The problem asks for the value of $\sin\left( -\frac{11\pi}{3} \right)$.

Step-by-Step Solution:

1. Simplifying the Angle: The angle $-\frac{11\pi}{3}$ is outside the standard range of $0$ to $2\pi$, so we need to simplify it using the periodicity of the sine function. The sine function has a period of $2\pi$, meaning $\sin(\theta) = \sin(\theta + 2n\pi)$, where $n$ is an integer.

   First, let's reduce $-\frac{11\pi}{3}$ by adding multiples of $2\pi$: $-\frac{11\pi}{3} + 2\pi = -\frac{11\pi}{3} + \frac{6\pi}{3} = -\frac{5\pi}{3}$ So, $\sin\left( -\frac{11\pi}{3} \right) = \sin\left( -\frac{5\pi}{3} \right)$.

2. Finding the Reference Angle: The angle $-\frac{5\pi}{3}$ lies in the fourth quadrant. To find the reference angle, subtract this from $2\pi$: $2\pi - \frac{5\pi}{3} = \frac{6\pi}{3} - \frac{5\pi}{3} = \frac{\pi}{3}$ So the reference angle is $\frac{\pi}{3}$.

3. Determining the Sine Value: In the fourth quadrant, sine is negative. The value of $\sin\left( \frac{\pi}{3} \right)$ is $\frac{\sqrt{3}}{2}$, so: $\sin\left( -\frac{5\pi}{3} \right) = -\sin\left( \frac{\pi}{3} \right) = -\frac{\sqrt{3}}{2}$

Thus, the value of $\sin\left( -\frac{11\pi}{3} \right)$ is $-\frac{\sqrt{3}}{2}$.

Would you like more details or have any questions?

Related Questions:

1. What is the value of $\cos\left( -\frac{11\pi}{3} \right)$?
2. How would you find $\tan\left( -\frac{11\pi}{3} \right)$?
3. What is the period of the sine function, and how is it used in simplification?
4. Can you explain the concept of reference angles in trigonometry?
5. How do you determine the sign of a trigonometric function in each quadrant?

Tip:

When simplifying angles in radians, always try to reduce them to their reference angles using the periodic properties of the trigonometric functions.