To solve for [tex]\(\sin(\theta)\)[/tex] and [tex]\(\tan(\theta)\)[/tex] given that [tex]\(\cos(\theta) = \frac{\sqrt{2}}{2}\)[/tex] and [tex]\(\frac{3\pi}{2} < \theta < 2\pi\)[/tex], we need to follow these steps:
1. Identify the Quadrant:
Since [tex]\(\frac{3\pi}{2} < \theta < 2\pi\)[/tex], we know that [tex]\(\theta\)[/tex] is in the fourth quadrant.
2. Sign of the Trigonometric Functions in the Fourth Quadrant:
In the fourth quadrant, cosine is positive and sine is negative.
3. Finding [tex]\(\sin(\theta)\)[/tex]:
We can use the Pythagorean identity:
[tex]\[
\sin^2(\theta) + \cos^2(\theta) = 1
\][/tex]
Substitute the given value of [tex]\(\cos(\theta)\)[/tex]:
[tex]\[
\sin^2(\theta) + \left(\frac{\sqrt{2}}{2}\right)^2 = 1
\][/tex]
[tex]\[
\sin^2(\theta) + \frac{2}{4} = 1
\][/tex]
[tex]\[
\sin^2(\theta) + \frac{1}{2} = 1
\][/tex]
[tex]\[
\sin^2(\theta) = 1 - \frac{1}{2}
\][/tex]
[tex]\[
\sin^2(\theta) = \frac{1}{2}
\][/tex]
Therefore,
[tex]\[
\sin(\theta) = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2}
\][/tex]
Since [tex]\(\theta\)[/tex] is in the fourth quadrant where sine is negative:
[tex]\[
\sin(\theta) = -\frac{\sqrt{2}}{2}
\][/tex]
4. Finding [tex]\(\tan(\theta)\)[/tex]:
The tangent function is defined as:
[tex]\[
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
\][/tex]
Substitute the values:
[tex]\[
\tan(\theta) = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}
\][/tex]
[tex]\[
\tan(\theta) = -1
\][/tex]
So, the values are:
[tex]\[
\sin(\theta) = -\frac{\sqrt{2}}{2}
\][/tex]
[tex]\[
\tan(\theta) = -1
\][/tex]
Thus, the correct answers are:
[tex]\(-\frac{\sqrt{2}}{2}\)[/tex] and [tex]\(-1\)[/tex].
