Câu 13: \(P=sin^4a+cos^4a\)
\(=\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a\)
\(=1^2-2\cdot\left(\dfrac{\sqrt{2}}{3}\right)^2=1-2\cdot\dfrac{2}{9}=1-\dfrac{4}{9}=\dfrac{5}{9}\)
=>Chọn C
Câu 14:
\(T=sin\alpha+cos\alpha=\dfrac{\sqrt{2}}{2}\cdot sin\left(\alpha+\dfrac{\Omega}{4}\right)\)
mà \(-\dfrac{\sqrt{2}}{2}< =sin\left(\alpha+\dfrac{\Omega}{4}\right)\cdot\dfrac{\sqrt{2}}{2}< =\dfrac{\sqrt{2}}{2}\)
nên \(-\dfrac{\sqrt{2}}{2}< =T< =\dfrac{\sqrt{2}}{2}\)
Dấu '=' xảy ra khi \(sin\left(\alpha+\dfrac{\Omega}{4}\right)=1\)
=>\(\alpha+\dfrac{\Omega}{4}=\dfrac{\Omega}{2}\)
=>\(a=\dfrac{\Omega}{4}=45^0\)
=>\(T=sin45+cos45=\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}=\sqrt{2}\)
=>Chọn A
13.
\(P=sin^4a+cos^4a=\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a\)
\(=1-2\left(sina.cosa\right)^2=1-2.\left(\dfrac{\sqrt{2}}{3}\right)^2=\dfrac{5}{9}\)
14.
\(T=sina+cosa\le\sqrt{2\left(sin^2a+cos^2a\right)}=\sqrt{2}\)
15.
\(S_{AMN}=\dfrac{1}{2}AM.AN.sin\widehat{xAy}=\dfrac{\sqrt{3}}{4}.AM.AN\le\dfrac{\sqrt{3}}{16}\left(AM+AN\right)^2=\dfrac{9\sqrt{3}}{4}\)
