Given cos a = 1/2 with a between 0 and pi, and sin b = 3/5 with b between -pi/2 and pi/2, find sin(a + b).

1. From cos a = 1/2, the Pythagorean identity gives sin a = sqrt(1 - 1/4) = sqrt(3/4) = sqrt(3)/2, positive because a lies between 0 and pi. 2. From sin b = 3/5, the same identity gives cos b = sqrt(1 - 9/25) = sqrt(16/25) = 4/5, positive because b lies between -pi/2 and pi/2. 3. Apply sin(a + b) = sin a cos b + cos a sin b. 4. The first product is (sqrt(3)/2)(4/5) = 4 sqrt(3) / 10. 5. The second product is (1/2)(3/5) = 3/10. 6. Adding gives (4 sqrt(3) + 3) / 10. 7. The option (4 + 3 sqrt(3)) / 10 swaps which term carries the surd and is a different number, close to 0.92 instead of about 0.99. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 9 "Trigonometric Identities and Equations", section 9.2 Sum and Difference Identities_