Cho tam giác ABC. CMR:
a) Nếu \(\dfrac{b^2-a^2}{2c}\)=2.b.cosA- a.cosB thì ABC cân tại C
b) Nếu \(\dfrac{sinB}{sinC}\)=2.cosA thì tam giác ABC cân tại B
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\(\dfrac{b^2-a^2}{2c}=b.\dfrac{\left(b^2+c^2-a^2\right)}{2bc}-a.\dfrac{\left(a^2+c^2-b^2\right)}{2ac}\)
\(\Leftrightarrow\dfrac{b^2-a^2}{2c}=\dfrac{b^2+c^2-a^2}{2c}-\dfrac{a^2+c^2-b^2}{2c}\)
\(\Leftrightarrow b^2-a^2=\left(b^2+c^2-a^2\right)-\left(a^2+c^2-b^2\right)\)
\(\Leftrightarrow3b^2=3a^2\Leftrightarrow a=b\)
Hay tam giác cân tại C
\(\overrightarrow{BA}=\left(3;0\right)\Rightarrow AB=3=AC\) ; \(\overrightarrow{AC}=\left(a-2;b+2\right)\) ; \(\overrightarrow{BC}=\left(a+1;b+2\right)\)
\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}=\dfrac{6\sqrt{5}}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-2\right)^2+\left(b+2\right)^2=9\\\left(a+1\right)^2+\left(b+2\right)^2=\dfrac{36}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(\dfrac{1}{5};-\dfrac{22}{5}\right)\\\left(a;b\right)=\left(\dfrac{1}{5};\dfrac{2}{5}\right)\end{matrix}\right.\)
a) Kẻ các đường cao \(AD;BE;CF\)
ta có : \(AD=AB.sinB\) và \(AD=AC.sinC\)
\(\Rightarrow AB.sinB=AC.sinC\Leftrightarrow c.sinB=b.sinC\Leftrightarrow\dfrac{c}{sinC}=\dfrac{b}{sinB}\)
làm tương tự ta có : \(\dfrac{b}{sinB}=\dfrac{a}{sinA}\) và \(\dfrac{a}{sinA}=\dfrac{c}{sinC}\)
\(\Rightarrow\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}\left(đpcm\right)\)
b) ta có : \(BC^2=BE^2+EC^2=AB^2-AE^2+\left(AC-AE\right)^2\)
\(\Leftrightarrow BC=AB^2-AE^2+AC^2-2AC.AE+AE^2\)
\(\Leftrightarrow BC^2=AB^2+AC^2-2AC.AB.cosA\)
\(\Leftrightarrow a^2=b^2+c^2-2bc.cosA\left(đpcm\right)\)
c) ta có : \(AB=BF+FA=BC.cosB+AC.cosA\)
\(\Leftrightarrow c=a.cosB+b.cosA\left(đpcm\right)\)
đặc \(M\) là chân đường trung tuyên kẻ từ \(A\) \(\left(m_a\right)\)
ta có : \(AM^2=AB^2+BM^2-2AB.BM.cosB\)
\(\Leftrightarrow AM^2=AB^2+BM^2-2AB.BM\dfrac{AB^2+BC^2-AC^2}{2AB.2BM}\)
\(\Leftrightarrow AM^2=AB^2+\left(\dfrac{BC}{2}\right)^2-\dfrac{AB^2+BC^2-AC^2}{2}\)
\(\Leftrightarrow AM^2=AB^2-\dfrac{AB^2+BC^2-AC^2}{2}+\dfrac{BC^2}{4}\)
\(\Leftrightarrow AM^2=\dfrac{2AB^2-AB^2-BC^2+AC^2}{2}+\dfrac{BC^2}{4}\) \(\Leftrightarrow AM^2=\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{2}+\dfrac{BC^2}{4}\) \(\Leftrightarrow AM^2=\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{4}\Leftrightarrow m_a^2=\dfrac{c^2+b^2}{2}-\dfrac{a^2}{4}\left(đpcm\right)\)(chú ý câu này sử dụng công thức ở câu \(b;c\) nha)
Ta có: A = \(sin\dfrac{A}{2}+sin\dfrac{B}{2}+sin\dfrac{C}{2}=cos\dfrac{B+C}{2}+2sin\dfrac{B+C}{4}cos\dfrac{B-C}{4}\)
\(\Leftrightarrow A-2sin\dfrac{B+C}{4}cos\dfrac{B-C}{4}-cos^2\dfrac{B+C}{4}+sin^2\dfrac{B+C}{4}=0\)\(\Leftrightarrow A-2sin\dfrac{B+C}{4}cos\dfrac{B-C}{4}+2sin^2\dfrac{B+C}{4}-1=0\)
Δ' = \(cos^2\dfrac{B-C}{4}-2\left(A-1\right)\ge0\)
\(\Rightarrow A-1\le\dfrac{1}{2}\Leftrightarrow A\le\dfrac{3}{2}\)
a) ta có : \(cos^2\left(a-b\right)-sin^2\left(a+b\right)\)
\(=\left(cosa.cosb+sina.sinb\right)^2-\left(sina.cosb+sinb.cosa\right)^2\)
\(=cos^2a.cos^2b+sin^2a.sin^2b-sin^2a.cos^2b-sin^2b.cos^2a\)
\(=cos^2a.cos^2b-sin^2a.cos^2b+sin^2a.sin^2b-sin^2b.cos^2a\)\(=cos^2b\left(cos^2a-sin^2a\right)-sin^2b\left(cos^2a-sin^2a\right)\)
\(=\left(cos^2b-sin^2b\right)\left(cos^2a-sin^2a\right)=cos2a.cos2b\left(đpcm\right)\)