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The problem asks to match trigonometric expressions with their corresponding non-trigonometric functions. Let's go step-by-step and solve it.

### Given expressions:

1. \( \sin(\theta) = \frac{x}{2} \) — Find \( \tan(\theta) \).
2. \( \sin(\theta) = \frac{x}{2} \) — Find \( \cos(\theta) \).
3. \( \sin(\theta) = \frac{x}{2} \) — Find \( \sin(2\theta) \).
4. \( \tan(\theta) = \frac{x}{\sqrt{4 - x^2}} \) — Find \( \sin(\theta) \).
5. \( \tan(\theta) = \frac{x}{2} \) — Find \( \cos(\theta) \).

### Using Trigonometric Identities:
- **Pythagorean identity:** \( \sin^2(\theta) + \cos^2(\theta) = 1 \).
- **Double angle identity for sine:** \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \).

### Step-by-step matching:

1. **A. \( \sin(\theta) = \frac{x}{2} \) — Find \( \tan(\theta) \)**:
   Using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \) and solving for \( \cos(\theta) \) from \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we have:
   \[
   \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left( \frac{x}{2} \right)^2} = \sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4 - x^2}}{2}
   \]
   So,
   \[
   \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{x}{2}}{\frac{\sqrt{4 - x^2}}{2}} = \frac{x}{\sqrt{4 - x^2}}.
   \]
   This matches option **4**.

2. **B. \( \sin(\theta) = \frac{x}{2} \) — Find \( \cos(\theta) \)**:
   We already solved \( \cos(\theta) \) in the previous step:
   \[
   \cos(\theta) = \frac{\sqrt{4 - x^2}}{2}.
   \]
   This matches option **3**.

3. **C. \( \sin(\theta) = \frac{x}{2} \) — Find \( \sin(2\theta) \)**:
   Using the double angle identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), we substitute:
   \[
   \sin(2\theta) = 2 \cdot \frac{x}{2} \cdot \frac{\sqrt{4 - x^2}}{2} = \frac{x\sqrt{4 - x^2}}{2}.
   \]
   This matches option **5**.

4. **D. \( \tan(\theta) = \frac{x}{\sqrt{4 - x^2}} \) — Find \( \sin(\theta) \)**:
   We use the identity \( \sin(\theta) = \frac{\tan(\theta)}{\sqrt{1 + \tan^2(\theta)}} \). Substituting \( \tan(\theta) = \frac{x}{\sqrt{4 - x^2}} \):
   \[
   \sin(\theta) = \frac{\frac{x}{\sqrt{4 - x^2}}}{\sqrt{1 + \left( \frac{x}{\sqrt{4 - x^2}} \right)^2}} = \frac{\frac{x}{\sqrt{4 - x^2}}}{\sqrt{ \frac{4}{4 - x^2}}} = \frac{x}{2}.
   \]
   This matches option **1**.

5. **E. \( \tan(\theta) = \frac{x}{2} \) — Find \( \cos(\theta) \)**:
   We know that \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). Using the Pythagorean identity for \( \cos(\theta) \), we have:
   \[
   \cos(\theta) = \frac{\sqrt{4 - x^2}}{2}.
   \]
   This matches option **2**.

### Final Answers:
1. \( A \) matches **4**.
2. \( B \) matches **3**.
3. \( C \) matches **5**.
4. \( D \) matches **1**.
5. \( E \) matches **2**.

Let me know if you'd like more details or have any questions! Here are 5 related questions to practice further:

1. How do you derive the double-angle identity for sine?
2. How can you solve for \( \sin(\theta) \) given a value for \( \tan(\theta) \)?
3. What is the difference between using the Pythagorean identity and the double-angle identity?
4. How do you find \( \cos(2\theta) \) using the identities?
5. What happens to \( \sin(\theta) \) and \( \cos(\theta) \) when \( \theta = 90^\circ \)?

**Tip:** When working with trigonometric identities, drawing a triangle and labeling the sides can be helpful for visualizing the relationships!

Match each of the trigonometric expressions below with the equivalent non-trigonometric function. Use the double-angle identity for sine: sin(2θ) = 2sin(θ)cos(θ).

Solve a problem involving matching trigonometric expressions with their equivalent non-trigonometric functions using identities like the double-angle identity and the Pythagorean identity. Suitable for high school and early college students.

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