Trong tam giác ABC có độ dài các cạnh BC = a; AC = b; AB = c. Chứng minh: (b² – c²).cos A = a(c.cos C – b.cos B)

Ta có: a(c.cos C – b.cos B) = accosC – abcosB

\[ = ac.\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} - ab.\frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\]

\[ = c.\frac{{{a^2} + {b^2} - {c^2}}}{{2b}} - b.\frac{{{a^2} + {c^2} - {b^2}}}{{2c}}\]

\[ = \frac{{{a^2}{c^2} + {c^2}{b^2} - {c^4} - {b^2}{a^2} - {b^2}{c^2} + {b^4}}}{{2bc}}\]

\[ = \frac{{{b^2}\left( {{b^2} + {c^2} - {a^2}} \right) - {c^2}\left( {{b^2} + {c^2} - {a^2}} \right)}}{{2bc}}\]

\[ = \left( {{b^2} - {c^2}} \right)\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]

\[ = \left( {{b^2} - {c^2}} \right)\cos A\]

Vậy (b² – c²).cos A = a(c.cos C – b.cos B)