To prove the trigonometric identity sin(B + C/2) = cos(A/2) for the interior angles of a triangle ABC, we can use the following steps:

1. Start with the given triangle ABC and its interior angles A, B, and C.
2. Use the fact that the sum of interior angles in a triangle is 180 degrees: A + B + C = 180°.
3. Divide both sides of the equation by 2 to get (A/2) + (B/2) + (C/2) = 90°.
4. Subtract (B/2) and (C/2) from both sides of the equation to obtain (A/2) = 90° – (B/2) – (C/2).
5. Take the cosine of both sides of the equation to get cos(A/2) = cos(90° – (B/2) – (C/2)).
6. Use the cosine identity cos(90° – θ) = sin(θ) to rewrite the equation as cos(A/2) = sin((B/2) + (C/2)).
7. Rearrange the right side of the equation to obtain sin((B/2) + (C/2)) = cos(A/2), which is the desired result.

Therefore, we have shown that sin(B + C/2) = cos(A/2) for the interior angles of triangle ABC.