Giải bài 4 trang 148 – SGK môn Đại số lớp 10

Hướng dẫn:

Áp dụng các công thức lượng giác cơ bản như:

$1) \sin^2\alpha +\cos^2\alpha =1\\ 2) \tan \alpha =\dfrac{\sin \alpha}{\cos \alpha}\\ 3)\cot \alpha =\dfrac{\cos \alpha}{\sin \alpha}\\ 4) \tan \alpha.\cot \alpha =1\\ 5) \dfrac{1}{\cos^2 \alpha}=1+\tan^2\alpha\\ 6) \dfrac{1}{\sin^2 \alpha}=1+\cot^2\alpha\\$

a) 

Vì $0<\alpha <\dfrac{\pi}{2}$ nên $\sin \alpha > 0$

Ta có: 

$\begin{aligned} & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1 \\ & \Leftrightarrow {{\sin }^{2}}\alpha =1-{{\cos }^{2}}\alpha =1-\dfrac{16}{169}=\dfrac{153}{169} \\ & \Rightarrow \sin \alpha =\dfrac{2\sqrt{17}}{13}\,\,\left( \text{vì}\,\sin \alpha >0 \right) \\ \end{aligned}$

Suy ra $\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }=\dfrac{3\sqrt{17}}{4};\cot \alpha =\dfrac{4}{3\sqrt{17}}$

b) Vì $\pi <\alpha <\dfrac{3\pi }{2}\Rightarrow \cos \alpha <0$

Ta có:

$\begin{aligned} & {{\cos }^{2}}\alpha =1-0,49=0,51 \\ & \Rightarrow \cos \alpha \approx -0,71 \\ \end{aligned}$

Suy ra $\tan \alpha \approx 0,99;\cot \alpha \approx 1,01$

c) 

Vì $\dfrac{\pi }{2}<\alpha <\pi \Rightarrow \left\{ \begin{aligned} & \sin \alpha >0 \\ & \cos \alpha <0 \\ \end{aligned} \right.$

Ta có:

$\begin{aligned} & \dfrac{1}{{{\cos }^{2}}\alpha }=1+{{\tan }^{2}}\alpha \\ & \Rightarrow {{\cos }^{2}}\alpha =\dfrac{1}{1+{{\tan }^{2}}\alpha }=\dfrac{49}{279} \\ & \Rightarrow \cos \alpha =-\dfrac{7}{\sqrt{274}} \\ \end{aligned}$

Do đó:

$\sin \alpha =\tan \alpha .\cos\alpha =\dfrac{15}{\sqrt{274}};\cot \alpha =-\dfrac{7}{15}$

d)

Vì $\dfrac{3\pi }{2}<\alpha <2\pi \Rightarrow \sin \alpha <0;\cos\alpha >0$

$\begin{aligned} & \dfrac{1}{{{\sin }^{2}}\alpha }=1+{{\cot }^{2}}\alpha \\ & \Rightarrow {{\sin }^{2}}\alpha =\dfrac{1}{1+{{\cot }^{2}}\alpha } \\ & \Rightarrow \sin \alpha =-\dfrac{1}{\sqrt{10}} \\ \end{aligned}$
Do đó:
$\cos \alpha =\cot \alpha .\sin\alpha =\dfrac{3}{\sqrt{10}};\tan \alpha =-\dfrac{1}{3}$