1. **State the problem:** Find the exact value of $\sin\left(-\frac{11\pi}{4}\right)$.\n\n2. **Recall the sine function properties:** The sine function is periodic with period $2\pi$, so $\sin(\theta) = \sin(\theta + 2k\pi)$ for any integer $k$. Also, $\sin(-\theta) = -\sin(\theta)$.\n\n3. **Simplify the angle:**\nWe want to find an equivalent angle between $0$ and $2\pi$.\n\nCalculate $-\frac{11\pi}{4} + 4\pi = -\frac{11\pi}{4} + \frac{16\pi}{4} = \frac{5\pi}{4}$.\nSo, $\sin\left(-\frac{11\pi}{4}\right) = \sin\left(\frac{5\pi}{4}\right)$.\n\n4. **Evaluate $\sin\left(\frac{5\pi}{4}\right)$:**\n$\frac{5\pi}{4}$ is in the third quadrant where sine is negative.\nThe reference angle is $\frac{5\pi}{4} - \pi = \frac{\pi}{4}$.\n\nSince $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, then\n$$\sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}.$$\n\n5. **Final answer:**\n$$\sin\left(-\frac{11\pi}{4}\right) = -\frac{\sqrt{2}}{2}.$$