Triangle Trigonometry: Evaluating \(\frac{\sin(A-B)}{\sin(A+B)}\)

We need to find the value of the expression \(\frac{\sin(A-B)}{\sin(A+B)}\) in a triangle \(ABC\), where \(a\), \(b\), and \(c\) are the side lengths opposite angles \(A\), \(B\), and \(C\) respectively.

Step 1: Use Triangle Angle Properties

In any triangle \(ABC\), the sum of angles is \(A+B+C = \pi\). Therefore, \(A+B = \pi - C\).

Using the sine property \(\sin(\pi - x) = \sin x\), we get:

\( \sin(A+B) = \sin(\pi - C) = \sin C \)

Step 2: Substitute into the Expression

Substitute \(\sin(A+B) = \sin C\) into the given expression:

\( \frac{\sin(A-B)}{\sin(A+B)} = \frac{\sin(A-B)}{\sin C} \)

Step 3: Apply Sine Expansion Formula

Use the sine subtraction formula \(\sin(A-B) = \sin A \cos B - \cos A \sin B\):

\( \frac{\sin(A-B)}{\sin C} = \frac{\sin A \cos B - \cos A \sin B}{\sin C} \)

Step 4: Apply the Sine Rule

The Sine Rule states \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\), where \(R\) is the circumradius. From this, we have \(\sin A = \frac{a}{2R}\), \(\sin B = \frac{b}{2R}\), and \(\sin C = \frac{c}{2R}\).

Substitute these into the expression:

\( \frac{\frac{a}{2R} \cos B - \cos A \frac{b}{2R}}{\frac{c}{2R}} \)

Multiply the numerator and denominator by \(2R\):

\( \frac{a \cos B - b \cos A}{c} \)

Step 5: Apply the Law of Cosines

The Law of Cosines gives:

• \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)
• \(\cos B = \frac{a^2 + c^2 - b^2}{2ac}\)

Substitute these cosine values into the expression \(\frac{a \cos B - b \cos A}{c}\):

\( \frac{a \left( \frac{a^2 + c^2 - b^2}{2ac} \right) - b \left( \frac{b^2 + c^2 - a^2}{2bc} \right)}{c} \)

Simplify the numerator:

\( \frac{\frac{a^2 + c^2 - b^2}{2c} - \frac{b^2 + c^2 - a^2}{2c}}{c} \)

\( \frac{1}{c} \left( \frac{(a^2 + c^2 - b^2) - (b^2 + c^2 - a^2)}{2c} \right) \)

\( \frac{1}{2c^2} (a^2 + c^2 - b^2 - b^2 - c^2 + a^2) \)

\( \frac{1}{2c^2} (2a^2 - 2b^2) \)

\( \frac{2(a^2 - b^2)}{2c^2} \)

\( \frac{a^2 - b^2}{c^2} \)

Conclusion

The expression \(\frac{\sin(A-B)}{\sin(A+B)}\) simplifies to \(\frac{a^2 - b^2}{c^2}\). This matches Option C.