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 $S = \boxed{\frac{91}{2}}.$
\frac{91}{2}
Level 4
Precalculus
Problem: Compute
\[\cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To compute the sum of squares of cosines from 0 to 90 degrees, we recognize that the cosine function is periodic and symmetric. The values of \(\cos^2 \theta\) for \(\theta\) from 0 to 90 degrees can be calculated directly or derived from known properties of trigonometric functions.

First, we note that \(\cos^2 \theta\) can be expressed using the identity:
\[\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}.\]
This allows us to rewrite the sum:
\[\cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ = \sum_{k=0}^{90} \cos^2 k^\circ = \sum_{k=0}^{90} \frac{1 + \cos(2k^\circ)}{2}.\]

This can be separated into two sums:
\[\sum_{k=0}^{90} \cos^2 k^\circ = \frac{1}{2} \sum_{k=0}^{90} 1 + \frac{1}{2} \sum_{k=0}^{90} \cos(2k^\circ).\]

The first sum, \(\sum_{k=0}^{90} 1\), counts the number of terms, which is 91:
\[\sum_{k=0}^{90} 1 = 91.\]

Thus, we have:
\[\frac{1}{2} \sum_{k=0}^{90} 1 = \frac{1}{2} \cdot 91 = \frac{91}{2}.\]

Next, we consider the second sum, \(\sum_{k=0}^{90} \cos(2k^\circ)\). This sum can be evaluated using the formula for the sum of cosines:
\[\sum_{k=0}^{n} \cos(k \theta) = \frac{\sin\left(\frac{(n+1)\theta}{2}\right) \cos\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}.\]
For our case, \(n = 90\) and \(\theta = 2^\circ\):
\[\sum_{k=0}^{90} \cos(2k^\circ) = \frac{\sin(91^\circ) \cos(90^\circ)}{\sin(1^\circ)}.\]

Since \(\cos(90^\circ) = 0\), this entire term evaluates to 0:
\[\sum_{k=0}^{90} \cos(2k^\circ) = 0.\]

Combining both parts, we find:
\[\sum_{k=0}^{90} \cos^2 k^\circ = \frac{91}{2} + 0 = \frac{91}{2}.\]

Thus, the final answer is \(\frac{91}{2}\).<|END_OF_LATENT|> The final answer is $\frac{91}{2}$.