$\frac{sinB+sinC}{cosB+cosC}$ =sin A chứng minh tam giác ABC vuông

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1. \[\begin{array}{l}
   \frac{{\sin B + \sin C}}{{\cos C + \cos B}} = \sin A\\
    \Leftrightarrow \frac{{\frac{b}{{2R}} + \frac{c}{{2R}}}}{{\frac{{{a^2} + {b^2} – {c^2}}}{{2ab}} + \frac{{{a^2} + {c^2} – {b^2}}}{{2ac}}}} = \frac{a}{{2R}}\\
    \Leftrightarrow \frac{{b + c}}{{2R}} = \frac{a}{{2R}}\left( {\frac{{{a^2} + {b^2} – {c^2}}}{{2ab}} + \frac{{{a^2} + {c^2} – {b^2}}}{{2ac}}} \right)\\
    \Leftrightarrow b + c = \frac{{{a^2} + {b^2} – {c^2}}}{{2b}} + \frac{{{a^2} + {c^2} – {b^2}}}{{2c}}\\
    \Leftrightarrow b + c = \frac{{{a^2}c + {b^2}c – {c^3} + {a^2}b + {c^2}b – {b^3}}}{{2bc}}\\
    \Leftrightarrow {a^2}c + {b^2}c – {c^3} + {a^2}b + {c^2}b – {b^3} = 2{b^2}c + 2b{c^2}\\
    \Leftrightarrow {a^2}c + {a^2}b = {b^2}c + {b^3} + {c^2}b + {c^3}\\
    \Leftrightarrow {a^2}\left( {b + c} \right) = {b^2}\left( {b + c} \right) + {c^2}\left( {b + c} \right)\\
    \Leftrightarrow {a^2} = {b^2} + {c^2}\left( {do\,b + c > 0} \right)\\
    \Rightarrow \Delta ABC\,vuong\,tai\,A
   \end{array}\]

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