 88^\circ = \ldots = \sin^2 44^\circ + \sin^2 46^\circ = 1
$$

Therefore, by pairing the terms of the sum in pairs of the form $\sin^2 x + \sin^2 (90 - x)$, we have

$$
S = 44 + \sin^2 45^\circ = 44 + \frac{1}{2} = \frac{89}{2}
$$