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:

$\sin(x) + \sin(2\pi - x) = 0$

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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 $2\pi$.