Given that [tex]\(\cos (\theta) = \frac{\sqrt{2}}{2}\)[/tex] and [tex]\(\frac{3\pi}{2} < \theta < 2\pi\)[/tex], we need to evaluate [tex]\(\sin (\theta)\)[/tex] and [tex]\(\tan (\theta)\)[/tex].
Firstly, let's determine in which quadrant [tex]\(\theta\)[/tex] lies. The interval [tex]\(\frac{3\pi}{2} < \theta < 2\pi\)[/tex] means that [tex]\(\theta\)[/tex] is in the fourth quadrant.
In the fourth quadrant:
- Cosine is positive.
- Sine is negative.
- Tangent is negative (since it is the ratio of sine to cosine).
To find [tex]\(\sin (\theta)\)[/tex], we use the Pythagorean identity:
[tex]\[
\sin^2 (\theta) + \cos^2 (\theta) = 1
\][/tex]
We know [tex]\(\cos (\theta) = \frac{\sqrt{2}}{2}\)[/tex], so:
[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]
Solving for [tex]\(\sin (\theta)\)[/tex]:
[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]
Next, we evaluate [tex]\(\tan (\theta)\)[/tex]:
[tex]\[
\tan (\theta) = \frac{\sin (\theta)}{\cos (\theta)}
\][/tex]
[tex]\[
\tan (\theta) = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}
\][/tex]
[tex]\[
\tan (\theta) = -1
\][/tex]
Thus, the evaluated values are:
[tex]\[
\sin (\theta) = -\frac{\sqrt{2}}{2}, \quad \tan (\theta) = -1
\][/tex]
To align with the given answer:
[tex]\[
\sin (\theta) = -0.7071067811865475 \quad \text{and} \quad \tan (\theta) = -0.9999999999999999
\][/tex]
which approximate to the numerical form given.
So, the final answer, in the simplified form, is:
[tex]\[
\sin (\theta) = -\frac{\sqrt{2}}{2} \quad \text{and} \quad \tan (\theta) = -1
\][/tex]
When choosing from the provided options:
[tex]\[
\sin (\theta) = -\frac{\sqrt{2}}{2}
\][/tex]
