\(\Leftrightarrow\dfrac{1-cos2A}{2}+\dfrac{1-cos2B}{2}+cos^2C=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{2}\left(cos2A+cos2B\right)+cos^2C=0\)
\(\Leftrightarrow\dfrac{1}{4}-cos\left(A+B\right)cos\left(A-B\right)+cos^2C=0\)
\(\Leftrightarrow\dfrac{1}{4}-cos\left(180^0-C\right).cos\left(A-B\right)+cos^2C=0\)
\(\Leftrightarrow1+4cosC.cos\left(A-B\right)+4cos^2C=0\)
\(\Leftrightarrow4cos^2C+4cosC.cos\left(A-B\right)+cos^2\left(A-B\right)+1-cos^2\left(A-B\right)=0\)
\(\Leftrightarrow\left[2cosC+cos\left(A-B\right)\right]^2+sin^2\left(A-B\right)=0\)
Do \(\left\{{}\begin{matrix}\left[2cosC+cos\left(A-B\right)\right]^2\ge0\\sin^2\left(A-B\right)\ge0\end{matrix}\right.\) ; \(\forall A;B;C\)
\(\Rightarrow\left[2cosC+cos\left(A-B\right)\right]^2+sin^2\left(A-B\right)\ge0\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}2cosC+cos\left(A-B\right)=0\\sin\left(A-B\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}A=B\\2cosC+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}A=B\\cosC=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C=120^0\\A=B=30^0\end{matrix}\right.\)
