Question

Cho DABC thỏa điều kiện : \(sin^2A+sin^2B+cos^2C+\frac{1}{4}=2sinA.sinB+cosC.\) Chứng minh rằng DABC đều.

Answer 1

\(sin^2A+sin^2B+cos^2C+\frac{1}{4}=2sinA.sinB+cosC\)

\(\Leftrightarrow sin^2A+sin^2B-2sinA.sinB+\frac{1}{4}\left(4cos^2C-4cosC+1\right)=0\)

\(\Leftrightarrow\left(sinA-sinB\right)^2+\frac{1}{4}\left(2cosC-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}sinA-sinB=0\\2cosC-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=B\\cosC=\frac{1}{2}\Rightarrow C=60^0\end{matrix}\right.\)

\(\Rightarrow A=B=C=60^0\Rightarrow\Delta ABC\) đều