To solve for [tex]\(\cos(\theta)\)[/tex] given that [tex]\(\sin(\theta) = -\frac{1}{3}\)[/tex] and [tex]\(\pi < \theta < \frac{3\pi}{2}\)[/tex], follow these steps:
1. Identify the Quadrant:
Since [tex]\( \pi < \theta < \frac{3\pi}{2} \)[/tex], [tex]\(\theta\)[/tex] is in the third quadrant. In this quadrant, both [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex] are negative.
2. Use the Pythagorean Identity:
Recall the Pythagorean identity:
[tex]\[
\sin^2(\theta) + \cos^2(\theta) = 1
\][/tex]
3. Calculate [tex]\(\sin^2(\theta)\)[/tex]:
Given [tex]\(\sin(\theta) = -\frac{1}{3}\)[/tex]:
[tex]\[
\sin^2(\theta) = \left(-\frac{1}{3}\right)^2 = \frac{1}{9}
\][/tex]
4. Calculate [tex]\(\cos^2(\theta)\)[/tex]:
Substitute [tex]\(\sin^2(\theta)\)[/tex] into the Pythagorean identity:
[tex]\[
\sin^2(\theta) + \cos^2(\theta) = 1
\][/tex]
[tex]\[
\frac{1}{9} + \cos^2(\theta) = 1
\][/tex]
Solving for [tex]\(\cos^2(\theta)\)[/tex]:
[tex]\[
\cos^2(\theta) = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}
\][/tex]
5. Determine [tex]\(\cos(\theta)\)[/tex]:
Taking the square root of [tex]\(\cos^2(\theta)\)[/tex]:
[tex]\[
\cos(\theta) = \pm \sqrt{\frac{8}{9}}
\][/tex]
[tex]\[
\cos(\theta) = \pm \frac{\sqrt{8}}{3}
\][/tex]
Since [tex]\(\theta\)[/tex] is in the third quadrant where [tex]\(\cos(\theta)\)[/tex] is negative:
[tex]\[
\cos(\theta) = -\frac{\sqrt{8}}{3}
\][/tex]
6. Simplify [tex]\(\cos(\theta)\)[/tex]:
[tex]\[
\cos(\theta) = -\frac{\sqrt{8}}{3} = -\frac{2\sqrt{2}}{3}
\][/tex]
Thus, the correct value of [tex]\(\cos(\theta)\)[/tex] is:
[tex]\[
-\frac{2\sqrt{2}}{3}
\][/tex]
So the answer is [tex]\(-\frac{2\sqrt{2}}{3}\)[/tex].
