Tinh : A=(sin40+cos10)2 -(cos40+sin10)2 +cos140
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Câu 5. Cho x,y dương thỏa mãn \(x+y=\dfrac{1}{2}\).Tìm giá trị nhỏ nhất của
\(P=\dfrac{1}{x}+\dfrac{1}{y}\)
Giải:
\(P=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{\dfrac{1}{2}}{xy}=\dfrac{2}{xy}\)
--> P nhỏ nhất khi \(xy\) lớn nhất
Ta có:
\(x^2+y^2\ge2xy\) ( BĐT AM-GM )
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow1\ge4xy\)
\(\Leftrightarrow xy\le\dfrac{1}{4}\)
\(\Rightarrow P\ge2:\dfrac{1}{4}=8\)
Vậy \(Min_P=8\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A= \(\frac{1}{2}\)[sin(-10)+sin90] +\(\frac{1}{2}\)(sin10+sin90)
A= \(\frac{1}{2}\)(-sin10 +1) +\(\frac{1}{2}\)(sin10 +1)
A=\(\frac{1}{2}\)(-sin10+sin10)+1
A= 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=cos10+cos170+cos40+cos140+cos70+cos110\)
\(A=cos10+cos\left(180-10\right)+cos40+cos\left(180-40\right)+cos70+cos\left(180-70\right)\)
\(A=cos10-cos10+cos40-cos40+cos70-cos70\)
\(A=0\)
\(B=sin5+sin355+sin10+sin350+...+sin175+sin185+sin360\)
\(B=sin5+sin\left(360-5\right)+sin10+sin\left(360-10\right)+...+sin175+sin\left(360-175\right)+sin360\)
\(B=sin5-sin5+sin10-sin10+...+sin175-sin175+sin360\)
\(B=sin360=0\)
\(C=cos^22+cos^288+cos^24+cos^284+...+cos^244+cos^246\)
\(C=cos^22+cos^2\left(90-2\right)+cos^24+cos^2\left(90-4\right)+...+cos^244+cos^2\left(90-44\right)\)
\(C=cos^22+sin^22+cos^24+sin^24+...+cos^244+sin^244\)
\(C=1+1+...+1\) (có \(\frac{44-2}{2}+1=22\) số 1)
\(\Rightarrow C=22\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\dfrac{\cos10^0-\sqrt{3}\sin10^0}{\sin10^0\cos10^0}\)
\(=\dfrac{4\left(\dfrac{1}{2}cos10^0-\dfrac{\sqrt{3}}{2}sin10^0\right)}{2sin10^0cos10^0}=\dfrac{4\left(s\text{in3}0^0cos10^0-cos30^0s\text{in}10^0\right)}{sin20^0}=\dfrac{4sin\left(30^0-10^0\right)}{s\text{in2}0^0}=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\sin10^0+\sin40^0-\cos50^0-\cos80^0\)
\(=\left(\sin10^0-\cos80^0\right)+\left(\sin40^0-\cos50^0\right)\)
\(=\left(\cos80^0-\cos80^0\right)+\left(\cos50^0-\cos50^0\right)\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=cos20.cos40.cos60.cos80\)
\(A.sin20=sin20.cos20.cos40.cos60.cos80\)
\(Asin20=\frac{1}{2}sin40.cos40.cos80.cos60\)
\(Asin20=\frac{1}{4}sin80.cos80.cos60\)
\(Asin20=\frac{1}{8}sin160.cos60\)
\(Asin20=\frac{1}{8}sin20.cos60\)
\(A=\frac{1}{8}cos60=\frac{1}{16}\)
\(B=sin10.cos40.cos20\)
\(Bcos10=sin10.cos10.cos20.cos40\)
\(Bcos10=\frac{1}{2}sin20.cos20.cos40\)
\(Bcos10=\frac{1}{4}sin40.cos40\)
\(Bcos10=\frac{1}{8}sin80=\frac{1}{8}cos10\)
\(B=\frac{1}{8}\)
A=sin240+cos210+2sin40cos10-cos240-sin210-2sin10cos40+cos(90+50)
A=(sin240-cos240)+(cos210-sin210)+2(sin40cos10-cos40sin10)-sin50
A=(sin40-cos40)(sin40+cos40)-(sin10-cos10)(sin10+cos10)+1-sin50
A=\(\sqrt{2}\) sin(40-\(\frac{\pi}{4}\))\(\sqrt{2}\) cos(40-\(\frac{\pi}{4}\))-\(\sqrt{2}\)sin(10-\(\frac{\pi}{4}\))\(\sqrt{2}\) cos(10-\(\frac{\pi}{4}\))+1-sin50
A=-2sin5cos5+2sin35cos35+1-sin50
A= - sin10+sin70+1-sin50
A= 2cos40sin30-sin(90-40)+1
A=cos40-cos40+1 =1