Answer:
To prove \(\sin(105^\circ) \cdot \cos(15^\circ) = \frac{1}{4}(2 + \sqrt{3})\), we can use trigonometric identities.
1. We know that \(\sin(105^\circ) = \sin(60^\circ + 45^\circ)\).
2. Applying the sum-to-product identity for sine, \(\sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B\), we get:
\[ \sin(105^\circ) = \sin(60^\circ) \cdot \cos(45^\circ) + \cos(60^\circ) \cdot \sin(45^\circ) \]
3. Using the fact that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), \(\cos(60^\circ) = \frac{1}{2}\), \(\cos(45^\circ) = \frac{\sqrt{2}}{2}\), and \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\), we substitute these values:
\[ \sin(105^\circ) = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \]
4. Simplifying, we get:
\[ \sin(105^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} \]
5. Now, we know that \(\cos(15^\circ) = \sin(75^\circ)\) using the complementary angle property.
6. Again applying the sum-to-product identity, we have:
\[ \sin(105^\circ) \cdot \cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} \cdot \frac{\sqrt{2} + \sqrt{6}}{4} \]
7. Multiplying the numerators and denominators, we get:
\[ \sin(105^\circ) \cdot \cos(15^\circ) = \frac{2 + 2\sqrt{3} + 2}{16} \]
8. Simplifying further, we arrive at:
\[ \sin(105^\circ) \cdot \cos(15^\circ) = \frac{1}{4}(2 + \sqrt{3}) \]
This concludes the proof.
