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a)Có \(b^2+c^2-a^2=cosA.2bc\)
\(S=\dfrac{1}{2}bc.sinA\)\(\Rightarrow4S=2bc.sinA\)
\(\Rightarrow\dfrac{b^2+c^2-a^2}{4S}=\dfrac{cosA.2bc}{2bc.sinA}=cotA\) (dpcm)
b) CM tương tự câu a \(\Rightarrow\dfrac{a^2+c^2-b^2}{4S}=\dfrac{cosB.2ac}{2ac.sinB}=cotB\); \(\dfrac{a^2+b^2-c^2}{4S}=\dfrac{cosC.2ab}{2ab.sinC}=cotC\)
Cộng vế với vế \(\Rightarrow cotA+cotB+cotC=\dfrac{b^2+c^2-a^2}{4S}+\dfrac{a^2+c^2-b^2}{4S}+\dfrac{a^2+b^2-c^2}{4S}\)\(=\dfrac{a^2+b^2+c^2}{4S}\) (dpcm)
c) Gọi ma;mb;mc là độ dài các đường trung tuyến kẻ từ đỉnh A;B;C của tam giác ABC
Có \(GA^2+GB^2+GC^2=\dfrac{4}{9}\left(m_a^2+m_b^2+m_b^2\right)\)\(=\dfrac{4}{9}\left[\dfrac{2\left(b^2+c^2\right)-a^2}{4}+\dfrac{2\left(a^2+c^2\right)-b^2}{4}+\dfrac{2\left(b^2+c^2\right)-a^2}{4}\right]\)
\(=\dfrac{4}{9}.\dfrac{3\left(a^2+b^2+c^2\right)}{4}=\dfrac{a^2+b^2+c^2}{3}\) (đpcm)
d) Có \(a\left(b.cosC-c.cosB\right)=ab.cosC-ac.cosB\)
\(=\dfrac{a^2+b^2-c^2}{2}-\dfrac{a^2+c^2-b^2}{2}\)
\(=b^2-c^2\) (dpcm)
\(a\left(c.cosC-b.cosB\right)=a\left(c.\dfrac{a^2+b^2-c^2}{2ab}-b.\dfrac{a^2+c^2-b^2}{2ac}\right)\)
\(=\dfrac{\left(a^2+b^2-c^2\right).c^2}{2bc}-\dfrac{\left(a^2+c^2-b^2\right).b^2}{2bc}\)
\(=\dfrac{b^4-c^4+a^2c^2-a^2b^2}{2bc}\)
\(=\dfrac{\left(b^2-c^2\right)\left(b^2+c^2-a^2\right)}{2bc}=\left(b^2-c^2\right).cosA\)
a, 3 đường trung tuyến cách nhau tại trọng tâm, khoảng cách từ trọng tâm đến đỉnh bằng \(\dfrac{2}{3}\) độ dài trung tuyến đi qua đỉnh đó
Từ định lí trên ta có \(\left\{{}\begin{matrix}m_a=\dfrac{2}{3}GA\\m_b=\dfrac{2}{3}GB\\m_c=\dfrac{2}{3}GC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m_a^2=\dfrac{4}{9}GA^2\\m_b^2=\dfrac{4}{9}GB^2\\m_c^2=\dfrac{4}{9}GB^2\end{matrix}\right.\)
Đặt D = GA2 + GB2 + GC2
⇒ D = ma2 + mb2 + mc2
⇒ D = \(\dfrac{2\left(a^2+b^2\right)-c^2+2\left(b^2+c^2\right)-a^2+2\left(a^2+c^2\right)-b^2}{4}\)
⇒ D = \(\dfrac{a^2+b^2+c^2}{3}\)
b, cotA = \(\dfrac{cosA}{sinA}=\dfrac{\dfrac{b^2+c^2-a^2}{2bc}}{\dfrac{a}{2R}}=R.\dfrac{b^2+c^2-a^2}{abc}\)
Tương tự ta có
cotB = \(R.\dfrac{a^2+c^2-b^2}{abc}\)
cotC = \(R.\dfrac{a^2+b^2-c^2}{abc}\)
Vậy cotA + cotB + cotC = \(R.\dfrac{a^2+b^2+c^2}{abc}\) (1)
