To evaluate $\sin(-\pi)$, you don't need to add or subtract $\pi$ or $2\pi$; you can directly compute it based on the properties of the sine function. Here's the explanation:

Step 1: Sine Function Properties

1. Sine is an odd function: This means $\sin(-x) = -\sin(x)$.
2. Periodicity: The sine function has a period of $2\pi$, so $\sin(x) = \sin(x + 2\pi n)$, where $n \in \mathbb{Z}$.

Step 2: Evaluate $\sin(-\pi)$

Using the odd property: $\sin(-\pi) = -\sin(\pi).$

From the unit circle:

• At $x = \pi$, $\sin(\pi) = 0$.

Thus: $\sin(-\pi) = -0 = 0.$

Final Answer:

$\sin(-\pi) = 0.$

Do You Need to Add or Subtract $2\pi$?

No, adding or subtracting $2\pi$ would yield the same result, as it does not change the sine value due to periodicity.

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Would you like additional examples or clarifications?

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Related Questions:

1. What is the value of $\sin(-2\pi)$?
2. Can you explain why sine is an odd function geometrically?
3. How does the periodicity of sine help simplify angle calculations?
4. What is $\cos(-\pi)$ and how is cosine different from sine in terms of odd/even properties?
5. Can you evaluate $\tan(-\pi)$ using sine and cosine?

Tip:

For trigonometric functions, always start with symmetry and periodicity properties—they simplify calculations significantly!