We are asked to evaluate the expression:

\[
\sin(x) + \sin(2\pi - x)
\]

### Step-by-Step Evaluation:

1. **Use the identity for \(\sin(2\pi - x)\)**:
   The sine function has the following property:
   \[
   \sin(2\pi - x) = -\sin(x)
   \]
   This is because sine is periodic and symmetric around \(\pi\), and \(\sin(2\pi - x)\) is in the fourth quadrant where sine is negative.

2. **Substitute the identity into the expression**:
   \[
   \sin(x) + \sin(2\pi - x) = \sin(x) + (-\sin(x)) = 0
   \]

### Final Answer:
\[
\sin(x) + \sin(2\pi - x) = 0
\]

Let me know if you'd like further details or have any questions!

Here are 5 related questions to explore:
1. What is the general formula for \(\sin(a \pm b)\)?
2. How does the sine function behave under different angle transformations?
3. What are the sine and cosine identities for angles greater than \(2\pi\)?
4. How does the unit circle help in understanding sine and cosine functions?
5. What is the periodicity of the sine and cosine functions?

**Tip:** Always remember that the sine and cosine functions are periodic, which means they repeat their values in regular intervals of \(2\pi\).

Explore how to evaluate sin(x) + sin(2pi - x) using trigonometric identities and properties. This detailed step-by-step solution covers the sine function's symmetry and periodicity. Suitable for students in Grades 10-12 looking to deepen their understanding of trigonometry.

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