Tìm GTLN của P = \(\sin^4x+\cos^4x+\sin x\cos x\)
Tìm GTLN, GTNN:
a, \(y=4\sin^2x-4\sin x+3\).
b, \(y=\cos^2x+2\sin x+2\).
c, \(y=\sin^4x-2\cos^2x+1\).
a.
Tìm min:
$y=(4\sin ^2x-4\sin x+1)+2=(2\sin x-1)^2+2$
Vì $(2\sin x-1)^2\geq 0$ với mọi $x$ nên $y=(2\sin x-1)^2+2\geq 0+2=2$
Vậy $y_{\min}=2$
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Mặt khác:
$y=4\sin x(\sin x+1)-8(\sin x+1)+11$
$=(\sin x+1)(4\sin x-8)+11$
$=4(\sin x+1)(\sin x-2)+11$
Vì $\sin x\in [-1;1]\Rightarrow \sin x+1\geq 0; \sin x-2<0$
$\Rightarrow 4(\sin x+1)(\sin x-2)\leq 0$
$\Rightarrow y=4(\sin x+1)(\sin x-2)+11\leq 11$
Vậy $y_{\max}=11$
b.
$y=\cos ^2x+2\sin x+2=1-\sin ^2x+2\sin x+2$
$=3-\sin ^2x+2\sin x$
$=4-(\sin ^2x-2\sin x+1)=4-(\sin x-1)^2\leq 4-0=4$
Vậy $y_{\max}=4$.
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Mặt khác:
$y=3-\sin ^2x+2\sin x = (1-\sin ^2x)+(2+2\sin x)$
$=(1-\sin x)(1+\sin x)+2(1+\sin x)=(1+\sin x)(1-\sin x+2)$
$=(1+\sin x)(3-\sin x)$
Vì $\sin x\in [-1;1]$ nên $1+\sin x\geq 0; 3-\sin x>0$
$\Rightarrow y=(1+\sin x)(3-\sin x)\geq 0$
Vậy $y_{\min}=0$
c.
$y=\sin ^4x-2\cos ^2x+1=\sin ^4x-2(1-\sin ^2x)+1$
$=\sin ^4x+2\sin ^2x-1$
$=(\sin ^4x-1)+(2\sin ^2x-2)+2$
$=(\sin ^2x-1)(\sin ^2x+1)+2(\sin ^2x-1)+2$
$=(\sin ^2x-1)(\sin ^2x+3)+2$
Vì $\sin x\in [-1;1]$ nên $\sin ^2x\leq 1$
$\Rightarrow (\sin ^2x-1)(\sin ^2x+3)\leq 0$
$\Rightarrow y=(\sin ^2x-1)(\sin ^2x+3)+2\leq 2$
Vậy $y_{\max}=2$
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$y=\sin ^4x+2\sin ^2x-1=\sin ^2x(\sin ^2x+2)-1$
Vì $\sin ^2x\geq 0$ nên $\sin ^2x(\sin ^2x+2)\geq 0$
$\Rightarrow y=\sin ^2x(\sin ^2x+2)-1\geq 0-1=-1$
Vậy $y_{\min}=-1$
tìm GTLN và GTNN
1. y=\(2\sin^3x+\sin x\)
2. y=\(\cos^2x-2\sin x\)
3. y=\(\sin^2x+\cos^4x\)
4. y=\(\sin^4x+\cos^4x+\sin x\times\cos x\)
1. Ta có: \(-1\le sinx\le1\)
\(\Rightarrow-3\le y\le3\) (hàm đã cho đồng biến trên \(\left[-\frac{\pi}{2};\frac{\pi}{2}\right]\)
\(y_{min}=-3\) khi \(sinx=-1\)
\(y_{max}=3\) khi \(sinx=1\)
2.
\(y=1-sin^2x-2sinx=2-\left(sinx+1\right)^2\)
Do \(-1\le sinx\le1\Rightarrow0\le sinx+1\le2\)
\(\Rightarrow-2\le y\le2\)
\(y_{min}=-2\) khi \(sinx=1\)
\(y_{max}=2\) khi \(sinx=-1\)
3.
\(y=1-cos^2x+cos^4x=\left(cos^2x-\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow y\ge\frac{3}{4}\Rightarrow y_{min}=\frac{3}{4}\) khi \(cos^2x=\frac{1}{2}\)
\(y=1+cos^2x\left(cos^2x-1\right)\le1\) do \(cos^2x-1\le0\)
\(\Rightarrow y_{max}=1\) khi \(\left[{}\begin{matrix}cos^2x=1\\cos^2x=0\end{matrix}\right.\)
4.
