Sum of Squares of Sines

Let's make use of the fact that $\sin(x) = \cos(90^\circ - x).$ For example, $\sin(1^\circ)=\cos(89^\circ)$ and so on. Thus, we can write

$-\sin^2{1^\circ} + \sin^2{2^\circ}-\sin^2{3^\circ} +...+\sin^2{88^\circ} - \sin^2{89^\circ} +\sin^2{90^\circ} \\ = -\sin^2{1^\circ} - \sin^2{89^\circ} + \sin^2{2^\circ} + \sin^2{88^\circ} - \sin^2{3^\circ} - \sin^2{87^\circ} +... \\ \quad ... -\sin^2{43^\circ} - \sin^2{47^\circ} +\sin^2{44^\circ} + \sin^2{46^\circ} -\sin^2{45^\circ} +\sin^2{90^\circ} \\ = -\sin^2{1^\circ} - \cos^2{1^\circ} + \sin^2{2^\circ} + \cos^2{2^\circ} -\sin^2{3^\circ} - \cos^2{3^\circ} +... \\ \quad ... -\sin^2{43^\circ} - \cos^2{43^\circ} +\sin^2{44^\circ} + \cos^2{44^\circ} -\sin^2{45^\circ} +\sin^2{90^\circ} \\ = -1+1-1...-1+1-\left( \frac{1}{\sqrt{2}} \right)^2 + 1 \\ = -22+22-\frac{1}{2}+1 = \boxed{\frac{1}{2}}$

As sin^2(A) = cos^2(90-A), and sin^2(A) + cos^2(A) = 1, only sin^2(45) is left

$sin(x) = cos(90-x)$ . Therefore, this can be rewritten as $-sin^2(1°) + sin^2(2°)-sin^2(3°)+...-sin^2(45°) -cos^2(1°)+cos^2(2°)-cos^2(3°)+...+cos^2(44°)$ Since $sin^2(x)+cos^2(x) = 1$ , we have $-1+1-1+...+1-sin^2(45°) +sin^2(90°) = -sin^2(45°) + sin^2(90°) = -(\frac{\sqrt{2}}{2})^2 + (1)^2 =-\frac{1}{2} + 1 =\frac{1}{2}$