cho tan\(\alpha\)= \(\dfrac{-7}{3}\) với \(\dfrac{3\pi}{2}< \alpha< 2\pi\). tính các giá trị lượng giác của\(\alpha\)
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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) 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 \(cos\alpha< 0;tan\alpha< 0;cot\alpha< 0\).
\(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{5}}{3}\);
\(tan\alpha=\dfrac{2}{3}:\dfrac{-\sqrt{5}}{3}=\dfrac{-2}{\sqrt{5}}\); \(cot\alpha=1:tan\alpha=\dfrac{-\sqrt{5}}{2}\).
`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`
Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha,cos\alpha< 0;tan\alpha,cot\alpha< 0\).
\(cos\left(\alpha-\dfrac{\pi}{2}\right)=cos\left(\dfrac{\pi}{2}-\alpha\right)=sin\alpha< 0\).
\(sin\left(\dfrac{\pi}{2}+\alpha\right)=cos\alpha< 0\).
\(tan\left(\dfrac{3\pi}{2}-\alpha\right)=tan\left(\dfrac{3\pi}{2}-\alpha-2\pi\right)\)\(=tan\left(-\dfrac{\pi}{2}-\alpha\right)\)\(=-tan\left(\dfrac{\pi}{2}+\alpha\right)=cot\left(\alpha\right)>0\).
\(cot\left(\alpha+\pi\right)=cot\left(\alpha\right)>0\).
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 =
sinα =cosα =
a) Ta có \({\cos ^2}\alpha + {\sin ^2}\alpha \,\,\, = \,1\)
mà \(\sin \alpha = \frac{{\sqrt {15} }}{4}\) nên \({\cos ^2}\alpha + {\left( {\frac{{\sqrt {15} }}{4}} \right)^2}\,\,\, = \,1 \Rightarrow {\cos ^2}\alpha = \frac{1}{{16}}\)
Lại có \(\frac{\pi }{2} < \alpha < \pi \) nên \(\cos \alpha < 0 \Rightarrow \cos \alpha = - \frac{1}{4}\)
Khi đó \(\tan \alpha = \frac{{\sin \alpha }}{{co{\mathop{\rm s}\nolimits} \alpha }} = - \sqrt {15} ;\cot \alpha = \frac{1}{{\tan \alpha }} = - \frac{1}{{\sqrt {15} }}\)
b)
Ta có \({\cos ^2}\alpha + {\sin ^2}\alpha \,\,\, = \,1\)
mà \(\cos \alpha = - \frac{2}{3}\) nên \({\sin ^2}\alpha + {\left( {\frac{{ - 2}}{3}} \right)^2}\,\,\, = \,1 \Rightarrow {\sin ^2}\alpha = \frac{5}{9}\)
Lại có \( - \pi < \alpha < 0\) nên \(\sin \alpha < 0 \Rightarrow \sin \alpha = - \frac{{\sqrt 5 }}{3}\)
Khi đó \(\tan \alpha = \frac{{\sin \alpha }}{{co{\mathop{\rm s}\nolimits} \alpha }} = \frac{{\sqrt 5 }}{2};\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{2}{{\sqrt 5 }}\)
c)
Ta có \(\tan \alpha = 3\) nên
\(\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{3}\)
\(\frac{1}{{{{\cos }^2}\alpha }} = 1 + {\tan ^2}\alpha \,\,\, = \,1 + {3^2} = 10\,\, \Rightarrow {\cos ^2}\alpha = \frac{1}{{10}}\)
Mà \({\cos ^2}\alpha + {\sin ^2}\alpha \,\,\, = \,1 \Rightarrow {\sin ^2}\alpha = \frac{9}{{10}}\)
Với \( - \pi < \alpha < 0\) thì \(\sin \alpha < 0 \Rightarrow \sin \alpha = - \sqrt {\frac{9}{{10}}} \)
Với \( - \pi < \alpha < - \frac{\pi }{2}\) thì \(\cos \alpha < 0 \Rightarrow \cos \alpha = - \sqrt {\frac{1}{{10}}} \)
và \( - \frac{\pi }{2} \le \alpha < 0\) thì \(\cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {\frac{1}{{10}}} \)
d)
Ta có \(\cot \alpha = - 2\) nên
\(\tan \alpha = \frac{1}{{\cot \alpha }} = - \frac{1}{2}\)
\(\frac{1}{{{{\sin }^2}\alpha }} = 1 + co{{\mathop{\rm t}\nolimits} ^2}\alpha \,\,\, = \,1 + {( - 2)^2} = 5\,\, \Rightarrow {\sin ^2}\alpha = \frac{1}{5}\)
Mà \({\cos ^2}\alpha + {\sin ^2}\alpha \,\,\, = \,1 \Rightarrow {\cos ^2}\alpha = \frac{4}{5}\)
Với \(0 < \alpha < \pi \) thì \(\sin \alpha > 0 \Rightarrow \sin \alpha = \sqrt {\frac{1}{5}} \)
Với \(0 < \alpha < \frac{\pi }{2}\) thì \(\cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {\frac{4}{5}} \)
