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Ta có : \(cos2A+2\sqrt{2}\left(cosB+cosC\right)=3\)
\(\Leftrightarrow1-2sin^2A+2\sqrt{2}.2.cos\left(\dfrac{B+C}{2}\right).cos\left(\dfrac{B-C}{2}\right)=3\)
\(\Leftrightarrow2sin^2A-4\sqrt{2}.sin\dfrac{A}{2}.cos\left(\dfrac{B-C}{2}\right)+2=0\)
\(\Leftrightarrow sin^2A-2\sqrt{2}.sin\dfrac{A}{2}.cos\left(\dfrac{B-C}{2}\right)+1=0\)
\(\Delta\) ABC không tù nên \(cos\dfrac{A}{2}\ge cos45^o=\dfrac{\sqrt{2}}{2}\)
Suy ra : VT \(\ge sin^2A-4.cos\dfrac{A}{2}.sin\dfrac{A}{2}.cos\left(\dfrac{B-C}{2}\right)+1=K\)
Thấy : \(K=sin^2A-2.sinA.cos\left(\dfrac{B-C}{2}\right)+cos\left(\dfrac{B-C}{2}\right)^2+1-cos\left(\dfrac{B-C}{2}\right)^2\)
\(=\left(sinA-cos\left(\dfrac{B-C}{2}\right)\right)^2+sin^2\left(\dfrac{B-C}{2}\right)\ge0\)
Suy ra : \(VT\ge K\ge0=VP\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}sinA=cos\left(\dfrac{B-C}{2}\right)\\sin\left(\dfrac{B-C}{2}\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sinA=cos0^o=1\\B=C\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=\dfrac{\pi}{2}\\B=C=\dfrac{\pi}{4}\end{matrix}\right.\) ( do \(A+B+C=\pi\) )
Vậy ...
\(\dfrac{cosA}{a}+\dfrac{cosB}{b}+\dfrac{cosC}{c}\)
\(=\dfrac{b^2+c^2-a^2}{2abc}+\dfrac{a^2+c^2-b^2}{2abc}+\dfrac{a^2+b^2-c^2}{2abc}\)
\(=\dfrac{a^2+b^2+c^2}{2abc}\) (đpcm)
a2 = b2 + c2 - 2bc.cosA
b2 = a2 + c2 - 2ac.cosB
c2 = a2 + b2 - 2ab.cosC
⇒ a2 + b2 + c2 = 2bc.cosA + 2ac.cosB + 2ab.cosC
⇒ VT = \(\dfrac{2bc.cosA}{2abc}+\dfrac{2ab.cosC}{2abc}+\dfrac{2ac.cosB}{2abc}\)
⇒ VT = \(\dfrac{cosA}{a}+\dfrac{cosB}{b}+\dfrac{cosC}{c}\)
1.
\(sinA+sinB-sinC=2sin\dfrac{A+B}{2}.cos\dfrac{A-B}{2}-sin\left(A+B\right)\)
\(=2sin\dfrac{A+B}{2}.cos\dfrac{A-B}{2}-2sin\dfrac{A+B}{2}.cos\dfrac{A+B}{2}\)
\(=2sin\dfrac{A+B}{2}.\left(cos\dfrac{A-B}{2}-cos\dfrac{A+B}{2}\right)\)
\(=2sin\dfrac{A+B}{2}.2sin\dfrac{A}{2}.sin\dfrac{B}{2}\)
\(=4sin\dfrac{A}{2}.sin\dfrac{B}{2}.cos\dfrac{C}{2}\)
Sao t lại đc như này v, ai check hộ phát
Đề bài sai, phản ví dụ:
Tam giác ABC vuông tại A với \(AB=1;AC=\sqrt{3};BC=2\)
Khi đó \(AM=\dfrac{1}{2}BC=1=AB\) thỏa mãn yêu cầu bài toán
Góc \(B=60^0;A=90^0\)
Khi đó: \(sinA=1\) trong khi \(2sin\left(B-A\right)=2sin\left(-30\right)=-1\)
Ta có : \(\dfrac{1}{2}\sqrt{\overrightarrow{AB}^2\overrightarrow{AC}^2-\left(\overrightarrow{AB}.\overrightarrow{AC}\right)^2}\)
\(=\dfrac{1}{2}.\sqrt{AB^2AC^2-\left(AB.AC.CosBAC\right)^2}\)
\(=\dfrac{1}{2}.\sqrt{AB^2AC^2-AB^2.AC^2.Cos^2BAC}\)
\(=\dfrac{1}{2}\sqrt{AB^2AC^2\left(1-Cos^2BAC\right)}\)
Thấy : \(Sin^2a+Cos^2a=1\)
\(\Rightarrow Sin^2a=1-Cos^2a\)
\(\Rightarrow\dfrac{1}{2}\sqrt{AB^2AC^2Sin^2BAC}=\dfrac{1}{2}\left|AB.AC.SinBAC\right|=\dfrac{1}{2}AB.AC.SinBAC=S\)
=> ĐPCM
Sao đề là lạ đoạn kia là \(\left(\overrightarrow{AB}.\overrightarrow{AC}\right)^2\)à
a)\(VT=sinA+sinB+sinC=2sin\frac{A+B}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)(đpcm)
Giúp e với ; plz
Ta có : \(cos^2A+cos^2B+cos^2C=1-2.cosA.cosB.cosC\)
Đặt cos A = a ; cos B = b ; cos C = c thì : \(a^2+b^2+c^2+2abc=1\)
Dự đoán : a = b = c = 1/2 nên ta đặt
a = \(\sqrt{\dfrac{xy}{\left(y+z\right)\left(z+x\right)}}\) ; \(b=\sqrt{\dfrac{yz}{\left(x+z\right)\left(x+y\right)}};c=\sqrt{\dfrac{xz}{\left(y+z\right)\left(x+y\right)}}\) ( x ; y ; z > 0 )
Khi đó : \(\Sigma\sqrt{\dfrac{cosA.cosB}{cosC}}=\Sigma\sqrt{\dfrac{y}{x+z}}\)
Cần c/m : \(\Sigma\sqrt{\dfrac{y}{x+z}}>2\) (*)
BĐT quen thuộc ; AD BĐT AM - GM ta được : \(\sqrt{\dfrac{x+z}{y}}\le\dfrac{1}{2}\left(\dfrac{x+y+z}{y}\right)\Rightarrow\sqrt{\dfrac{y}{x+z}}\ge\dfrac{2y}{x+y+z}\)
Suy ra : \(\Sigma\sqrt{\dfrac{y}{x+z}}\ge\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
" = " ko xảy ra nên hiển nhiên (*) đúng
Hoàn tất c/m