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?

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.