To find the exact value of [tex]\(\cos \frac{17\pi}{8}\)[/tex], let's go through the steps systematically.
First, we use the periodicity property of cosine:
[tex]\[
\cos \theta = \cos (2\pi k + \theta) \text{ for any integer } k.
\][/tex]
We will reduce [tex]\(\frac{17\pi}{8}\)[/tex] into an angle in the interval [tex]\([0, 2\pi)\)[/tex]:
[tex]\[
\frac{17\pi}{8} = 2\pi - \frac{\pi}{8}.
\][/tex]
Using the property that [tex]\(\cos(2\pi - \theta) = \cos(\theta)\)[/tex], we can write:
[tex]\[
\cos \frac{17\pi}{8} = \cos (2\pi - \frac{\pi}{8}) = \cos (-\frac{\pi}{8}).
\][/tex]
Since cosine is an even function ([tex]\(\cos(-\theta) = \cos(\theta)\)[/tex]), we have:
[tex]\[
\cos \left(- \frac{\pi}{8}\right) = \cos \left(\frac{\pi}{8}\right).
\][/tex]
Now, we need to find [tex]\(\cos \frac{\pi}{8}\)[/tex]. We use the half-angle identity for cosine:
[tex]\[
\cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}}.
\][/tex]
Let [tex]\(\theta = \frac{\pi}{4}\)[/tex], then:
[tex]\[
\cos \frac{\pi}{8} = \sqrt{\frac{1 + \cos \frac{\pi}{4}}{2}}.
\][/tex]
We know:
[tex]\[
\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}.
\][/tex]
Substituting this value:
[tex]\[
\cos \frac{\pi}{8} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}.
\][/tex]
Simplify the expression inside the square root:
[tex]\[
\frac{1 + \frac{\sqrt{2}}{2}}{2} = \frac{2 + \sqrt{2}}{4} \quad \text{(combining fractions)}.
\][/tex]
Taking the square root:
[tex]\[
\cos \frac{\pi}{8} = \sqrt{\frac{2 + \sqrt{2}}{4}}.
\][/tex]
Therefore, the exact value of [tex]\(\cos \frac{17\pi}{8}\)[/tex] is:
[tex]\[
\cos \frac{17\pi}{8} = \sqrt{\frac{2 + \sqrt{2}}{4}}.
\][/tex]
The correct answer is:
C. [tex]\(\sqrt{\frac{2+\sqrt{2}}{4}}\)[/tex].
