1. **State the problem:** Find the exact value of $\cos(-510^\circ)$.\n\n2. **Recall the cosine function property:** Cosine is an even function, so $\cos(-\theta) = \cos(\theta)$. Therefore, $\cos(-510^\circ) = \cos(510^\circ)$.\n\n3. **Reduce the angle to within $0^\circ$ to $360^\circ$:** Since cosine is periodic with period $360^\circ$, subtract multiples of $360^\circ$:\n$$\cos(510^\circ) = \cos(510^\circ - 360^\circ) = \cos(150^\circ)$$\n\n4. **Evaluate $\cos(150^\circ)$:** $150^\circ$ is in the second quadrant where cosine is negative.\n\nUse the reference angle $180^\circ - 150^\circ = 30^\circ$.\n\nSo, $\cos(150^\circ) = -\cos(30^\circ)$.\n\n5. **Recall $\cos(30^\circ)$:** $\cos(30^\circ) = \frac{\sqrt{3}}{2}$.\n\n6. **Final answer:**\n$$\cos(-510^\circ) = \cos(150^\circ) = -\frac{\sqrt{3}}{2}$$