Law of Sines:The Law of Sines states that in any triangle, the ratio of a side length to the sine of the angle opposite that side is constant. We can write it as:a / sin(A) = b / sin(B) = c / sin(C)We don't need the side lengths in this problem, but the concept helps us derive the required relationship.Expressing sin(B + C):Our goal is to find sin((B + C)/2). We can use the angle subtraction identity for sine:sin(x - y) = sin(x)cos(y) - cos(x)sin(y)Here, we can rewrite it as:sin(B + C) = sin(B)cos(C) + cos(B)sin(C) (Equation 1)Relate sin(B + C) to sin((B + C)/2):We can use the double angle identity for sine:sin(2x) = 2sin(x)cos(x)Rewriting for our case:sin((B + C)) = 2sin((B + C)/2)cos((B + C)/2) (Equation 2)Combine Equations:Substitute Equation 1 into Equation 2:2sin((B + C)/2)cos((B + C)/2) = sin(B)cos(C) + cos(B)sin(C)Simplify:Factor out sin((B + C)/2):sin((B + C)/2) [2cos((B + C)/2)] = sin(B)cos(C) + cos(B)sin(C)Relate A to B and C:Since A, B, and C are angles in a triangle, we know their sum is 180°. Therefore:A + B + C = 180°A = 180° - (B + C)Substitute and Manipulate:Substitute A from step 6 into the equation from step 5:sin((B + C)/2) [2cos((B + C)/2)] = sin(180° - A)cos(C) + cos(180° - A)sin(C)