Chứng minh rằng tam giác ABC vuông tại A nếu sin A = sin B + 2sin C / 2cos B + cos C

\(\sin A\left( {2\cos B + \cos C} \right) = \sin B + 2\sin C\)

Thay \[\left\{ \begin{array}{l}\sin A = \frac{a}{{2R}};\sin B = \frac{b}{{2R}};\sin C = \frac{c}{{2R}}\\\cos B = \frac{{{a^2} + c{}^2 - {b^2}}}{{2ac}};\cos C = \frac{{{a^2} + b{}^2 - {c^2}}}{{2ab}}\end{array} \right.\]

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

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

\[2{a^2}b + 2c{}^2b - 2{b^3} + {a^2}c + b{}^2c - {c^3} = 2{b^2}c + 4b{c^2}\]

\[\begin{array}{l}2{a^2}b - 2c{}^2b - 2{b^3} + {a^2}c - b{}^2c - {c^3} = 0\\\left( {2{a^2}b + {a^2}c} \right) - \left( {2c{}^2b + {c^3}} \right) - \left( {2{b^3} + b{}^2c} \right) = 0\end{array}\]

\[{a^2}\left( {2b + c} \right) - c{}^2\left( {2b + c} \right) - b{}^2\left( {2b + c} \right) = 0\]

\( \Leftrightarrow \left( {2b + c} \right)\left( {{a^2} - {c^2} - {b^2}} \right) = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}2b + c = 0\left( {loai{\rm{ }}vi{\rm{ }}b,c > 0} \right)\\{a^2} = {c^2} + {b^2}\end{array} \right.\). Vậy tam giác \(ABC\) vuông tại \(A\).