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.
Similar Questions
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2. If b cos B = c cos C and if the triangle ABC is not right-angled, then ABC must be isosceles.
Which of the above statements is/are correct?
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Important Questions from Properties of Triangles
Consider the following statements:
1. If ABC is a right-angled triangle, right-angled at A, and if sin \(\rm B = \frac 1 3,\) then cosec C = 3.
2. If b cos B = c cos C and if the triangle ABC is not right-angled, then ABC must be isosceles.
Which of the above statements is/are correct?
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