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 \(\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha }=-\dfrac{1}{4}:\dfrac{\sqrt{15}}{4}=-\dfrac{\sqrt{15}}{15}. \)