Compute
\[\cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ.\]
Let $S = \cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ.$  Then
\begin{align*}
S &= \cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ \\
&= \cos^2 90^\circ + \cos^2 89^\circ + \cos^2 88^\circ + \dots + \cos^2 0^\circ \\
&= \sin^2 0^\circ + \sin^2 1^\circ + \sin^2 2^\circ + \dots + \sin^2 90^\circ,
\end{align*}so
\begin{align*}
2S &= (\cos^2 0^\circ + \sin^2 0^\circ) + (\cos^2 1^\circ + \sin^2 1^\circ) + (\cos^2 2^\circ + \sin^2 2^\circ) + \dots + (\cos^2 90^\circ + \sin^2 90^\circ) \\
&= 91,
\end{align*}which means $S = \boxed{\frac{91}{2}}.$
Let $S = \cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ.$  Then
\begin{align*}
S &= \cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ \\
&= \cos^2 90^\circ + \cos^2 89^\circ + \cos^2 88^\circ + \dots + \cos^2 0^\circ \\
&= \sin^2 0^\circ + \sin^2 1^\circ + \sin^2 2^\circ + \dots + \sin^2 90^\circ,
\end{align*}so
\begin{align*}
2S &= (\cos^2 0^\circ + \sin^2 0^\circ) + (\cos^2 1^\circ + \sin^2 1^\circ) + (\cos^2 2^\circ + \sin^2 2^\circ) + \dots + (\cos^2 90^\circ + \sin^2 90^\circ) \\
&= 91,
\end{align*}which means $S = \boxed{\frac{91}{2}}.$
The final answer is $$\boxed{\frac{91}{2}}$$. I hope it is correct.