To solve the problem of evaluating [tex]\(\sin \left(\arccos\left(-\frac{4}{\sqrt{25}}\right)\right)\)[/tex], let's break it down step-by-step.
1. Simplify the argument of the inverse cosine (arccos) function:
[tex]\[
-\frac{4}{\sqrt{25}}
\][/tex]
Note that [tex]\(\sqrt{25} = 5\)[/tex]. So,
[tex]\[
-\frac{4}{\sqrt{25}} = -\frac{4}{5}
\][/tex]
Thus, we need to evaluate [tex]\(\arccos\left(-\frac{4}{5}\right)\)[/tex].
2. Geometric interpretation of [tex]\( \arccos \)[/tex]:
The [tex]\(\arccos\left(-\frac{4}{5}\right)\)[/tex] is the angle [tex]\( \theta \)[/tex] such that [tex]\(\cos(\theta) = -\frac{4}{5}\)[/tex].
3. Use the Pythagorean identity to find [tex]\(\sin(\theta)\)[/tex]. Recall that [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex].
Given [tex]\(\cos(\theta) = -\frac{4}{5}\)[/tex],
[tex]\[
\cos^2(\theta) = \left( -\frac{4}{5} \right)^2 = \frac{16}{25}
\][/tex]
Therefore,
[tex]\[
\sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25}
\][/tex]
4. Evaluate [tex]\(\sin(\theta)\)[/tex]:
Since we have [tex]\(\sin^2(\theta) = \frac{9}{25}\)[/tex],
[tex]\[
\sin(\theta) = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5}
\][/tex]
We need to determine which of [tex]\( \pm \frac{3}{5} \)[/tex] is correct. Given the range of the [tex]\(\arccos\)[/tex] function, [tex]\(0 \leq \theta \leq \pi\)[/tex], the [tex]\( \sin(\theta) \)[/tex] in this range is positive. Therefore,
[tex]\[
\sin\left( \arccos\left(-\frac{4}{5}\right) \right) = \frac{3}{5}
\][/tex]
5. Verify the options:
- [tex]\(-\frac{\sqrt{25}}{\sqrt{128}}\)[/tex]
- [tex]\(\frac{3}{\sqrt{25}}\)[/tex]
- [tex]\(\frac{\sqrt{25}}{\sqrt{128}}\)[/tex]
- [tex]\(-\frac{3}{\sqrt{25}}\)[/tex]
Simplify the correct answer by checking that [tex]\(\frac{3}{\sqrt{25}} = \frac{3}{5}\)[/tex].
Thus, the correct answer is [tex]\(\frac{3}{\sqrt{25}}\)[/tex].
