Let's solve these problems step by step.

Problem (a): $\theta$ is in the second quadrant.

Given: $\cos \theta = -\frac{5}{\sqrt{315}}$

1. Understanding the Quadrant:

   • In the second quadrant, $\cos \theta$ is negative, and $\sin \theta$ is positive.

2. Find $\sin \theta$:

   • Use the Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$
   • Substitute the given value of $\cos \theta$: $\sin^2 \theta + \left(-\frac{5}{\sqrt{315}}\right)^2 = 1$ $\sin^2 \theta + \frac{25}{315} = 1$ $\sin^2 \theta = 1 - \frac{25}{315} = \frac{315}{315} - \frac{25}{315} = \frac{290}{315}$
   • Therefore, $\sin \theta = \sqrt{\frac{290}{315}}$.

   Since $\theta$ is in the second quadrant, $\sin \theta$ is positive, so: $\sin \theta = \frac{\sqrt{290}}{\sqrt{315}}$

   Simplify further if necessary, but this is your answer in the required format: $\text{Answer: } \frac{\sqrt{290}}{\sqrt{315}}$

Problem (b): $\theta$ is in the third quadrant.

Given: $\cos \theta = -\frac{5}{\sqrt{315}}$

1. Understanding the Quadrant:

   • In the third quadrant, both $\sin \theta$ and $\cos \theta$ are negative.

2. Find $\sin \theta$:

   • Using the same Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$ $\sin^2 \theta = \frac{290}{315}$
   • Since $\theta$ is in the third quadrant, $\sin \theta$ will be negative: $\sin \theta = -\frac{\sqrt{290}}{\sqrt{315}}$

This is the answer for $\sin \theta$ when $\theta$ is in the third quadrant: $\text{Answer: } -\frac{\sqrt{290}}{\sqrt{315}}$

Let me know if you want further details or have any questions.

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Questions:

1. What is the Pythagorean identity used in trigonometry?
2. How do trigonometric signs change across the four quadrants?
3. What is the cosine value in the second quadrant?
4. How do you simplify square roots in trigonometric expressions?
5. What are some strategies for solving trigonometric identities?

Tip: Remember to keep track of the signs of trigonometric functions in different quadrants when solving problems like these.