Use a sum or difference formula to find the exact value of the trigonometric function. \( \sin \frac{11 \pi}{12} \) \( \sin \frac{11 \pi}{12}=\square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

To find \( \sin \frac{11 \pi}{12} \), we can rewrite \( \frac{11 \pi}{12} \) as \( \frac{\pi}{3} + \frac{\pi}{4} \), because \( \frac{\pi}{3} = \frac{4 \pi}{12} \) and \( \frac{\pi}{4} = \frac{3 \pi}{12} \). Now, we can use the sine sum formula: \[ \sin(a + b) = \sin a \cos b + \cos a \sin b \] Letting \( a = \frac{\pi}{3} \) and \( b = \frac{\pi}{4} \), we have: \[ \sin \frac{11\pi}{12} = \sin \left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \sin \frac{\pi}{3} \cos \frac{\pi}{4} + \cos \frac{\pi}{3} \sin \frac{\pi}{4} \] Now substituting the values: \[ \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \quad \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{\pi}{3} = \frac{1}{2}, \quad \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \] Putting it all together: \[ \sin \frac{11\pi}{12} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \] \[ = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \] \[ = \frac{\sqrt{6} + \sqrt{2}}{4} \] Thus, the exact value of \( \sin \frac{11 \pi}{12} \) is: \[ \sin \frac{11 \pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4} \]