Giải bài 3 trang 155 – SGK môn Đại số lớp 10

Hướng dẫn:

Áp dụng các công thức lượng giác cơ bản:

$\sin^2 \alpha +\cos^2 \alpha =1$

$1+\tan ^2 \alpha=\dfrac 1 {\cos^2 \alpha}\\ 1+\cot ^2 \alpha=\dfrac 1 {\sin^2 \alpha}$

a) Ta có: 

$\begin{align} & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1 \\ & \Leftrightarrow {{\sin }^{2}}\alpha +{{\left( -\dfrac{\sqrt{2}}{3} \right)}^{2}}=1 \\ & \Leftrightarrow {{\sin }^{2}}\alpha =\dfrac{7}{9} \\ & \Leftrightarrow \sin \alpha =\pm \dfrac{\sqrt{7}}{3} \\ \end{align}$

Do $\dfrac{\pi }{2}<\alpha <\pi$ nên $\sin \alpha >0$. Vậy $\sin\alpha =\dfrac{\sqrt{7}}{3}.$

b) Ta có: 

$\begin{align} & 1+{{\tan }^{2}}\alpha =\dfrac{1}{{{\cos }^{2}}\alpha } \\ & \Leftrightarrow \dfrac{1}{{{\cos }^{2}}\alpha }=1+{{\left( 2\sqrt{2} \right)}^{2}} \\ & \Leftrightarrow {{\cos }^{2}}\alpha =\dfrac{1}{9} \\ & \Leftrightarrow \cos \alpha =\pm \dfrac{1}{3} \\ \end{align}$

Do $\pi <\alpha <\dfrac{3\pi }{2}$ nên $\cos \alpha <0$. Vậy $\cos \alpha =-\dfrac{1}{3}.$

c) Ta có: 

$\begin{align} & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1 \\ & \Leftrightarrow {{\left( -\dfrac{2}{3} \right)}^{2}}+{{\cos }^{2}}=1 \\ & \Leftrightarrow {{\cos }^{2}}\alpha =\dfrac{5}{9} \\ & \Leftrightarrow \cos \alpha =\pm \dfrac{\sqrt{5}}{3} \\ \end{align}$

Do $\dfrac{3\pi }{2}<\alpha <2\pi$  nên $\cos \alpha >0$. Vậy $\cos \alpha =\dfrac{\sqrt{5}}{3}.$

Suy ra $\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }=-\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=-\dfrac{2\sqrt{5}}{5} .$

d) Ta có: 

$\begin{align} & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1 \\ & \Leftrightarrow {{\sin }^{2}}\alpha +{{\left( -\dfrac{1}{4} \right)}^{2}}=1 \\ & \Leftrightarrow {{\sin }^{2}}\alpha =\dfrac{15}{16} \\ & \Leftrightarrow \sin \alpha =\pm \dfrac{\sqrt{15}}{4} \\ \end{align}$

Do $\dfrac{\pi }{2}<\alpha <\pi$ nên $\sin \alpha >0$. Vậy $\sin\alpha =\dfrac{\sqrt{15}}{4}$.

Suy ra $\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha }=-\dfrac{1}{4}:\dfrac{\sqrt{15}}{4}=-\dfrac{\sqrt{15}}{15}.$