Step 1: Understanding the Concept:
This problem requires simplifying a sum of sine functions using conditional identities. We will use sum-to-product formulas and the given condition \( A+B+C=4S \).
Step 2: Key Formula or Approach:
\( \sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2} \)
\( \sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2} \)
Step 3: Detailed Explanation:
Let the expression be \( E \). Group the first two terms and the last two terms:
\[ E = [\sin(2S-A) + \sin(2S-B)] + [\sin(2S-C) - \sin 2S] \]
First Bracket:
Sum of angles: \( (2S-A) + (2S-B) = 4S - (A+B) = C \) (since \( 4S = A+B+C \)).
Diff of angles: \( (2S-A) - (2S-B) = B-A = -(A-B) \).
\[ \text{Term}_1 = 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A-B}{2}\right) \]
Second Bracket:
Sum of angles: \( (2S-C) + 2S = 4S - C = A+B \).
Diff of angles: \( (2S-C) - 2S = -C \).
\[ \text{Term}_2 = 2\cos\left(\frac{A+B}{2}\right)\sin\left(-\frac{C}{2}\right) = -2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{C}{2}\right) \]
Combine Terms:
\[ E = 2\sin\frac{C}{2} \left[ \cos\frac{A-B}{2} - \cos\frac{A+B}{2} \right] \]
Using \( \cos X - \cos Y = 2\sin\frac{X+Y}{2}\sin\frac{Y-X}{2} \):
\[ \cos\frac{A-B}{2} - \cos\frac{A+B}{2} = 2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \]
Substitute back:
\[ E = 2\sin\frac{C}{2} \left[ 2\sin\frac{A}{2} \sin\frac{B}{2} \right] = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \]
Step 4: Final Answer:
The correct option is (D).