và \(\frac{\pi }{2} \le \alpha < \pi \) thì \(\cos \alpha < 0 \Rightarrow \cos \alpha = - \sqrt {\frac{4}{5}} \)
a) Vì \(0<\alpha <\frac{\pi }{2} \) nên \(\sin \alpha > 0\). Mặt khác, từ \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) suy ra
\(\sin \alpha = \sqrt {1 - {{\cos }^2}a} = \sqrt {1 - \frac{1}{{25}}} = \frac{{2\sqrt 6 }}{5}\)
Do đó, \(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{{2\sqrt 6 }}{5}}}{{\frac{1}{5}}} = 2\sqrt 6 \) và \(\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{{\frac{1}{5}}}{{\frac{{2\sqrt 6 }}{5}}} = \frac{{\sqrt 6 }}{{12}}\)
b) Vì \(\frac{\pi }{2} < \alpha < \pi\) nên \(\cos \alpha < 0\). Mặt khác, từ \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) suy ra
\(\cos \alpha = \sqrt {1 - {{\sin }^2}a} = \sqrt {1 - \frac{4}{9}} = -\frac{{\sqrt 5 }}{3}\)
Do đó, \(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{2}{3}}}{{-\frac{{\sqrt 5 }}{3}}} = -\frac{{2\sqrt 5 }}{5}\) và \(\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{{-\frac{{\sqrt 5 }}{3}}}{{\frac{2}{3}}} = -\frac{{\sqrt 5 }}{2}\)
c) Ta có: \(\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{{\sqrt 5 }}\)
Ta có: \({\tan ^2}\alpha + 1 = \frac{1}{{{{\cos }^2}\alpha }} \Rightarrow {\cos ^2}\alpha = \frac{1}{{{{\tan }^2}\alpha + 1}} = \frac{1}{6} \Rightarrow \cos \alpha = \pm \frac{1}{{\sqrt 6 }}\)
Vì \(\pi < \alpha < \frac{{3\pi }}{2} \Rightarrow \sin \alpha < 0\;\) và \(\,\,\cos \alpha < 0 \Rightarrow \cos \alpha = -\frac{1}{{\sqrt 6 }}\)
Ta có: \(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} \Rightarrow \sin \alpha = \tan \alpha .\cos \alpha = \sqrt 5 .(-\frac{1}{{\sqrt 6 }}) = -\sqrt {\frac{5}{6}} \)
d) Vì \(\cot \alpha = - \frac{1}{{\sqrt 2 }}\;\,\) nên \(\,\,\tan \alpha = \frac{1}{{\cot \alpha }} = - \sqrt 2 \)
Ta có: \({\cot ^2}\alpha + 1 = \frac{1}{{{{\sin }^2}\alpha }} \Rightarrow {\sin ^2}\alpha = \frac{1}{{{{\cot }^2}\alpha + 1}} = \frac{2}{3} \Rightarrow \sin \alpha = \pm \sqrt {\frac{2}{3}} \)
Vì \(\frac{{3\pi }}{2} < \alpha < 2\pi \Rightarrow \sin \alpha < 0 \Rightarrow \sin \alpha = - \sqrt {\frac{2}{3}} \)
Ta có: \(\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }} \Rightarrow \cos \alpha = \cot \alpha .\sin \alpha = \left( { - \frac{1}{{\sqrt 2 }}} \right).\left( { - \sqrt {\frac{2}{3}} } \right) = \frac{{\sqrt 3 }}{3}\)
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 \(sin\alpha>0;tan\alpha< 0;cot\alpha< 0\).
\(sin\alpha=\sqrt{1-cos^2\alpha}=\dfrac{\sqrt{51}}{10}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{51}}{10}:\left(-0,7\right)=-\dfrac{\sqrt{51}}{7}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-7}{\sqrt{51}}\).
\(\left\{{}\begin{matrix}tan\alpha=-\dfrac{7}{3}\\sin^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{sin\alpha}{cos\alpha}=-\dfrac{7}{3}\\sin^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\sin^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\\dfrac{49}{9}cos^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\cos^2\alpha=\dfrac{9}{58}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\cos\alpha=\dfrac{3}{\sqrt{58}}\end{matrix}\right.\) (Vì \(\dfrac{3\pi}{2}< \alpha< 2\pi\Rightarrow cos\alpha>0\))
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{\sqrt{58}}\\cos\alpha=\dfrac{3}{\sqrt{58}}\end{matrix}\right.\)
\(cot\alpha=\dfrac{1}{tan\alpha}=-\dfrac{3}{7}\)