\(y=\sin^4x+\cos^4x\\ =1-2\sin^2x\cdot\cos^2x\\ =1-\dfrac{1}{2}\sin^22x\\ 0\le\sin^22x\le1\\ \Leftrightarrow\dfrac{1}{2}\le y\le1\\ y_{min}=\dfrac{1}{2}\Leftrightarrow\sin^22x=1\Leftrightarrow x=\dfrac{k\pi}{2}\pm\dfrac{\pi}{4}\\ y_{max}=1\Leftrightarrow\sin^22x=0\Leftrightarrow x=k\pi\)
\(y=3\sin x+4\cos x\\ =5\left(\dfrac{3\sin x}{5}+\dfrac{4\cos x}{5}\right)\\ =5\cos\left(x-a\right),\forall\cos a=\dfrac{4}{5},\sin a=\dfrac{3}{5}\\ -1\le\cos\left(x-a\right)\le1\\ \Leftrightarrow-5\le y\le5\\ y_{min}=-5\Leftrightarrow\cos\left(x-a\right)=-1\\ y_{max}=5\Leftrightarrow\cos\left(x-a\right)=1\)
\(y=sin^4x+cos^4x\)
Ta có: \(0\le sin^4x\le1\)
\(0\le cos^4x\le1\)
\(0\le sin^4x+cos^4x\le2\)
Vây GTNN là 0, GTLN là 2
y=3sinx+4cosx
\(-3\le3sinx\le3\\ -4\le4cosx\le4\\ -7\le3sinx+4cosx\le7\)
Vậy GTNN là -7, GTLN là 7
Lời giải:
1.
$y=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$
$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}.\sin ^22x$
Vì $\sin 2x\in [-1;1]\Rightarrow \sin ^22x\in [0;1]$
$\Rightarrow y\in [\frac{1}{2};1]$
Vậy $y_{\min}=\frac{1}{2}$ và $y_{\max}=1$
2.
Áp dụng BĐT Bunhiacopxky:
$y^2=(3\sin x+4\cos x)^2\leq (\sin ^2x+\cos ^2x)(3^2+4^2)$
$\Leftrightarrow y^2\leq 25$
$\Leftrightarrow -5\leq y\leq 5$
Vậy $y_{\min}=-5$ và $y_{\max}=5$
3/
\(y=\cos (2x+\frac{\pi}{4})-\cos (2x-\frac{\pi}{4})=-2\sin 2x\sin \frac{\pi}{4}=-\sqrt{2}\sin 2x\)
Vì $\sin 2x\in [-1;1]$
$\Rightarrow y\in [-\sqrt{2}; \sqrt{2}]$
Vậy $y_{\min}=-\sqrt{2}; y_{\max}=\sqrt{2}$