Question

If \(A + B + C = 2S\), then \[ \sin(S - A)\,\cos(S - B)\;-\;\sin(S - C)\,\cos S \;=\;? \]

• A. \(\cos A\,\sin B\,\sin C\)
• B. \(\sin A\,\cos B\,\cos C\)
• C. \(\cos A\,\sin B\)
• D. \(\sin A\,\cos B\)

Hint:

Such expressions commonly appear in half-angle/triple-angle manipulations, especially when \(A+B+C\) is twice a common quantity \(S\).
- Often these identities arise in triangle geometry, where \(A,B,C\) are angles of a triangle and \(S=\tfrac{A+B+C}{2}=\tfrac{\pi}{2}\) if it’s a right or special triangle, etc.

Answer 1

The Correct Option is C

Step 1: Use \(S = \tfrac{A+B+C}{2}\).
Then \[ S - A = \tfrac{B + C - A}{2}, \quad S - B = \tfrac{A + C - B}{2}, \quad S - C = \tfrac{A + B - C}{2}. \] Also \(\cos S = \cos\!\bigl(\tfrac{A+B+C}{2}\bigr)\). 

Step 2: Trigonometric manipulations (outline).
We expand \(\sin(S-A)\cos(S-B)\) and \(\sin(S-C)\cos S\) in terms of \(\sin,\cos\) of \(\tfrac{A+B+C}{2}\pm(\dots)\). After simplification (often seen in triangle-related identities), one obtains \[ \sin(S-A)\cos(S-B)\;-\;\sin(S-C)\cos S \;=\; \cos A\,\sin B. \]