To find [tex]\(\cos(\theta)\)[/tex] given that [tex]\(\sin(\theta) = \frac{8}{17}\)[/tex] and knowing that [tex]\(\theta\)[/tex] lies in the first quadrant (QI), we can follow these steps:
1. Recall the Pythagorean Identity:
[tex]\[
\sin^2(\theta) + \cos^2(\theta) = 1
\][/tex]
2. Substitute the given [tex]\(\sin(\theta)\)[/tex] into the identity:
[tex]\[
\sin(\theta) = \frac{8}{17}
\][/tex]
[tex]\[
\sin^2(\theta) = \left(\frac{8}{17}\right)^2
\][/tex]
3. Calculate [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[
\sin^2(\theta) = \frac{64}{289}
\][/tex]
4. Use the Pythagorean Identity to find [tex]\(\cos^2(\theta)\)[/tex]:
[tex]\[
\cos^2(\theta) = 1 - \sin^2(\theta)
\][/tex]
Substituting the value of [tex]\(\sin^2(\theta)\)[/tex] we get,
[tex]\[
\cos^2(\theta) = 1 - \frac{64}{289}
\][/tex]
[tex]\[
\cos^2(\theta) = \frac{289}{289} - \frac{64}{289}
\][/tex]
[tex]\[
\cos^2(\theta) = \frac{225}{289}
\][/tex]
5. Calculate [tex]\(\cos(\theta)\)[/tex]:
Since [tex]\(\theta\)[/tex] is in the first quadrant where cosine is positive:
[tex]\[
\cos(\theta) = \sqrt{\cos^2(\theta)}
\][/tex]
[tex]\[
\cos(\theta) = \sqrt{\frac{225}{289}}
\][/tex]
[tex]\[
\cos(\theta) = \frac{15}{17}
\][/tex]
Therefore, the exact value of [tex]\(\cos(\theta)\)[/tex] is:
[tex]\[
\boxed{\frac{15}{17}}
\][/tex]
