Cho góc x có cos x = -1/2 . tính giá trị biểu thức S= 4sin²x + 8tan²x
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Cho biết \(cosx=-\dfrac{1}{2}\)
\(sin^2x+cos^2x=1\Rightarrow sin^2x=1-cos^2x\)
\(\Rightarrow sin^2x=1-\dfrac{1}{4}=\dfrac{3}{4}\)
\(S=4sin^2x+8tan^2x\)
\(\Rightarrow S=4\left(sin^2x+2\dfrac{sin^2x}{cos^2x}\right)\)
\(\Rightarrow S=4\left(\dfrac{3}{4}+2\dfrac{\dfrac{3}{4}}{\dfrac{1}{4}}\right)\)
\(\Rightarrow S=4\left(\dfrac{3}{4}+6\right)\)
\(\Rightarrow S=4.\dfrac{27}{4}=27\)
4.
Gọi H là chân đường cao kẻ từ C xuống đường thẳng d.
Ta có: \(CH=d\left(C;d\right)=\dfrac{\left|-3.2-4.5+4\right|}{\sqrt{3^2+4^2}}=\dfrac{22}{5}\)
Khi đó: \(S_{ABC}=\dfrac{1}{2}CH.AB=\dfrac{1}{2}.\dfrac{22}{5}.AB=15\Rightarrow AB=\dfrac{75}{11}\)
\(\Rightarrow IA=IB=\dfrac{75}{22}\)
Gọi \(A=\left(4m;3m+1\right)\) là điểm cần tìm.
Ta có: \(IA=\dfrac{75}{22}\Leftrightarrow\sqrt{\left(4m-2\right)^2+\left(3m-\dfrac{3}{2}\right)^2}=\dfrac{75}{22}\)
\(\Leftrightarrow\sqrt{25m^2-25m+\dfrac{25}{4}}=\dfrac{75}{22}\)
\(\Leftrightarrow\left|m-\dfrac{1}{2}\right|=\dfrac{15}{22}\)
\(\Leftrightarrow\left[{}\begin{matrix}m-\dfrac{1}{2}=\dfrac{15}{22}\\m-\dfrac{1}{2}=-\dfrac{15}{22}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{13}{11}\\m=-\dfrac{2}{11}\end{matrix}\right.\)
\(m=\dfrac{13}{11}\Rightarrow A=\left(\dfrac{52}{11};\dfrac{50}{11}\right)\Rightarrow B=\left(-\dfrac{8}{11};\dfrac{5}{11}\right)\)
Vậy \(A=\left(\dfrac{52}{11};\dfrac{50}{11}\right);B=\left(-\dfrac{8}{11};\dfrac{5}{11}\right)\)
1.
\(P=\left(m;m+1\right)\) là điểm cần tìm
\(\Rightarrow NP=\sqrt{\left(m-3\right)^2+m^2}=\sqrt{2m^2-6m+9}\)
Ta có: \(NM=NP\)
\(\Leftrightarrow\sqrt{\left(-1-3\right)^2+\left(2-1\right)^2}=\sqrt{2m^2-6m+9}\)
\(\Leftrightarrow m^2-3m-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=4\\m=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}P=\left(4;5\right)\\P=\left(-1;0\right)\end{matrix}\right.\)
Vậy \(P=\left(4;5\right)\) hoặc \(P=\left(-1;0\right)\)
\(0< =sin^2x< =1\)
=>\(-2< =sin^2x-2< =-1\)
=>\(sin^2x-2< 0\)
\(0< =cos^2x< =1\)
=>\(-2< =cos^2x-2< =-1\)
\(\Leftrightarrow cos^2x-2< 0\)
\(\sqrt{sin^4x+4cos^2x}+\sqrt{cos^4x+4\cdot sin^2x}\)
\(=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\cdot\left(1-cos^2x\right)}\)
\(=\sqrt{sin^4x-4sin^xx+4}+\sqrt{cos^4x-4\cdot cos^2x+4}\)
\(=\sqrt{\left(sin^2x-2\right)^2}+\sqrt{\left(cos^2x-2\right)^2}\)
\(=\left|sin^2x-2\right|+\left|cos^2x-2\right|\)
\(=2-sin^2x+2-cos^2x\)
\(=4-\left(sin^2x+cos^2x\right)=4-1=3\)
\(B=cos^2x+sin^2x+tan^2x\)
\(=1+tan^2x\)
\(=\dfrac{1}{cos^2x}=1:\dfrac{1}{4}=4\)
\(tana-cota=2\sqrt{3}\Rightarrow\left(tana-cota\right)^2=12\)
\(\Rightarrow\left(tana+cota\right)^2-4=12\Rightarrow\left(tana+cota\right)^2=16\)
\(\Rightarrow P=4\)
\(sinx+cosx=\dfrac{1}{5}\Rightarrow\left(sinx+cosx\right)^2=\dfrac{1}{25}\)
\(\Rightarrow1+2sinx.cosx=\dfrac{1}{25}\Rightarrow sinx.cosx=-\dfrac{12}{25}\)
\(P=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}=\dfrac{1}{-\dfrac{12}{25}}=-\dfrac{25}{12}\)
Ta có:
\(cos^2x+sin^2x=1\Leftrightarrow sinx=\sqrt{1-cos^2x}=\sqrt{1-\left(-\dfrac{1}{2}\right)^2}=\sqrt{1-\dfrac{1}{4}}=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow tanx=\dfrac{sinx}{cosx}=\dfrac{\dfrac{\sqrt{3}}{2}}{-\dfrac{1}{2}}=-\sqrt{3}\)
Ta tính được S:
\(S=4\cdot\left(\dfrac{\sqrt{3}}{2}\right)^2+8\cdot\left(-\sqrt{3}\right)^2=27\)
\(4sin^2x+8tan^2x=4\left(1-cos^2x\right)+\dfrac{8sin^2x}{cos^2x}\\ =4\left(1-cos^2x\right)+\dfrac{8\left(1-cos^2x\right)}{cos^2x}=4\left[1-\left(-\dfrac{1}{2}\right)^2\right]+\dfrac{8\left[1-\left(-\dfrac{1}{2}\right)^2\right]}{\left(-\dfrac{1}{2}\right)^2}=27\)