Answer:
[tex]cos \beta = -\sqrt\dfrac{2}{3}[/tex]
Step-by-step explanation:
It is given that [tex]\angle \beta[/tex] lies in the third quadrant.
In 3rd quadrant, sine and cosine both are negative.
Also given that:
[tex]sin \beta =-\dfrac{\sqrt3}{3}[/tex]
We know that identity between sine and cosine as:
[tex]sin^2\theta + cos^2\theta = 1[/tex]
Here, [tex]\angle \theta\ is\ \angle \beta[/tex]
Therefore, the identity can be written as:
[tex]sin^2\beta + cos^2\beta= 1[/tex]
Putting the value of [tex]sin\beta[/tex]:
[tex](-\dfrac{\sqrt3}{3})^2+ cos^2\beta= 1\\\Rightarrow cos^2\beta = 1 -\dfrac{3}{9}\\\Rightarrow cos^2\beta = \dfrac{9-3}{9}\\\Rightarrow cos^2\beta = \dfrac{6}{9}\\\Rightarrow cos\beta = \pm \sqrt{\dfrac{6}{9}}\\\Rightarrow cos\beta = \pm \sqrt{\dfrac{2}{3}}[/tex]
But, it is given that [tex]\beta[/tex] is in 3rd quadrant. That means, value of cos[tex]\beta[/tex] will be negative.
Therefore, the correct answer is:
[tex]cos \beta = -\sqrt\dfrac{2}{3}[/tex]