Theo công thức tính diện tích
S = \(\dfrac{abc}{4R}\) ⇒ abc = 4 . S . R
Thế vào (1) ta có
cotA + cotB + cotC = \(R.\dfrac{a^2+b^2+c^2}{4.S.R}=\dfrac{a^2+b^2+c^2}{4S}\)
a, \(\overrightarrow{GA}=-\dfrac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(\Rightarrow GA^2=\dfrac{1}{9}\left(AB^2+AC^2+2AB.AC.cosA\right)\)
\(=\dfrac{1}{9}\left(c^2+b^2+2bc.cosA\right)\)
\(=\dfrac{1}{9}\left(c^2+b^2+b^2+c^2-a^2\right)=\dfrac{2b^2+2c^2-a^2}{9}\)
Tương tự \(GB^2=\dfrac{2a^2+2c^2-b^2}{9}\); \(GC^2=\dfrac{2a^2+2b^2-c^2}{9}\)
\(\Rightarrow GA^2+GB^2+GC^2=\dfrac{a^2+b^2+c^2}{3}\)
b, \(cotA+cotB+cotC=\dfrac{cosA}{sinA}+\dfrac{cosB}{sinB}+\dfrac{cosC}{sinC}\)
\(=\dfrac{b^2+c^2-a^2}{2bcsinA}+\dfrac{a^2+c^2-b^2}{2acsinB}+\dfrac{a^2+b^2-c^2}{2absinC}\)
\(=\dfrac{b^2+c^2-a^2}{2bcsinA}+\dfrac{a^2+c^2-b^2}{2ac.\dfrac{b}{a}sinA}+\dfrac{a^2+b^2-c^2}{2ab.\dfrac{c}{a}sinA}\)
\(=\dfrac{a}{2sinA}\left(\dfrac{b^2+c^2-a^2}{abc}+\dfrac{a^2+c^2-b^2}{abc}+\dfrac{a^2+b^2-c^2}{abc}\right)\)
\(=\dfrac{a^2+b^2+c^2}{2bcsinA}=\dfrac{a^2+b^2+c^2}{4.S}\)
Áp dụng định lí cosin trong tam giác ABC ta có:
\({a^2} = {b^2} + {c^2} - 2bc.\cos A\)\( \Rightarrow \cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\)
Mà \(\sin A = \sqrt {1 - {{\cos }^2}A} \).
\( \Rightarrow \sin A = \sqrt {1 - {{\left( {\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right)}^2}} = \sqrt {\frac{{{{(2bc)}^2} - {{({b^2} + {c^2} - {a^2})}^2}}}{{{{(2bc)}^2}}}} \)
\( \Leftrightarrow \sin A = \frac{1}{{2bc}}\sqrt {{{(2bc)}^2} - {{({b^2} + {c^2} - {a^2})}^2}} \)
Đặt \(M = \sqrt {{{(2bc)}^2} - {{({b^2} + {c^2} - {a^2})}^2}} \)
\(\begin{array}{l} \Leftrightarrow M = \sqrt {(2bc + {b^2} + {c^2} - {a^2})(2bc - {b^2} - {c^2} + {a^2})} \\ \Leftrightarrow M = \sqrt {\left[ {{{(b + c)}^2} - {a^2}} \right].\left[ {{a^2} - {{(b - c)}^2}} \right]} \\ \Leftrightarrow M = \sqrt {(b + c - a)(b + c + a)(a - b + c)(a + b - c)} \end{array}\)
Ta có: \(a + b + c = 2p\)\( \Rightarrow \left\{ \begin{array}{l}b + c - a = 2p - 2a = 2(p - a)\\a - b + c = 2p - 2b = 2(p - b)\\a + b - c = 2p - 2c = 2(p - c)\end{array} \right.\)
