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max, min :
15 tháng 9 2017
dùng ông thức hạ bậc
cos2a=\(\dfrac{1+cos2a}{2}\)
pt<=>1+cos(4x+\(\dfrac{2\Pi}{3}\))-3sin(2x+\(\dfrac{5\Pi}{6}\))+1=0
<=>-\(\dfrac{1}{2}\)cos4x-\(\dfrac{\sqrt{3}}{2}\)sin4x+\(\dfrac{3\sqrt{3}}{2}\)sin2x-\(\dfrac{3}{2}\)cos2x+2=0
<=>(-\(\dfrac{1}{2}\)cos4x+\(\dfrac{3\sqrt{3}}{2}\)sin2x+2)+(-\(\sqrt{3}\)sin2x.cos2x-\(\dfrac{3}{2}\)cos2x)=0
<=>[-\(\dfrac{1}{2}\)(1-2sin22x)+\(\dfrac{3\sqrt{3}}{2}\)sin2x+2)-cos2x.(\(\sqrt{3}\)sin2x+\(\dfrac{3}{2}\))=0
<=>(sin22x+\(\dfrac{3\sqrt{3}}{2}\)sin2x+\(\dfrac{3}{2}\))-cos2x.(\(\sqrt{3}\)sin2x+\(\dfrac{3}{2}\))=0
<=>(sin2x+\(\dfrac{\sqrt{3}}{2}\))(sin2x+\(\sqrt{3}\))-cos2x.(sin2x+\(\dfrac{\sqrt{3}}{2}\))=0
<=>(sin2x+\(\dfrac{\sqrt{3}}{2}\))(sin2x-cos2x+\(\sqrt{3}\))=0
tới đây bạn tự giải nhé
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 3 2023
Đặt \(x=\sqrt{\dfrac{a}{bc}}\) ; \(y=\sqrt{\dfrac{b}{ca}}\) ; \(z=\sqrt{\dfrac{c}{ab}}\)
\(\Rightarrow a=\dfrac{1}{yz}\) ; \(b=\dfrac{1}{zx}\) ; \(c=\dfrac{1}{xy}\)
\(\Rightarrow xy+yz+zx=1\)
Khi đó, tồn tại một tam giác ABC sao cho:
\(x=tan\dfrac{A}{2}\) ; \(y=tan\dfrac{B}{2}\) ; \(z=tan\dfrac{C}{2}\)
Thay vào bài toán:
\(A=\dfrac{x^2}{1+x^2}+\sqrt{3}\left(\dfrac{y^2}{1+y^2}+\dfrac{z^2}{1+z^2}\right)\)
\(=\dfrac{tan^2\dfrac{A}{2}}{1+tan^2\dfrac{A}{2}}+\sqrt{3}\left(\dfrac{tan^2\dfrac{B}{2}}{1+tan^2\dfrac{B}{2}}+\dfrac{tan^2\dfrac{C}{2}}{1+tan^2\dfrac{C}{2}}\right)\)
\(=sin^2\dfrac{A}{2}+\sqrt{3}\left(sin^2\dfrac{B}{2}+sin^2\dfrac{C}{2}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}cosA+\dfrac{\sqrt{3}}{2}\left(2-cosB-cosC\right)\)
\(=\dfrac{1+2\sqrt{3}}{2}-\dfrac{1}{2}\left(cosA+\sqrt{3}cosB+\sqrt{3}cosC\right)\)
Xét \(B=cosA+\sqrt{3}\left(cosB+cosC\right)=cosA+2\sqrt{3}cos\dfrac{B+C}{2}cos\dfrac{B-C}{2}\)
\(\le cosA+2\sqrt{3}cos\dfrac{B+C}{2}=-2sin^2\dfrac{A}{2}+2\sqrt{3}sin\dfrac{A}{2}+1\)
Xét hàm \(f\left(t\right)=-2t^2+2\sqrt{3}sint+1\) với \(t\in\left(0;1\right)\)
\(f'\left(t\right)=-4t+2\sqrt{3}=0\Rightarrow t=\dfrac{\sqrt{3}}{2}\)
\(f\left(0\right)=1\) ; \(f\left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{5}{2}\) ; \(f\left(1\right)=2\sqrt{3}-1\)
\(\Rightarrow B_{max}=\dfrac{5}{2}\)
\(\Rightarrow A\ge\dfrac{1+2\sqrt{3}}{2}-\dfrac{5}{4}=\dfrac{4\sqrt{3}-3}{4}\)
HP
29 tháng 8 2021
\(y=2sin^2x+3sinx.cosx+cos^2x\)
\(=-\left(1-2sin^2x\right)+\dfrac{3}{2}sin2x+\dfrac{1}{2}\left(2cos^2x-1\right)+\dfrac{1}{2}\)
\(=-cos2x+\dfrac{3}{2}sin2x+\dfrac{1}{2}cos2x+\dfrac{1}{2}\)
\(=\dfrac{3}{2}sin2x-\dfrac{1}{2}cos2x+\dfrac{1}{2}\)
\(=\dfrac{\sqrt{10}}{2}\left(\dfrac{3}{\sqrt{10}}sin2x-\dfrac{1}{\sqrt{10}}cos2x\right)+\dfrac{1}{2}\)
\(=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\)
Vì \(sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)\in\left[-1;1\right]\)
\(\Rightarrow y=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\in\left[-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2};\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\right]\)
\(\Rightarrow y_{min}=-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=-1\Leftrightarrow...\)
\(y_{max}=\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=1\Leftrightarrow...\)
18 tháng 8 2021
Tham khảo: tìm GTLN - GTNN của hàm số : y=sinx cosx sinxcosx - Hoc24
Đặt
Xét hàm
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2021
Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=t\Rightarrow t\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)
\(\Rightarrow y=t+\dfrac{t^2-1}{2}=\dfrac{1}{2}t^2+t-\dfrac{1}{2}\)
Xét hàm \(y=f\left(t\right)=\dfrac{1}{2}t^2+t-\dfrac{1}{2}\) trên \(\left[-\sqrt{2};\sqrt{2}\right]\)
\(-\dfrac{b}{2a}=-1\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(f\left(-\sqrt{2}\right)=\dfrac{1-2\sqrt{2}}{2}\) ; \(f\left(-1\right)=-1\) ; \(f\left(\sqrt{2}\right)=\dfrac{1+2\sqrt{2}}{2}\)
\(\Rightarrow y_{min}=-1\) ; \(y_{max}=\dfrac{1+2\sqrt{2}}{2}\)
HP
18 tháng 9 2021
\(y=sin\left(x+\dfrac{\pi}{3}\right)-sinx\)
\(=\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx-sinx\)
\(=\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx\)
\(=cos\left(x+\dfrac{\pi}{6}\right)\in\left[-1;1\right]\)
\(\Rightarrow\left\{{}\begin{matrix}y_{mịn}=-1\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\\y_{max}=1\Leftrightarrow x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
NL
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