For arbitrary angles A, B, C and D, what is the exact value of the expression sin(B − C) cos(A − D) + sin(A − B) cos(C − D) + sin(C − A) cos(B − D)?

Introduction / Context:
This trigonometry question presents a symmetric expression involving four angles A, B, C and D, combined through sine and cosine of differences. The task is to simplify the entire expression to a constant. The problem tests skill with sum and difference identities for sine and recognition of cancellation patterns.

Given Data / Assumptions:
- Expression: sin(B − C) cos(A − D) + sin(A − B) cos(C − D) + sin(C − A) cos(B − D).
- A, B, C and D are arbitrary angles where the trigonometric functions are defined.

Concept / Approach:
We use the identity sin α cos β = (1/2)[sin(α + β) + sin(α − β)]. Applying this identity to each term converts the expression into a sum of sine terms that can be grouped and simplified. Because the angles appear in a cyclic and symmetric fashion, many terms cancel out when properly rearranged.

Step-by-Step Solution:
Step 1: Apply sin α cos β = (1/2)[sin(α + β) + sin(α − β)] to the first term with α = B − C and β = A − D.Step 2: This gives (1/2)[sin((B − C) + (A − D)) + sin((B − C) − (A − D))].Step 3: Similarly expand sin(A − B) cos(C − D) and sin(C − A) cos(B − D) using the same identity.Step 4: Collect all resulting sine terms. Because of the symmetric structure, many arguments appear in positive and negative pairs, such as sin(X) and sin(−X), which sum to zero.Step 5: After grouping, every sine term cancels with another term in the sum, leaving the overall expression equal to 0.

Verification / Alternative check:
To validate the result, you can substitute random numerical values for A, B, C and D, for example in degrees or radians, and evaluate the original expression numerically. For many different choices, the computed value will consistently be extremely close to zero, confirming the algebraic simplification.

Why Other Options Are Wrong:
The options 1, 3/2 and −3 would imply that the expression has a fixed nonzero value independent of the angles, which contradicts both algebraic manipulation and numerical checks. Such values usually arise from partial simplification where only some terms are combined and the cancellations are not followed through completely.

Common Pitfalls:
Students may hesitate to use the identity sin α cos β = (1/2)[sin(α + β) + sin(α − β)] because it doubles the number of terms, but in symmetric problems that is often the key to seeing cancellations. Another pitfall is incorrect handling of signs when expanding and grouping terms, which can obscure the fact that every term has a counterpart with the opposite argument.

Final Answer:
The value of the expression for all permissible A, B, C and D is 0.