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a:
2: pi/2<a<pi
=>sin a>0 và cosa<0
tan a=-2
1+tan^2a=1/cos^2a=1+4=5
=>cos^2a=1/5
=>\(cosa=-\dfrac{1}{\sqrt{5}}\)
\(sina=\sqrt{1-\dfrac{1}{5}}=\dfrac{2}{\sqrt{5}}\)
cot a=1/tan a=-1/2
3: pi<a<3/2pi
=>cosa<0; sin a<0
1+cot^2a=1/sin^2a
=>1/sin^2a=1+9=10
=>sin^2a=1/10
=>\(sina=-\dfrac{1}{\sqrt{10}}\)
\(cosa=-\dfrac{3}{\sqrt{10}}\)
tan a=1:cota=1/3
b;
tan x=-2
=>sin x=-2*cosx
\(A=\dfrac{2\cdot sinx+cosx}{cosx-3sinx}\)
\(=\dfrac{-4cosx+cosx}{cosx+6cosx}=\dfrac{-3}{7}\)
2: tan x=-2
=>sin x=-2*cosx
\(B=\dfrac{-4cosx+3cosx}{-6cosx-2cosx}=\dfrac{1}{8}\)
a)\(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\sin^2\alpha=1-\cos^2\alpha\)
\(\Rightarrow1-2^2=-3\) \(\Rightarrow\cos=-\sqrt{3}\left(0< \alpha< \dfrac{\pi}{2}\right)\)
b) \(\tan\alpha\times\cot\alpha=1\Rightarrow\tan\alpha=\dfrac{1}{\cot\alpha}\Rightarrow\tan=\dfrac{1}{4}\)
a)Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
\(cos\alpha=2sin\alpha\)(1)
Nếu \(sin\alpha=0\Rightarrow cos\alpha\) (vô lý).
Vì vậy \(sin\alpha\ne0\) . Từ (1) \(\Rightarrow\dfrac{cos\alpha}{sin\alpha}=2\)\(\Leftrightarrow cot\alpha=2\).
Suy ra: \(tan\alpha=\dfrac{1}{2}\).
\(sin\alpha=\sqrt{\dfrac{1}{1+cot^2\alpha}}=\dfrac{1}{\sqrt{3}}\).
\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{\dfrac{2}{3}}\).
a) Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha< 0;cot\alpha>0;tan\alpha>0\).
Vì vậy: \(sin\alpha=-\sqrt{1-cos^2\alpha}=\dfrac{-\sqrt{15}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\sqrt{15}}{4}:\dfrac{-1}{4}=\sqrt{15}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{\sqrt{15}}\).
b) Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(cos\alpha< 0;tan\alpha< 0;cot\alpha< 0\).
\(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{5}}{3}\);
\(tan\alpha=\dfrac{2}{3}:\dfrac{-\sqrt{5}}{3}=\dfrac{-2}{\sqrt{5}}\); \(cot\alpha=1:tan\alpha=\dfrac{-\sqrt{5}}{2}\).
\(\left\{{}\begin{matrix}tan\alpha=-\dfrac{7}{3}\\sin^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{sin\alpha}{cos\alpha}=-\dfrac{7}{3}\\sin^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\sin^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\\dfrac{49}{9}cos^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\cos^2\alpha=\dfrac{9}{58}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\cos\alpha=\dfrac{3}{\sqrt{58}}\end{matrix}\right.\) (Vì \(\dfrac{3\pi}{2}< \alpha< 2\pi\Rightarrow cos\alpha>0\))
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{\sqrt{58}}\\cos\alpha=\dfrac{3}{\sqrt{58}}\end{matrix}\right.\)
\(cot\alpha=\dfrac{1}{tan\alpha}=-\dfrac{3}{7}\)
b) Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
Vì vậy:
\(cos\alpha=\sqrt{1-0,6^2}=\dfrac{4}{5}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=0,6:\dfrac{4}{5}=0,75;cot\alpha=1:tan\alpha=\dfrac{4}{3}\).
Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(sin\alpha>0;tan\alpha< 0;cot\alpha< 0\).
\(sin\alpha=\sqrt{1-cos^2\alpha}=\dfrac{\sqrt{51}}{10}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{51}}{10}:\left(-0,7\right)=-\dfrac{\sqrt{51}}{7}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-7}{\sqrt{51}}\).
a) Do 0 < α < nên sinα > 0, tanα > 0, cotα > 0
sinα =
cotα = ; tanα =
b) π < α < nên sinα < 0, cosα < 0, tanα > 0, cotα > 0
cosα = -√(1 - sin2 α) = -√(1 - 0,49) = -√0,51 ≈ -0,7141
tanα ≈ 0,9802; cotα ≈ 1,0202.
c) < α < π nên sinα > 0, cosα < 0, tanα < 0, cotα < 0
cosα = ≈ -0,4229.
sinα =
cotα = -
d) Vì < α < 2π nên sinα < 0, cosα > 0, tanα < 0, cotα < 0
Ta có: tanα =
cosα =
Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha,cos\alpha< 0;tan\alpha,cot\alpha< 0\).
\(cos\left(\alpha-\dfrac{\pi}{2}\right)=cos\left(\dfrac{\pi}{2}-\alpha\right)=sin\alpha< 0\).
\(sin\left(\dfrac{\pi}{2}+\alpha\right)=cos\alpha< 0\).
\(tan\left(\dfrac{3\pi}{2}-\alpha\right)=tan\left(\dfrac{3\pi}{2}-\alpha-2\pi\right)\)\(=tan\left(-\dfrac{\pi}{2}-\alpha\right)\)\(=-tan\left(\dfrac{\pi}{2}+\alpha\right)=cot\left(\alpha\right)>0\).
\(cot\left(\alpha+\pi\right)=cot\left(\alpha\right)>0\).
Với 0 < α < :
a) sin(α - π) < 0; b) cos( - α) < 0;
c) tan(α + π) > 0; d) cot(α + ) < 0
Vì \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
a) \(sin\left(\alpha-\pi\right)=-sin\left(\pi-\alpha\right)=-sin\alpha< 0\).
b) \(cos\left(\dfrac{3\pi}{2}-\alpha\right)=cos\left(\dfrac{3\pi}{2}-\alpha-2\pi\right)=cos\left(-\dfrac{\pi}{2}-\alpha\right)\)
\(=cos\left(\dfrac{\pi}{2}+\alpha\right)=-sin\alpha< 0\).
c) \(tan\left(\alpha+\pi\right)=tan\alpha>0\).
d) \(cot\left(\alpha+\dfrac{\pi}{2}\right)=-tan\alpha< 0\).
`0<α<π/2 => {(sinα>0),(cosα>0):}`
`tanα+cotα=2`
Mà `tanα.cotα=1`
`=>` \(\left\{{}\begin{matrix}tan\text{α}=1\\cot\text{α}=1\end{matrix}\right.\)
`=> α=π/4`
`=> sinα=cosα=\sqrt2/2`