3) Use the sum or difference formula to find the exact value of \( \sin \frac{7 \pi}{12} \). a) \( \frac{\sqrt{6}+\sqrt{2}}{2} \) b) \( \frac{\sqrt{6}-\sqrt{2}}{2} \) c) \( \frac{\sqrt{6}-\sqrt{2}}{4} \) d) \( \frac{\sqrt{6}+\sqrt{2}}{4} \)

To find \( \sin \frac{7 \pi}{12} \), we can express \( \frac{7\pi}{12} \) as the sum of \( \frac{\pi}{3} \) and \( \frac{\pi}{4} \). This gives us \( \sin \frac{7\pi}{12} = \sin \left( \frac{\pi}{3} + \frac{\pi}{4} \right) \). Using the sine addition formula, \( \sin(a + b) = \sin a \cos b + \cos a \sin b \), we substitute: \[ \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \quad \cos \frac{\pi}{3} = \frac{1}{2}, \quad \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}. \] Now plug in the values: \[ \sin \frac{7\pi}{12} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}. \] Thus, the answer is \( d) \frac{\sqrt{6}+\sqrt{2}}{4} \). For a fun mathematical twist, did you know that the angles used in trigonometric calculations often come from the classic degrees used in geometry? The values of \( \frac{\pi}{3} \) and \( \frac{\pi}{4} \) correspond to 60 degrees and 45 degrees, which are super handy in various geometric principles, paving the way to explore elegant shapes like equilateral triangles and squares! Now, if you ever find yourself confused by the sum and difference formulas, remember that practice is key! A common mistake people make is mixing up sine and cosine functions. When adding angles, keep track of which function goes with which part of the formula. Regularly revisiting the unit circle can also help reinforce these relationships—it's like your backstage pass to the world of trigonometry!