1. Let's consider the problem: Find the exact value of $\sin(45^\circ)$ using trigonometric identities. 2. The formula we use is the sine of a sum identity: $$\sin(a+b) = \sin a \cos b + \cos a \sin b$$ 3. Since $45^\circ = 30^\circ + 15^\circ$, we can write: $$\sin(45^\circ) = \sin(30^\circ + 15^\circ) = \sin 30^\circ \cos 15^\circ + \cos 30^\circ \sin 15^\circ$$ 4. We know the exact values: $$\sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}$$ 5. Next, use half-angle formulas to find $\sin 15^\circ$ and $\cos 15^\circ$: $$\sin 15^\circ = \sin(45^\circ/3) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$$ 6. Using known values: $$\sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}$$ 7. Calculate $\sin 15^\circ$: $$\sin 15^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$$ 8. Similarly, calculate $\cos 15^\circ$ using the cosine difference identity: $$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$ 9. Substitute back into step 3: $$\sin 45^\circ = \frac{1}{2} \times \frac{\sqrt{6} + \sqrt{2}}{4} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{6} - \sqrt{2}}{4}$$ 10. Simplify: $$= \frac{\sqrt{6} + \sqrt{2}}{8} + \frac{\sqrt{3}(\sqrt{6} - \sqrt{2})}{8} = \frac{\sqrt{6} + \sqrt{2}}{8} + \frac{\sqrt{18} - \sqrt{6}}{8}$$ 11. Since $\sqrt{18} = 3\sqrt{2}$, combine terms: $$= \frac{\sqrt{6} + \sqrt{2} + 3\sqrt{2} - \sqrt{6}}{8} = \frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2}$$ 12. Therefore, the exact value is: $$\boxed{\sin 45^\circ = \frac{\sqrt{2}}{2}}$$ This matches the well-known value for $\sin 45^\circ$.