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To solve for \(\sin(\theta)\) given that \(\cos(\theta) = -\frac{1}{6}\) and \(\theta\) is in the third quadrant, follow these steps:

### Step 1: Use the Pythagorean Identity
The Pythagorean identity states:
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]
Substitute \(\cos(\theta) = -\frac{1}{6}\) into the equation:
\[
\sin^2(\theta) + \left(-\frac{1}{6}\right)^2 = 1
\]
\[
\sin^2(\theta) + \frac{1}{36} = 1
\]

### Step 2: Isolate \(\sin^2(\theta)\)
Subtract \(\frac{1}{36}\) from both sides:
\[
\sin^2(\theta) = 1 - \frac{1}{36}
\]
\[
\sin^2(\theta) = \frac{36}{36} - \frac{1}{36}
\]
\[
\sin^2(\theta) = \frac{35}{36}
\]

### Step 3: Solve for \(\sin(\theta)\)
Take the square root of both sides:
\[
\sin(\theta) = \pm \sqrt{\frac{35}{36}}
\]
\[
\sin(\theta) = \pm \frac{\sqrt{35}}{6}
\]

### Step 4: Determine the Sign of \(\sin(\theta)\) in the Third Quadrant
Since \(\theta\) is in the third quadrant, where both sine and cosine are negative, we select the negative value:
\[
\sin(\theta) = -\frac{\sqrt{35}}{6}
\]

### Final Answer
\[
\sin(\theta) = -\frac{\sqrt{35}}{6}
\]

Would you like further details on this solution, or do you have any questions?

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Here are 5 related questions that expand on this topic:

1. How would you find \(\tan(\theta)\) given \(\cos(\theta) = -\frac{1}{6}\) in the third quadrant?
2. If \(\sin(\theta) = -\frac{\sqrt{35}}{6}\), what is the value of \(\sin(2\theta)\)?
3. How does the sign of trigonometric functions change in each quadrant?
4. How can the Pythagorean identity be used to find the other trigonometric functions if \(\cos(\theta)\) is known?
5. Why does \(\theta\) being in the third quadrant affect the sign of \(\sin(\theta)\) and \(\cos(\theta)\)?

**Tip:** When working with trigonometric functions in different quadrants, remember ASTC: All Students Take Calculus, which helps to remember the signs of trigonometric functions in each quadrant.

If cos(θ) = -1/6 and θ is in the 3rd quadrant, find the exact value for sin(θ).

Learn how to find the exact value of sin(θ) when cos(θ) = -1/6 and θ is in the third quadrant. This solution uses the Pythagorean identity to solve for sin(θ) and explains the significance of the quadrant for determining the sign. Suitable for high school trigonometry students.

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