Using the half-angle formula for sine with cos 30 degrees = sqrt(3)/2, find sin 15 degrees.

1. Since 15 degrees is half of 30 degrees, use sin(a/2) = plus or minus the square root of (1 - cos a) / 2. 2. Substituting cos 30 = sqrt(3)/2 gives the square root of (1 - sqrt(3)/2) / 2. 3. Combine the numerator over 2 to get (2 - sqrt(3)) / 2, then divide by 2 again to get (2 - sqrt(3)) / 4. 4. Taking the square root gives sqrt(2 - sqrt(3)) / 2. 5. Only the positive root is kept because 15 degrees lies in quadrant I, where sine is positive. 6. The option with 2 + sqrt(3) inside would come from the cosine half-angle formula instead. 7. The option with 1 - sqrt(3) inside is negative under the root, so it is not even a real number. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 9 "Trigonometric Identities and Equations", section 9.3 Double-Angle, Half-Angle, and Reduction Formulas_