Chứng minh rằng trong tam giác ABC với AB=c, AC=b, BC=a ta luôn có: a) a*2 – b*2 = c(aCosB – bCosA) b) Nếu m_a= b thì Sin*2 A= 2(Sin*2C – Sin*2B) với

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
c.\left( {a.\cos B – b.\cos A} \right)\\
 = ca.\cos B – cb.\cos A\\
 = ca.\frac{{{a^2} + {c^2} – {b^2}}}{{2ca}} – cb.\frac{{{c^2} + {b^2} – {a^2}}}{{2bc}}\\
 = \frac{{{a^2} + {c^2} – {b^2}}}{2} – \frac{{{c^2} + {b^2} – {a^2}}}{2}\\
 = \frac{{2{a^2} – 2{b^2}}}{2} = {a^2} – {b^2}\\
 \Rightarrow {a^2} – {b^2} = c\left( {a.\cos B – b.\cos A} \right)\\
b,\\
\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R \Rightarrow \left\{ \begin{array}{l}
\sin A = \frac{a}{{2R}}\\
\sin B = \frac{b}{{2R}}\\
\sin C = \frac{c}{{2R}}
\end{array} \right.\\
{m_a}^2 = \frac{{{b^2} + {c^2}}}{2} – \frac{{{a^2}}}{4}\\
 \Leftrightarrow {b^2} = \frac{{{b^2} + {c^2}}}{2} – \frac{{{a^2}}}{4}\\
 \Leftrightarrow 4{b^2} = 2{b^2} + 2{c^2} – {a^2}\\
 \Leftrightarrow {a^2} = 2{c^2} – 2{b^2}\\
 \Leftrightarrow \frac{{{a^2}}}{{4{R^2}}} = 2.\left( {\frac{{{c^2}}}{{4{R^2}}} – \frac{{{b^2}}}{{4{R^2}}}} \right)\\
 \Leftrightarrow {\left( {\frac{a}{{2R}}} \right)^2} = 2.\left[ {{{\left( {\frac{c}{{2R}}} \right)}^2} – {{\left( {\frac{b}{{2R}}} \right)}^2}} \right]\\
 \Leftrightarrow {\sin ^2}A = 2.\left( {{{\sin }^2}C – {{\sin }^2}B} \right)
\end{array}\)