Question

Answer 1

\(sin^6\alpha+cos^6\alpha=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha-sin^2\alpha.cos^2\alpha+cos^4\alpha\right)=sin^4\alpha+2sin^2\alpha.cos^2\alpha+cos^4\alpha-3sin^2\alpha.cos^2\alpha=\left(sin^2\alpha+cos^2\alpha\right)^2-3sin^2\alpha.cos^2\alpha=1-3sin^2\alpha.cos^2\alpha\ge1-3.\dfrac{\left(sin^2\alpha+cos^2\alpha\right)^2}{4}=1-3.\dfrac{1}{4}=\dfrac{1}{4}\left(đpcm\right)\)Dấu "=" xảy ra \(\Leftrightarrow sin^2\alpha=cos^2\alpha=\dfrac{\sqrt{2}}{2}\Leftrightarrow\alpha=45^0\)