[TEX]\frac{1}{sin^22A}[/TEX]+[TEX]\frac{1}{sin^22B}[/TEX]+[TEX]\frac{1}{sin^22C}[/TEX]=[TEX]\frac{1}{2cosAcosBcosC}[/TEX]
tìm các góc của tam giác A,B,C
[TEX]\frac{1}{sin^22A}[/TEX]+[TEX]\frac{1}{sin^22B}[/TEX]+[TEX]\frac{1}{sin^22C}[/TEX]=[TEX]\frac{1}{2cosAcosBcosC}[/TEX]
[tex]\Leftrightarrow \frac{1}{sin^22A}+\frac{1}{sin^22B}+\frac{1}{sin^22C} =\frac{4sinAsinBsinC}{2sinAcosA2sinBcosB2sinCcosC}[/tex]
[tex]=\frac{sin2A+sin2B+sin2C}{sin2Asin2Bsin2C}(*)[/tex]
Đặt [tex]\left{\begin{x=sin2A}\\{y=sin2B}\\{z=sin2C}[/tex]
Thì [tex](*)[/tex] tương đương [tex]\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{x+y+z}{xyz}[/tex]
[tex]\Rightarrow \frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y} =x+y+z[/tex]
[tex]\Leftrightarrow x^2y^2+y^2z^2+z^2x^2=xyz.(x+y+z)=x^2yz+y^2xz+z^2xy(2)[/tex]
Đặt [tex]\left{\begin{q=xy}\\{p=yz}\\{r=zx}[/tex]
[tex]\Rightarrow (2) \Leftrightarrow p^2+q^2+r^2 =pq+qr+rp[/tex]
[tex](p-q)^2+(q-r)^2+(r-p)^2=0[/tex]
[tex]\Rightarrow \left{\begin{p=q}\\{q=r}\\{r=p} \Leftrightarrow p=q=r[/tex]
[tex]\Rightarrow x=y=z \Rightarrow sin2A=sin2B=sin2B \Rightarrow a=b=c \Rightarrow A=B=C=60^0[/tex]
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