Question

Cho tam giác ABC có ba cạnh a,b,c. Chứng minh rằng:

\(\dfrac{a^2+b^2+c^2}{2abc}=\dfrac{cosA}{a}+\dfrac{cosB}{b}+\dfrac{cosC}{c}\)

Answer 1

\(\dfrac{cosA}{a}+\dfrac{cosB}{b}+\dfrac{cosC}{c}\)

\(=\dfrac{b^2+c^2-a^2}{2abc}+\dfrac{a^2+c^2-b^2}{2abc}+\dfrac{a^2+b^2-c^2}{2abc}\)

\(=\dfrac{a^2+b^2+c^2}{2abc}\) (đpcm)

Answer 2

a² = b² + c² - 2bc.cosA

b² = a² + c² - 2ac.cosB

c² = a² + b² - 2ab.cosC

⇒ a² + b² + c² = 2bc.cosA + 2ac.cosB + 2ab.cosC

⇒ VT =  \(\dfrac{2bc.cosA}{2abc}+\dfrac{2ab.cosC}{2abc}+\dfrac{2ac.cosB}{2abc}\)

⇒ VT = \(\dfrac{cosA}{a}+\dfrac{cosB}{b}+\dfrac{cosC}{c}\)