Use a sum-to-product formula to find the exact value. Write your answer as a simplified fraction and rationalize th \[ \cos 345^{\circ}-\cos 75^{\circ}=\square \]

To find the exact value of \(\cos 345^{\circ} - \cos 75^{\circ}\) using a sum-to-product formula, we can use the following identity: \[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \] In this case, let \(A = 345^{\circ}\) and \(B = 75^{\circ}\). First, we calculate \(A + B\) and \(A - B\): \[ A + B = 345^{\circ} + 75^{\circ} = 420^{\circ} \] \[ A - B = 345^{\circ} - 75^{\circ} = 270^{\circ} \] Now, we can substitute these values into the sum-to-product formula: \[ \cos 345^{\circ} - \cos 75^{\circ} = -2 \sin\left(\frac{420^{\circ}}{2}\right) \sin\left(\frac{270^{\circ}}{2}\right) \] Calculating the angles: \[ \frac{420^{\circ}}{2} = 210^{\circ} \] \[ \frac{270^{\circ}}{2} = 135^{\circ} \] Now we find the sine values: \[ \sin(210^{\circ}) = -\frac{1}{2} \] \[ \sin(135^{\circ}) = \frac{\sqrt{2}}{2} \] Substituting these values back into the equation: \[ \cos 345^{\circ} - \cos 75^{\circ} = -2 \left(-\frac{1}{2}\right) \left(\frac{\sqrt{2}}{2}\right) \] Simplifying this expression: \[ = -2 \cdot -\frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 1 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \] Thus, the exact value of \(\cos 345^{\circ} - \cos 75^{\circ}\) is: \[ \boxed{\frac{\sqrt{2}}{2}} \]