\(y=\left(sin^2x+cos^2x\right)^2-2\left(sinx.cosx\right)^2+sinx.cosx\)
\(y=1-\frac{1}{2}sin^22x+\frac{1}{2}sin2x\)
\(y=\frac{9}{8}-\frac{1}{2}\left(sinx-\frac{1}{2}\right)^2\le\frac{9}{8}\)
\(y_{max}=\frac{9}{8}\) khi \(sinx=\frac{1}{2}\)
\(y=\frac{1}{2}\left(sinx+1\right)\left(2-sinx\right)\ge0;\forall x\)
\(\Rightarrow y_{min}=0\) khi \(sinx=-1\)
tìm GTLN và GTNN
1.y=\(3\sin^2x-2\)
2.y=\(2\sin^3x+\sin x\)
3.y=\(\cos^2x-2\sin x\)
4.y=\(\sin^2x+\cos^4x\)
5.y=\(\sin^4x+\cos^4x+\sin x\times\cos x\)
Tìm GTLN - GTNN
1 . \(y=S\times\left(1-\frac{S^2-1}{2}\right)\)
2. \(y=\sin^4x+\cos^4x\)
3.\(y=\sin^6+\cos^6\)
4.\(y=\frac{\cos x+2\sin x+3}{2\cos x-\sin x+4}\)
Tìm GTNN và GTLN của hàm số sau:
1.\(y=cosx+cos\left(x-\dfrac{\pi}{3}\right)\)
2.\(y=sin^4x+cos^4x\)
3.\(y=3-2\left|sinx\right|\)
2.
$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$
$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$
Vì: $0\leq \sin ^22x\leq 1$
$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$
Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$
3.
$0\leq |\sin x|\leq 1$
$\Rightarrow 3\geq 3-2|\sin x|\geq 1$
Vậy $y_{\min}=1; y_{\max}=3$
1.
\(y=\cos x+\cos (x-\frac{\pi}{3})=\cos x+\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)
\(=\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)
\(y^2=(\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)^2\leq (\cos ^2x+\sin ^2x)(\frac{9}{4}+\frac{3}{4})\)
\(\Leftrightarrow y^2\leq 3\Rightarrow -\sqrt{3}\leq y\leq \sqrt{3}\)
Vậy $y_{\min}=-\sqrt{3}; y_{max}=\sqrt{3}$
tìm gtnn của (x nhọn)
A=sin^4x+cos ^4x
B=sin^8x+cos^8x
Tìm GTLN - GTNN
1. \(y=S\times\left(1-\frac{S^2-1}{2}\right)\)
2. \(y=\sin^4x+\cos^4x\)
3.\(y=\sin^6+\cos^6\)
4.\(y=\frac{\cos x+2\sin x+3}{2\cos x-\sin x+4}\)
Câu 1:
\(y=S\left(\frac{3-S^2}{2}\right)=\frac{3}{2}S-\frac{1}{2}S^3\)
Khi \(S\rightarrow+\infty\) thì \(y\rightarrow-\infty\)
Khi \(S\rightarrow-\infty\) thì \(y\rightarrow+\infty\)
Hàm số không có GTLN và GTNN
Câu 2:
\(y=sin^4x+cos^4x+2sin^2x.cos^2x-2sin^2x.cos^2x\)
\(y=\left(sin^2x+cos^2x\right)^2-\frac{1}{2}\left(2sinx.cosx\right)^2\)
\(y=1-\frac{1}{2}sin^22x\)
Do \(0\le sin^22x\le1\)
\(\Rightarrow y_{max}=1\) khi \(sin2x=0\)
\(y_{min}=\frac{1}{2}\) khi \(sin2x=\pm1\)
Câu 3:
\(y=sin^6x+cos^6x+3sin^2x.cos^2x\left(sin^2x+cos^2x\right)-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)
\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\)
\(y=1-\frac{3}{4}sin^22x\)
Do \(0\le sin^22x\le1\)
\(\Rightarrow y_{max}=1\) khi \(sin2x=0\)
\(y_{min}=\frac{1}{4}\) khi \(sin2x=\pm1\)
Câu 4:
\(y=\frac{cosx+2sinx+3}{2cosx-sinx+4}\)
\(\Leftrightarrow2y.cosx-y.sinx+4y=cosx+2sinx+3\)
\(\Leftrightarrow\left(y+2\right)sinx+\left(1-2y\right)cosx=4y-3\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(\left(y+2\right)^2+\left(1-2y\right)^2\ge\left(4y-3\right)^2\)
\(\Leftrightarrow11y^2-24y+4\le0\)
\(\Leftrightarrow\frac{2}{11}\le y\le2\)
Chứng minh rằng với \(0^0\le x\le180^0\) ta có :
a) \(\left(\sin x+\cos x\right)^2=1+2\sin x\cos x\)
b) \(\left(\sin x-\cos x\right)^2=1-2\sin x\cos x\)
c) \(\sin^4x+\cos^4x=1-2\sin^2x\cos^2x\)
a) \(\left(sinx+cosx\right)^2=sin^2x+2sinxcosx+cos^2x\)\(=1+2sinxcosx\).
b) \(\left(sinx-cosx\right)^2=sin^2x-2sinxcosx+cos^2x\)\(=1-2sinxcosx\).
c) \(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\)
\(=1-2sin^2xcos^2x\).
Biết . \(\sin x+\cos x=\sqrt{2}\). Hỏi giá trị của \(\sin^4x+\cos^4x\)
\(sinx+cosx=\sqrt{2}\)
\(\Leftrightarrow\left(sinx+cosx\right)^2=2\)
\(\Leftrightarrow sin^2x+cos^2x+2.sinx.cosx=2\)
\(\Leftrightarrow1+2.sinx.cosx=2\)
\(\Leftrightarrow2.sinx.cosx=1\)
Khi đó \(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2.sinx.cosx=1^2-1=0\)