1. **State the problem:** Given that $\sec \theta = 3$, find the values of $\cos \theta$ and $\sin \theta$. 2. **Recall the definition:** $\sec \theta = \frac{1}{\cos \theta}$. 3. **Find $\cos \theta$:** Since $\sec \theta = 3$, then $$\cos \theta = \frac{1}{3}.$$ 4. **Find $\sin \theta$:** Use the Pythagorean identity $$\sin^2 \theta + \cos^2 \theta = 1.$$ Substitute $\cos \theta = \frac{1}{3}$: $$\sin^2 \theta + \left(\frac{1}{3}\right)^2 = 1,$$ $$\sin^2 \theta + \frac{1}{9} = 1,$$ $$\sin^2 \theta = 1 - \frac{1}{9} = \frac{8}{9},$$ so $$\sin \theta = \pm \sqrt{\frac{8}{9}} = \pm \frac{2\sqrt{2}}{3}.$$ 5. **Final answer:** $$\cos \theta = \frac{1}{3}, \quad \sin \theta = \pm \frac{2\sqrt{2}}{3}.$$