\(\begin{array}{l} \Leftrightarrow M = \sqrt {2(p - a).2p.2(p - b).2(p - c)} \\ \Leftrightarrow M = 4\sqrt {(p - a).p.(p - b).(p - c)} \\ \Rightarrow \sin A = \frac{1}{{2bc}}.4\sqrt {p(p - a)(p - b)(p - c)} \\ \Leftrightarrow \sin A = \frac{2}{{bc}}.\sqrt {p(p - a)(p - b)(p - c)} \end{array}\)
b) Ta có: \(S = \frac{1}{2}bc\sin A\)
Mà \(\sin A = \frac{2}{{bc}}\sqrt {p(p - a)(p - b)(p - c)} \)
\(\begin{array}{l} \Rightarrow S = \frac{1}{2}bc.\left( {\frac{2}{{bc}}\sqrt {p(p - a)(p - b)(p - c)} } \right)\\ \Leftrightarrow S = \sqrt {p(p - a)(p - b)(p - c)} .\end{array}\)
Giả thiết tương đương:
\(a^4+b^4+c^4+2b^2c^2=2a^2\left(b^2+c^2\right)+2b^2c^2\)
\(\Leftrightarrow a^4+\left(b^2+c^2\right)^2=2a^2\left(b^2+c^2\right)+2b^2c^2\)
\(\Leftrightarrow\left(b^2+c^2-a^2\right)^2=2b^2c^2\)
\(\Leftrightarrow b^2+c^2-a^2=\pm\sqrt{2}bc\)
\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{\pm\sqrt{2}bc}{2bc}=\pm\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}A=45^0\\A=135^0\end{matrix}\right.\)
a/ \(b^2-c^2=ab.cosC-ac.cosB\)
Ta có: \(b.cosC-c.cosB=ab.\dfrac{a^2+b^2-c^2}{2ab}-ac.\dfrac{a^2+c^2-b^2}{2ac}\)
\(=\dfrac{a^2+b^2-c^2}{2}-\dfrac{a^2+c^2-b^2}{2}=\dfrac{2b^2-2c^2}{2}=b^2-c^2\) (đpcm)
b/ \(ac.cosC-ab.cosB=ac.\dfrac{a^2+b^2-c^2}{2ab}-ab.\dfrac{a^2+c^2-b^2}{2ac}\)
\(=\dfrac{c^2\left(a^2+b^2-c^2\right)-b^2\left(a^2+c^2-b^2\right)}{2bc}=\dfrac{\left(ac\right)^2-\left(ab\right)^2+b^4-c^4}{2bc}\)
\(=\dfrac{-a^2\left(b^2-c^2\right)+\left(b^2-c^2\right)\left(b^2+c^2\right)}{2bc}=\left(b^2-c^2\right).\dfrac{\left(b^2+c^2-a^2\right)}{2bc}\)
\(=\left(b^2-c^2\right).cosA\) (đpcm)
c/ \(cotA+cotB+cotC=\dfrac{cosA}{sinA}+\dfrac{cosB}{sinB}+\dfrac{cosC}{sinC}=\dfrac{2R.cosA}{a}+\dfrac{2R.cosB}{b}+\dfrac{2R.cosC}{c}\)
\(=2R\left(\dfrac{b^2+c^2-a^2}{2abc}+\dfrac{a^2+c^2-b^2}{2abc}+\dfrac{a^2+b^2-c^2}{2abc}\right)\)
\(=2R\left(\dfrac{a^2+b^2+c^2}{2abc}\right)=\dfrac{a^2+b^2+c^2}{abc}.R\) (đpcm)
\(a.\left(c.cosC-b.cosB\right)=a.\left(c.\dfrac{a^2+b^2-c^2}{2ab}-b.\dfrac{a^2+c^2-b^2}{3ac}\right)\)
\(=\dfrac{\left(a^2+b^2-c^2\right)c^2}{2bc}-\dfrac{\left(a^2+c^2-b^2\right)b^2}{2bc}\)
\(=\dfrac{\left(b^2-c^2\right)\left(b^2+c^2-a^2\right)}{2bc}=\left(b^2-c^2\right)cosA\)