Given that [tex]\(\theta\)[/tex] is in Quadrant III and [tex]\(\tan \theta = \frac{3}{4}\)[/tex], we need to find [tex]\(\sin \theta\)[/tex].
1. Understand the given information:
- [tex]\(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}\)[/tex]
- In Quadrant III, both sine and cosine are negative.
2. Trigonometric identities and relationships:
- We know that [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex].
- Using the Pythagorean identity: [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex].
3. Express [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] using a common variable:
- Let [tex]\(\sin \theta = 3k\)[/tex] and [tex]\(\cos \theta = 4k\)[/tex] for some value [tex]\(k\)[/tex].
- Since [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3k}{4k} = \frac{3}{4}\)[/tex], our choice of [tex]\(3k\)[/tex] and [tex]\(4k\)[/tex] is consistent with [tex]\(\tan \theta\)[/tex].
4. Find [tex]\(k\)[/tex] using the Pythagorean identity:
- Substitute [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] into [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[
(3k)^2 + (4k)^2 = 1 \implies 9k^2 + 16k^2 = 1 \implies 25k^2 = 1 \implies k^2 = \frac{1}{25} \implies k = \frac{1}{5}
\][/tex]
5. Calculate [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
- [tex]\(\sin \theta = 3k = 3 \left( \frac{1}{5} \right) = \frac{3}{5}\)[/tex]
- [tex]\(\cos \theta = 4k = 4 \left( \frac{1}{5} \right) = \frac{4}{5}\)[/tex]
6. Adjust the sign based on Quadrant III:
- In Quadrant III, both sine and cosine are negative.
- Hence, [tex]\(\sin \theta = -\frac{3}{5}\)[/tex] and [tex]\(\cos \theta = -\frac{4}{5}\)[/tex].
Therefore, [tex]\(\sin \theta = -\frac{3}{5}\)[/tex].
