1. **State the problem:** Find the values of $\sin 135^\circ$ and $\tan 15^\circ$. 2. **Recall formulas and important rules:** - $\sin(180^\circ - \theta) = \sin \theta$. - $\tan(45^\circ - \theta) = \frac{\tan 45^\circ - \tan \theta}{1 + \tan 45^\circ \tan \theta}$. 3. **Calculate $\sin 135^\circ$:** Since $135^\circ = 180^\circ - 45^\circ$, $$\sin 135^\circ = \sin(180^\circ - 45^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2}.$$ 4. **Calculate $\tan 15^\circ$:** Using the formula for tangent difference, $$\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}.$$ Recall $\tan 45^\circ = 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$. Substitute values: $$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \times \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}.$$ Multiply numerator and denominator by $\sqrt{3}$ to rationalize: $$\tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}.$$ Multiply numerator and denominator by $\sqrt{3} - 1$: $$\tan 15^\circ = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}.$$ **Final answers:** $$\sin 135^\circ = \frac{\sqrt{2}}{2}, \quad \tan 15^\circ = 2 - \sqrt{3}.$$