Câu 6.56 trang 206 SBT Đại số 10 Nâng cao

LG a

\(\sin A = \dfrac{{\cos B + \cos C}}{{\sin B + \sin C}}\) thì tam giác ABC là tam giác vuông;

Lời giải chi tiết:

Vì \(\sin A = 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}\) và

\(\begin{array}{l}\dfrac{{\cos B + \cos C}}{{\sin B + \sin C}} = \dfrac{{2\cos \dfrac{{B + C}}{2}\cos \dfrac{{B - C}}{2}}}{{2\sin \dfrac{{B + C}}{2}\cos \dfrac{{B - C}}{2}}}\\ = \dfrac{{\cos \left( {\dfrac{\pi }{2} - \dfrac{A}{2}} \right)}}{{\sin \left( {\dfrac{\pi }{2} - \dfrac{A}{2}} \right)}} = \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}\end{array}\)

nên dễ thấy

\(\begin{array}{l}\sin A = \dfrac{{\cos B + \cos C}}{{\sin B + \sin C}}\\ \Leftrightarrow 2{\cos ^2}\dfrac{A}{2} = 1 \Leftrightarrow \cos A = 0\end{array}\)

\( \Leftrightarrow \widehat A\) là góc vuông.

LG b

 \(\dfrac{{\sin A}}{{\sin B}} = \dfrac{{\cos B + \cos C}}{{\cos C + \cos A}}\) thì tam giác ABC là một tam giác vuông hoặc một tam giác cân.

Lời giải chi tiết:

 Cách 1

\(\begin{array}{l}\dfrac{{\sin A}}{{\sin B}} = \dfrac{{\cos B + \cos C}}{{\cos C + \cos A}}\\ \Leftrightarrow \dfrac{{\sin \dfrac{A}{2}\cos \dfrac{A}{2}}}{{\sin \dfrac{B}{2}\cos \dfrac{B}{2}}} = \dfrac{{\sin \dfrac{A}{2}\cos \dfrac{{B - C}}{2}}}{{\sin \dfrac{B}{2}\cos \dfrac{{C - A}}{2}}}\\ \Leftrightarrow \cos \dfrac{A}{2}\cos \dfrac{{C - A}}{2} = \cos \dfrac{B}{2}\cos \dfrac{{B - C}}{2}\\ \Leftrightarrow \cos \dfrac{C}{2} + \cos \left( {A - \dfrac{C}{2}} \right)\\ = \cos \left( {B - \dfrac{C}{2}} \right) + \cos \dfrac{C}{2}\\ \Leftrightarrow \cos \left( {A - \dfrac{C}{2}} \right) = \cos \left( {B - \dfrac{C}{2}} \right)\\ \Leftrightarrow \left| {A - \dfrac{C}{2}} \right| = \left| {B - \dfrac{C}{2}} \right|\\ \Leftrightarrow \left[ \begin{array}{l}\widehat A = \widehat B\\\widehat A + \widehat B = \widehat C.\end{array} \right.\end{array}\)

Cách 2

\(\begin{array}{l}\dfrac{{\sin A}}{{\sin B}} = \dfrac{{\cos B + \cos C}}{{\cos C + \cos A}}\\ \Leftrightarrow \sin A\cos A - \sin B\cos B\\ = \cos C\left( {\sin B - \sin A} \right)\\ \Leftrightarrow \dfrac{1}{2}\left( {\sin 2A - \sin 2B} \right)\\ = \cos C\left( {\sin B - \sin A} \right)\\ \Leftrightarrow \cos \left( {A + B} \right)\sin \left( {A - B} \right)\\ = 2\cos C\cos \dfrac{{B + A}}{2}\sin \dfrac{{B - A}}{2}\\ \Leftrightarrow  - \cos C\sin \dfrac{{A - B}}{2}\cos \dfrac{{A - B}}{2}\\ =  - \cos C\sin \dfrac{{A - B}}{2}\cos \dfrac{{A + B}}{2}\\ \Leftrightarrow \cos C\sin \dfrac{{A - B}}{2}\left( {\cos \dfrac{{A + B}}{2} - \cos \dfrac{{A - B}}{2}} \right)\\ = 0\\ \Leftrightarrow \cos C\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{{A - B}}{2} = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos C = 0\\\sin \dfrac{{A - B}}{2} = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\widehat C\,\,vuông\\\widehat A = \widehat B\end{array} \right.\end{array}\)

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