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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 \(cos\alpha< 0;tan\alpha< 0;cot\alpha< 0\).
\(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{5}}{3}\);
\(tan\alpha=\dfrac{2}{3}:\dfrac{-\sqrt{5}}{3}=\dfrac{-2}{\sqrt{5}}\); \(cot\alpha=1:tan\alpha=\dfrac{-\sqrt{5}}{2}\).
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 =
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 \(sin\alpha>0;tan\alpha< 0;cot\alpha< 0\).
\(sin\alpha=\sqrt{1-cos^2\alpha}=\dfrac{\sqrt{51}}{10}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{51}}{10}:\left(-0,7\right)=-\dfrac{\sqrt{51}}{7}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-7}{\sqrt{51}}\).
Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha,cos\alpha< 0;tan\alpha,cot\alpha< 0\).
\(cos\left(\alpha-\dfrac{\pi}{2}\right)=cos\left(\dfrac{\pi}{2}-\alpha\right)=sin\alpha< 0\).
\(sin\left(\dfrac{\pi}{2}+\alpha\right)=cos\alpha< 0\).
\(tan\left(\dfrac{3\pi}{2}-\alpha\right)=tan\left(\dfrac{3\pi}{2}-\alpha-2\pi\right)\)\(=tan\left(-\dfrac{\pi}{2}-\alpha\right)\)\(=-tan\left(\dfrac{\pi}{2}+\alpha\right)=cot\left(\alpha\right)>0\).
\(cot\left(\alpha+\pi\right)=cot\left(\alpha\right)>0\).
\(\dfrac{\pi}{2}< a< \pi\) => sina > 0, cosa < 0
cos2a = \(\pm\sqrt{1-sin^22a}=\pm\sqrt{1-\left(\dfrac{5}{9}\right)^2}=\pm\dfrac{2\sqrt{14}}{9}\)
Nếu cos2a thì \(\dfrac{2\sqrt{14}}{9}\) thì
sina \(=\sqrt{\dfrac{1-cos2a}{2}}=\sqrt{\dfrac{1-\dfrac{2\sqrt{14}}{9}}{2}}=\dfrac{\sqrt{9-2\sqrt{14}}}{3\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}}{3\sqrt{2}}=\dfrac{\sqrt{7}-\sqrt{2}}{3\sqrt{2}}=\dfrac{\sqrt{14}-2}{6}\)
Nếu cos2a \(=-\dfrac{2\sqrt{14}}{9}\)
thì sina \(=\sqrt{\dfrac{1cos2a}{2}}=\sqrt{\dfrac{1+\dfrac{2\sqrt{14}}{9}}{2}}=\dfrac{2\sqrt{14}}{6}\)
cosa \(=-\sqrt{\dfrac{1+cos2a}{2}}=-\sqrt{\dfrac{9-2\sqrt{14}}{18}}=\dfrac{2-\sqrt{14}}{6}\)
ta có \(sin^2a+cos^2a=1\Rightarrow sina=\pm\sqrt{1-cos^2a}=\pm\sqrt{1-\left(\dfrac{-\sqrt{5}}{3}\right)^2}=\pm\dfrac{2}{3}\)
vì \(\Pi< a< \dfrac{3\Pi}{2}\Rightarrow sina< 0\) \(\Rightarrow sina=\dfrac{-2}{3}\)
lại có \(tana=\dfrac{sina}{cosa}=\dfrac{\dfrac{-2}{3}}{\dfrac{-\sqrt{5}}{3}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)
Vì \(\pi< a< \dfrac{3\pi}{2}\) nên \(\sin a< 0\) và \(\tan a>0\)
Và \(\cos a=-\dfrac{\sqrt{5}}{3}\) nên \(\sin a=-\dfrac{2}{3}\)
Vậy \(\tan a=\dfrac{2}{\sqrt{5}}\)
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: \(tan\alpha=\dfrac{1}{2}\).
\(sin\alpha=\sqrt{\dfrac{1}{1+cot^2\alpha}}=\dfrac{1}{\sqrt{3}}\).
\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{\dfrac{2}{3}}\).