Bài 3 trang 19

Đề bài

Tính các giá trị lượng giác của góc α, nếu:

a)    $\sin \alpha =\frac{5}{13}$ và $\frac{\pi }{2}<\alpha <\pi$

b)    $\cos \alpha =\frac{2}{5}$ và $0<\alpha <{90}^{\circ }$

c)    $\tan \alpha =\sqrt{3}$ và $\pi <\alpha <\frac{3\pi }{2}$

d)    $\cot \alpha =-\frac{1}{2}$ và ${270}^{\circ }<\alpha <{360}^{\circ }$

Lời giải chi tiết

a)    Ta có: ${\sin }^2\alpha +{\cos }^2\alpha =1\Rightarrow {\left(\frac{5}{13}\right)}^2+{\cos }^2\alpha =1\Rightarrow \cos \alpha =\pm \frac{12}{13}$

Do $\frac{\pi }{2}<\alpha <\pi \Rightarrow \cos \alpha =\frac{-12}{13}$$\Rightarrow \{\begin{array}{c}\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{-5}{12}\\ \cot \alpha =\frac{\cos \alpha }{\sin \alpha }=\frac{-12}{5}\end{array}$

b)    Ta có: ${\sin }^2\alpha +{\cos }^2\alpha =1\Rightarrow {\left(\frac{2}{5}\right)}^2+{\sin }^2\alpha =1\Rightarrow \sin \alpha =\pm \frac{\sqrt{21}}{5}$

Do $0<\alpha <{90}^{\circ }\Rightarrow \sin \alpha =\frac{\sqrt{21}}{5}$$\Rightarrow \{\begin{array}{c}\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\sqrt{21}}{2}\\ \cot \alpha =\frac{\cos \alpha }{\sin \alpha }=\frac{2\sqrt{21}}{21}\end{array}$

c)    Ta có: $1+{\tan }^2\alpha =\frac{1}{{\cos }^2\alpha }\Rightarrow 1+{\left(\sqrt{3}\right)}^2=\frac{1}{{\cos }^2\alpha }\Rightarrow \cos \alpha =\pm \frac{1}{2}$

Do $\pi <\alpha <\frac{3\pi }{2}\Rightarrow \cos \alpha =\frac{-1}{2}$ $\Rightarrow \{\begin{array}{c}\cot \alpha .\tan \alpha =1\\ \tan \alpha =\frac{\sin \alpha }{\cos \alpha }\end{array}\Rightarrow \{\begin{array}{c}\cot \alpha =\frac{\sqrt{3}}{3}\\ \sin \alpha =\frac{-\sqrt{3}}{2}\end{array}$

d)    Ta có: $1+{\cot }^2\alpha =\frac{1}{{\sin }^2\alpha }\Rightarrow 1+{\left(-\frac{1}{2}\right)}^2=\frac{1}{{\sin }^2\alpha }\Rightarrow \sin \alpha =\pm \frac{2\sqrt{5}}{5}$

Do ${270}^{\circ }<\alpha <{360}^{\circ }\Rightarrow \sin \alpha =\frac{-2\sqrt{5}}{5}.$

Ta có: $\cot \alpha =-\frac{1}{2}\Rightarrow \tan \alpha =\frac{1}{\cot \alpha }=-2$

Lại có:

$1+{\tan }^2\alpha =\frac{1}{{\cos }^2\alpha }\iff 1+4=\frac{1}{{\cos }^2\alpha }$$\iff {\cos }^2\alpha =\frac{1}{5}\iff \cos \alpha =\pm \frac{\sqrt{5}}{5}$

Do ${270}^o<\alpha <{360}^o\Rightarrow \cos \alpha >0\Rightarrow \cos \alpha =\frac{\sqrt{5}}{5}$