Question

In \( \triangle ABC \), prove that: \[ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{a^2 + b^2 + c^2}{2abc}. \]

Hint:

To derive summation identities in triangles, use the cosine rule: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}. \]

Answer 1

Step 1: Apply the Cosine Rule
The cosine of the angles in a triangle can be expressed as: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab}. \] Step 2: Set Up the Summation
The desired summation is: \[ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c}. \] Substituting the cosine formulas: \[ = \frac{1}{a} \cdot \frac{b^2 + c^2 - a^2}{2bc} + \frac{1}{b} \cdot \frac{a^2 + c^2 - b^2}{2ac} + \frac{1}{c} \cdot \frac{a^2 + b^2 - c^2}{2ab}. \] Step 3: Simplify the Expression
\[ = \frac{b^2 + c^2 - a^2}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{a^2 + b^2 - c^2}{2abc}. \] Now, combine the terms: \[ = \frac{(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2) + (a^2 + b^2 - c^2)}{2abc}. \] Simplifying further: \[ = \frac{a^2 + b^2 + c^2}{2abc}. \]