To find [tex]\(\sin \theta\)[/tex] given that [tex]\(\cos \theta = \frac{15}{17}\)[/tex], we can use the Pythagorean identity:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
Given [tex]\(\cos \theta = \frac{15}{17}\)[/tex], let's denote this value for clarity:
[tex]\[
\cos \theta = \frac{15}{17}
\][/tex]
First, let's square [tex]\(\cos \theta\)[/tex]:
[tex]\[
\cos^2 \theta = \left(\frac{15}{17}\right)^2 = \frac{225}{289}
\][/tex]
Next, substitute [tex]\(\cos^2 \theta\)[/tex] into the Pythagorean identity:
[tex]\[
\sin^2 \theta + \frac{225}{289} = 1
\][/tex]
Now, solve for [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[
\sin^2 \theta = 1 - \frac{225}{289}
\][/tex]
To proceed, we need to subtract [tex]\(\frac{225}{289}\)[/tex] from 1. We can convert 1 to a fraction with the same denominator:
[tex]\[
1 = \frac{289}{289}
\][/tex]
Thus:
[tex]\[
\sin^2 \theta = \frac{289}{289} - \frac{225}{289} = \frac{64}{289}
\][/tex]
Now, taking the square root of both sides to find [tex]\(\sin \theta\)[/tex]:
[tex]\[
\sin \theta = \sqrt{\frac{64}{289}} = \frac{\sqrt{64}}{\sqrt{289}} = \frac{8}{17}
\][/tex]
Given that [tex]\(\cos \theta\)[/tex] is positive and considering standard trigonometric functions in different quadrants, [tex]\(\sin \theta\)[/tex] should also be positive in this case.
Therefore, the value of [tex]\(\sin \theta\)[/tex] is:
[tex]\[
\boxed{\frac{8}{17}}
\][/tex]
