You are told that cosθ=−5/√315    If θθ is in the **second** quadrant, then the exact value of sinθsin⁡θ is interpreted as14−−√7147Your answer is incorrect.

**Note:** In this question we require you input your answer without decimals and without entering the words sin, cos or tan. For example, if your answer is 5√17√517, then enter **sqrt(5)/sqrt(17)**

Submit part


Your answer is incorrect.

You scored **0** marks for this part.

**Score: 0/1**Your answer is incorrect

Answered


b)

If θθ is in the **third** quadrant, then the exact value of sinθsin⁡θ is interpreted as−3–√3−33Your answer is incorrect.

**Note:** In this question we require you input your answer without decimals and without entering the words sin, cos or tan. For example, if your answer is 5√17√517, then enter **sqrt(5)/sqrt(17)**

Submit part






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.

---

**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.

Learn how to find the exact value of sin(theta) when given cos(theta) in the second and third quadrants of the unit circle. This problem involves applying the Pythagorean identity and understanding trigonometric signs across quadrants. Suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation