18: tính giá trị lượng giác của góc a, biết:

19: tính giá trị lượng giác của góc a, biết:

18
a
Vì $0 < \alpha < \frac{\pi}{2}$ nên $\sin\alpha > 0$, $\tan\alpha > 0$, $\cot\alpha > 0$.
Ta có $\sin^2\alpha + \cos^2\alpha = 1$.
$\sin^2\alpha = 1 - \cos^2\alpha = 1 - \left(\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25}$
$\sin\alpha = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}$.
$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{2\sqrt{6}/5}{1/5} = 2\sqrt{6}$
$\cot\alpha = \frac{1}{\tan\alpha} = \frac{1}{2\sqrt{6}} = \frac{\sqrt{6}}{12}$

b
Vì $\frac{\pi}{2} < \alpha < \pi$ nên $\cos\alpha < 0$, $\tan\alpha < 0$, $\cot\alpha < 0$.
Ta có $\sin^2\alpha + \cos^2\alpha = 1$.
$\cos^2\alpha = 1 - \sin^2\alpha = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9}$
$\cos\alpha = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}$.
$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{2/3}{-\sqrt{5}/3} = -\frac{2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}$
$\cot\alpha = \frac{1}{\tan\alpha} = -\frac{\sqrt{5}}{2}$

c)
Vì $\pi < \alpha < \frac{3\pi}{2}$ nên $\sin\alpha < 0$, $\cos\alpha < 0$.
Ta có $1 + \tan^2\alpha = \frac{1}{\cos^2\alpha}$.
$1 + (\sqrt{5})^2 = \frac{1}{\cos^2\alpha}$
$1 + 5 = \frac{1}{\cos^2\alpha} \Leftrightarrow 6 = \frac{1}{\cos^2\alpha}$.
$\cos^2\alpha = \frac{1}{6}$.
Vì $\cos\alpha < 0$ nên $\cos\alpha = -\sqrt{\frac{1}{6}} = -\frac{1}{\sqrt{6}} = -\frac{\sqrt{6}}{6}$.
$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} \Rightarrow \sin\alpha = \tan\alpha \cdot \cos\alpha$
$\sin\alpha = \sqrt{5} \cdot \left(-\frac{\sqrt{6}}{6}\right) = -\frac{\sqrt{30}}{6}$.
$\cot\alpha = \frac{1}{\tan\alpha} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$.

d)
Vì $\frac{3\pi}{2} < \alpha < 2\pi$ nên $\sin\alpha < 0$, $\cos\alpha > 0$.
Ta có $1 + \cot^2\alpha = \frac{1}{\sin^2\alpha}$.
$1 + \left(-\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{\sin^2\alpha}$
$1 + \frac{1}{2} = \frac{1}{\sin^2\alpha} \Leftrightarrow \frac{3}{2} = \frac{1}{\sin^2\alpha}$.
$\sin^2\alpha = \frac{2}{3}$
Vì $\sin\alpha < 0$ nên $\sin\alpha = -\sqrt{\frac{2}{3}} = -\frac{\sqrt{2}}{\sqrt{3}} = -\frac{\sqrt{6}}{3}$.
$\cot\alpha = \frac{\cos\alpha}{\sin\alpha} \Rightarrow \cos\alpha = \cot\alpha \cdot \sin\alpha$.
$\cos\alpha = \left(-\frac{1}{\sqrt{2}}\right) \cdot \left(-\frac{\sqrt{6}}{3}\right) = \frac{\sqrt{6}}{3\sqrt{2}} = \frac{\sqrt{3}}{3}$.
$\tan\alpha = \frac{1}{\cot\alpha} = \frac{1}{-1/\sqrt{2}} = -\sqrt{2}$.

19)
a.
Vì $-\frac{\pi}{2} < \alpha < 0$ nên $\cos\alpha > 0$, $\tan\alpha < 0$, $\cot\alpha < 0$.
Ta có $\cos^2\alpha = 1 - \sin^2\alpha = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9}$
$\cos\alpha = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.
$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1/3}{2\sqrt{2}/3} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$. ($\tan\alpha < 0$ nên có dấu -)
$\tan\alpha = -\frac{\sqrt{2}}{4}$.
$\cot\alpha = \frac{1}{\tan\alpha} = -2\sqrt{2}$

b
Vì $\pi < \alpha < \frac{3\pi}{2}$ nên $\sin\alpha < 0$, $\tan\alpha > 0$, $\cot\alpha > 0$.
Ta có $\sin^2\alpha = 1 - \cos^2\alpha = 1 - (-0.7)^2 = 1 - 0.49 = 0.51$.
$\sin\alpha = -\sqrt{0.51} = -\frac{\sqrt{51}}{10}$.
$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{-\sqrt{51}/10}{-0.7} = \frac{\sqrt{51}/10}{7/10} = \frac{\sqrt{51}}{7}$.
$\cot\alpha = \frac{1}{\tan\alpha} = \frac{7}{\sqrt{51}} = \frac{7\sqrt{51}}{51}$.

c.
Vì $0 < \alpha < \frac{\pi}{2}$ nên $\sin\alpha > 0$, $\cos\alpha > 0$.
Ta có $1 + \tan^2\alpha = \frac{1}{\cos^2\alpha}$.
$1 + 2^2 = \frac{1}{\cos^2\alpha} \Leftrightarrow 5 = \frac{1}{\cos^2\alpha}$.
$\cos^2\alpha = \frac{1}{5}$.
$\cos\alpha = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$.
$\sin\alpha = \tan\alpha \cdot \cos\alpha = 2 \cdot \frac{\sqrt{5}}{5} = \frac{2\sqrt{5}}{5}$.
$\cot\alpha = \frac{1}{\tan\alpha} = \frac{1}{2}$.

`d)`
Vì $\pi < \alpha < \frac{3\pi}{2}$ nên $\sin\alpha < 0$, $\cos\alpha < 0$.
Ta có $1 + \cot^2\alpha = \frac{1}{\sin^2\alpha}$.
$1 + \left(\frac{7}{3}\right)^2 = \frac{1}{\sin^2\alpha}$
$1 + \frac{49}{9} = \frac{1}{\sin^2\alpha} \Leftrightarrow \frac{58}{9} = \frac{1}{\sin^2\alpha}$.
$\sin^2\alpha = \frac{9}{58}$.
$\sin\alpha = -\sqrt{\frac{9}{58}} = -\frac{3}{\sqrt{58}} = -\frac{3\sqrt{58}}{58}$.
$\cos\alpha = \cot\alpha \cdot \sin\alpha = \frac{7}{3} \cdot \left(-\frac{3\sqrt{58}}{58}\right) = -\frac{7\sqrt{58}}{58}$.
$\tan\alpha = \frac{1}{\cot\alpha} = \frac{3}{7}$