Answer:
[tex]cos(\beta)=\frac{\sqrt{6}}{3}}[/tex]
Step-by-step explanation:
we know that
If the terminal ray of β lies in the fourth quadrant
then
sin (β) is negative and cos (β) is positive
Remember the trigonometric identity
[tex]sin^2(\beta)+cos^2(\beta)=1[/tex]
we have
[tex]sin(\beta)=-\frac{\sqrt{3}}{3}[/tex]
substitute
[tex](-\frac{\sqrt{3}}{3})^2+cos^2(\beta)=1[/tex]
[tex]\frac{3}{9}+cos^2(\beta)=1[/tex]
[tex]cos^2(\beta)=1-\frac{3}{9}[/tex]
[tex]cos^2(\beta)=\frac{6}{9}[/tex]
apply root square both sides
[tex]cos(\beta)=\frac{\sqrt{6}}{3}}[/tex]
