1. **State the problem:** Find the exact value of $\cos(-510^\circ)$. 2. **Recall the cosine function property:** Cosine is an even function, so $\cos(-\theta) = \cos(\theta)$. Therefore, $\cos(-510^\circ) = \cos(510^\circ)$. 3. **Reduce the angle:** Angles can be reduced by subtracting multiples of $360^\circ$ because cosine has a period of $360^\circ$. $$\cos(510^\circ) = \cos(510^\circ - 360^\circ) = \cos(150^\circ)$$ 4. **Evaluate $\cos(150^\circ)$:** $150^\circ$ is in the second quadrant where cosine is negative. It is $30^\circ$ away from $180^\circ$, so $$\cos(150^\circ) = -\cos(30^\circ)$$ 5. **Recall $\cos(30^\circ)$:** $\cos(30^\circ) = \frac{\sqrt{3}}{2}$. 6. **Final answer:** $$\cos(-510^\circ) = -\frac{\sqrt{3}}{2}$$