Chapter 8.3, Problem 58AYU

To determine

Prove the identity $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^2+b^2+c^2}{2abc}$ .

Answer to Problem 58AYU

  $\boxed{\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{C}=\frac{a^2+b^2+c^2}{2abc}}$

Explanation of Solution

Given information:

Use the Law of Cosines to prove the identity

  $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^2+b^2+c^2}{2abc}$

Calculation:

Let us cosider the triangle having sides a,b,c and corresponding opposite angle A,B,C. By the Law of Cosies, we get

  $\cos A=\frac{b^2+c^2-a^2}{2bc}\mathrm{.....}\left(1\right)$

  $\cos B=\frac{a^2+c^2-b^2}{2ac}\mathrm{......}\left(2\right)$

and $\cos C=\frac{a^2+b^2-c^2}{2ab}\mathrm{.......}\left(3\right)$

From (1),(2) and (3), we get

  $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{b^2+c^2-a^2}{2abc}+\frac{a^2+c^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}$

  $=\frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2abc}$

  $=\frac{a^2+b^2+c^2}{2abc}$

Hence, $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{C}=\frac{a^2+b^2+c^2}{2abc}$