1. **Problem:** Evaluate the trigonometric function $\sin\left(-\frac{8\pi}{3}\right)$ using its period. 2. **Recall the period of sine:** The sine function has a period of $2\pi$, meaning $\sin(x) = \sin(x + 2\pi k)$ for any integer $k$. 3. **Reduce the angle:** $$ -\frac{8\pi}{3} + 2\pi = -\frac{8\pi}{3} + \frac{6\pi}{3} = -\frac{2\pi}{3} $$ Since adding $2\pi$ (one full period) does not change the value, we have: $$ \sin\left(-\frac{8\pi}{3}\right) = \sin\left(-\frac{2\pi}{3}\right) $$ 4. **Use the odd property of sine:** $$ \sin(-x) = -\sin(x) $$ So, $$ \sin\left(-\frac{2\pi}{3}\right) = -\sin\left(\frac{2\pi}{3}\right) $$ 5. **Evaluate $\sin\left(\frac{2\pi}{3}\right)$:** $\frac{2\pi}{3}$ is in the second quadrant where sine is positive. $$ \sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} $$ 6. **Combine results:** $$ \sin\left(-\frac{8\pi}{3}\right) = -\frac{\sqrt{3}}{2} $$ **Final answer:** $$ \boxed{-\frac{\sqrt{3}}{2}} $$