Cho tam giác ABC với ba cạnh a, b, c. Chứng minh rằng: cosA/a + cosB/b + cosC/c = a^2+  b^2  + c^2 / 2abc

Lời giải

Theo định lí côsin: a² = b² + c² – 2bccosA

⇒ cosA = \(\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}\)

⇒ \(\frac{{{\rm{cosA}}}}{{\rm{a}}}\)=\(\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{abc}}}}\).

Tương tự ta có:

\(\frac{{{\rm{cosB }}}}{{\rm{b}}}\)=\(\frac{{{{\rm{a}}^2} + {{\rm{c}}^2} - {{\rm{b}}^2}}}{{2{\rm{abc}}}}\) và \(\frac{{{\rm{cosC}}}}{{\rm{c}}}\)=\(\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{abc}}}}\)

Như vậy: \(\frac{{{\mathop{\rm cosA}\nolimits} }}{{\rm{a}}} + \frac{{{\rm{cosB}}}}{{\rm{b}}} + \frac{{{\rm{cosC}}}}{{\rm{c}}}\) =\(\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2a{\rm{bc}}}}\) + \(\frac{{{{\rm{a}}^2} + {{\rm{c}}^2} - {{\rm{b}}^2}}}{{2{\rm{abc}}}}\) + \(\frac{{{{\rm{a}}^2} + {{\rm{c}}^2} - {{\rm{b}}^2}}}{{2{\rm{abc}}}}\)

⇒ \(\frac{{{\mathop{\rm cosA}\nolimits} }}{{\rm{a}}} + \frac{{{\rm{cosB}}}}{{\rm{b}}} + \frac{{{\rm{cosC}}}}{{\rm{c}}} = \frac{{{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{ + }}{{\rm{c}}^{\rm{2}}}}}{{{\rm{2abc}}}}\). ( ĐPCM ).