Question

cho tam giác ABC. c/m: cos2A+cos2B-cos2C<=\(\frac{3}{2}\)

Answer 1

\(cos2A+cos2B-cos2C\le\frac{3}{2}\)

\(\Leftrightarrow2cos\left(A+B\right).cos\left(A-B\right)-2cos^2C+1\le\frac{3}{2}\)

\(\Leftrightarrow-cos\left(C\right).cos\left(A-B\right)-cos^2C\le\frac{1}{4}\)

\(\Leftrightarrow4cos^2C+4cosC.cos\left(A-B\right)+1\ge0\)

\(\Leftrightarrow4cos^2C+4cosC.cos\left(A-B\right)+cos^2\left(A-B\right)+sin^2\left(A-B\right)\ge0\)

\(\Leftrightarrow\left(2cosC+cos\left(A-B\right)\right)^2+sin^2\left(A-B\right)\ge0\)(đúng)

Ta có ĐPCM