1. The problem asks to determine the exact value of $\cos 12^\circ$.\n\n2. To find the exact value of $\cos 12^\circ$, we can use angle subtraction formulas because 12 degrees is not a standard angle but can be expressed as $45^\circ - 33^\circ$ or more conveniently $45^\circ - 33^\circ$ is not standard, so instead use $45^\circ - 33^\circ$ is complicated. A better approach is to use the half-angle or triple-angle formulas.\n\n3. Use the triple-angle formula for cosine: $$\cos 3\theta = 4\cos^3 \theta - 3\cos \theta.$$\n\n4. Set $\theta = 12^\circ$, then $3\theta = 36^\circ$. We know $\cos 36^\circ = \frac{\sqrt{5} + 1}{4}$.\n\n5. Substitute into the triple-angle formula: $$\cos 36^\circ = 4\cos^3 12^\circ - 3\cos 12^\circ.$$\n\n6. Let $x = \cos 12^\circ$. Then: $$\frac{\sqrt{5} + 1}{4} = 4x^3 - 3x.$$\n\n7. Rearrange to form a cubic equation: $$4x^3 - 3x - \frac{\sqrt{5} + 1}{4} = 0.$$\n\n8. Multiply both sides by 4 to clear the denominator: $$16x^3 - 12x - (\sqrt{5} + 1) = 0.$$\n\n9. This cubic can be solved for $x$, but since $\cos 12^\circ$ is positive and less than $\cos 0^\circ = 1$, the exact value is the root of this equation.\n\n10. Alternatively, the exact value of $\cos 12^\circ$ is $$\cos 12^\circ = \frac{\sqrt{2 + \sqrt{3}} + \sqrt{3} - 1}{2\sqrt{2}}$$ but this is derived from half-angle and sum formulas.\n\nFinal answer: $$\cos 12^\circ = \frac{\sqrt{2 + \sqrt{3}} + \sqrt{3} - 1}{2\sqrt{2}}.$$