Derive an expression for \( \sin(a+b) \) using the sum formulas, where \( a = 60^{\circ} \) and \( b = 45^{\circ} \).

To derive the expression for \( \sin(a+b) \) when \( a = 60^{\circ} \) and \( b = 45^{\circ} \), we use the sum formula for sine: \[ \sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b). \] First, we need to find \( \sin(60^{\circ}) \), \( \cos(60^{\circ}) \), \( \sin(45^{\circ}) \), and \( \cos(45^{\circ}) \): - \( \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \) - \( \cos(60^{\circ}) = \frac{1}{2} \) - \( \sin(45^{\circ}) = \frac{\sqrt{2}}{2} \) - \( \cos(45^{\circ}) = \frac{\sqrt{2}}{2} \) Now substitute these values into the sum formula: \[ \sin(60^{\circ} + 45^{\circ}) = \sin(60^{\circ}) \cos(45^{\circ}) + \cos(60^{\circ}) \sin(45^{\circ}). \] Plugging in the known values: \[ \sin(60^{\circ} + 45^{\circ}) = \left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}\right) + \left(\frac{1}{2} \times \frac{\sqrt{2}}{2}\right). \] Calculating each term: \[ \frac{\sqrt{3} \cdot \sqrt{2}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}. \] Thus, the expression for \( \sin(105^{\circ}) \) is: \[ \sin(105^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}. \]