Trig Sum How

We'll be using the Pythagorean identity that states $\sin^2(x)+\cos^2(x) = 1$ and the cofunction identity that says $\cos(90 - x) = \sin x$

Using these identities, we see that $\cos^2(1) = \sin^2(89), \cos^2(2) = \sin^2(88), \cos^2(3) = \sin^2(87) ...$

Therefore, ignoring the $\sin(90)$ , we find that there are 44 pairs of Pythagorean identities that each sum to 1.

Now we are just left with $\sin^2(45)$ and $\sin(90)^2$ .

Well this is easy! $\sin^2(45) = (\frac{1}{\sqrt{2}})^2 =\frac{1}{2}$ and $\sin(90)^2 = \frac{1}{1}^2 =1$

Therefore, we have $44+1+0.5 = \boxed{45.5}$

$\cos x = \sin (90^\circ-x)$ The equation thus becomes $\sin^2 1^\circ + \cos^2 1^\circ + \sin^2 2^\circ + \cos^2 2^\circ + \ldots + \sin^2 44^\circ + \cos^2 44^\circ + \sin^2 45^\circ + \sin^2 90^\circ.$

We know that $\sin^2 x + \cos^2 x = 1$ for all real $x$ , thus we have the given expression as $1+1+1+1+1+...+1 + \sin^2 45^\circ + \sin^2 90^\circ = 44 + 0.5 + 1 = 45.5. \square$