Use a sum or difference formula to find the exact value of the trigonometric function. \( \cos \frac{\pi}{12} \) \( \cos \frac{\pi}{12}=\square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

To find \( \cos \frac{\pi}{12} \), we can use the cosine of a sum formula, specifically \( \frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4} \). Using the formula \( \cos(a - b) = \cos a \cos b + \sin a \sin b \), we substitute \( a = \frac{\pi}{3} \) and \( b = \frac{\pi}{4} \): \[ \cos \frac{\pi}{12} = \cos \left( \frac{\pi}{3} - \frac{\pi}{4} \right) = \cos \frac{\pi}{3} \cos \frac{\pi}{4} + \sin \frac{\pi}{3} \sin \frac{\pi}{4} \] Now, we calculate the values: - \( \cos \frac{\pi}{3} = \frac{1}{2} \) - \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \) - \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \) - \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \) Substituting these into the formula gives: \[ \cos \frac{\pi}{12} = \left(\frac{1}{2} \cdot \frac{\sqrt{2}}{2}\right) + \left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} \] Combining the fractions: \[ \cos \frac{\pi}{12} = \frac{\sqrt{2} + \sqrt{6}}{4} \] So, the exact value is: \[ \cos \frac{\pi}{12} = \frac{\sqrt{2} + \sqrt{6}}{4} \]