Find sin θ, given that cos θ = (2/3) and tan θ < 0.

To find \( \sin \theta \) given that \( \cos \theta = \frac{2}{3} \) and \( \tan \theta < 0 \), we can follow these steps:

### Step 1: Use the Pythagorean identity
The Pythagorean identity relates \( \sin \theta \) and \( \cos \theta \):
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
Substitute \( \cos \theta = \frac{2}{3} \) into the equation:
\[
\sin^2 \theta + \left( \frac{2}{3} \right)^2 = 1
\]
\[
\sin^2 \theta + \frac{4}{9} = 1
\]
Subtract \( \frac{4}{9} \) from both sides:
\[
\sin^2 \theta = 1 - \frac{4}{9} = \frac{9}{9} - \frac{4}{9} = \frac{5}{9}
\]
Thus:
\[
\sin^2 \theta = \frac{5}{9}
\]
Taking the square root of both sides:
\[
\sin \theta = \pm \frac{\sqrt{5}}{3}
\]

### Step 2: Determine the sign of \( \sin \theta \)
We are given that \( \tan \theta < 0 \), and since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), this implies that \( \sin \theta \) and \( \cos \theta \) must have opposite signs.

Given that \( \cos \theta = \frac{2}{3} \) is positive, \( \sin \theta \) must be negative for \( \tan \theta \) to be negative.

Thus:
\[
\sin \theta = -\frac{\sqrt{5}}{3}
\]

### Final Answer:
\[
\sin \theta = -\frac{\sqrt{5}}{3}
\]

Would you like more details on this solution or have any questions?

Here are 5 related questions for further exploration:
1. How would you find \( \tan \theta \) if \( \sin \theta \) and \( \cos \theta \) are known?
2. In which quadrant is \( \theta \) located, given that \( \cos \theta > 0 \) and \( \tan \theta < 0 \)?
3. How does the Pythagorean identity help relate the three trigonometric functions?
4. What happens to the signs of trigonometric functions in each quadrant?
5. How do you find the angle \( \theta \) if you know \( \sin \theta \) or \( \cos \theta \)?

**Tip**: Always consider the quadrant when determining the signs of trigonometric values.

Discover how to find sin θ given that cos θ = 2/3 and tan θ < 0 using trigonometric identities and quadrant rules. This solution covers key trigonometric concepts and is suitable for Grades 10-12 students.

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