1. The problem is to find the exact value of $\sin 15^\circ$. 2. We use the angle subtraction formula for sine: $$\sin(a - b) = \sin a \cos b - \cos a \sin b$$ 3. We can write $15^\circ$ as $45^\circ - 30^\circ$. 4. Substitute $a = 45^\circ$ and $b = 30^\circ$ into the formula: $$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$$ 5. Use known exact values: $$\sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 30^\circ = \frac{1}{2}$$ 6. Substitute these values: $$\sin 15^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$$ 7. Factor out $\frac{1}{4}$: $$\sin 15^\circ = \frac{1}{4} (\sqrt{6} - \sqrt{2})$$ Final answer: $$\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$$