Question

If the terminal ray of β lies in the third quadrant and the value of sin(β) is shown below. Determine cos(β) in the simplest form. Please show your calculations that lead to your answer.
sin(B)=-√3/3

Answer 1

Answer:

$cos \beta = -\sqrt\dfrac{2}{3}$

Step-by-step explanation:

It is given that $\angle \beta$ lies in the third quadrant.

In 3rd quadrant, sine and cosine both are negative.

Also given that:

$sin \beta =-\dfrac{\sqrt3}{3}$

We know that identity between sine and cosine as:

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

Here, $\angle \theta\ is\ \angle \beta$

Therefore, the identity can be written as:

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

Putting the value of $sin\beta$:

$(-\dfrac{\sqrt3}{3})^2+ cos^2\beta= 1\\\Rightarrow cos^2\beta = 1 -\dfrac{3}{9}\\\Rightarrow cos^2\beta = \dfrac{9-3}{9}\\\Rightarrow cos^2\beta = \dfrac{6}{9}\\\Rightarrow cos\beta = \pm \sqrt{\dfrac{6}{9}}\\\Rightarrow cos\beta = \pm \sqrt{\dfrac{2}{3}}$

But, it is given that $\beta$ is in 3rd quadrant. That means, value of cos$\beta$ will be negative.

Therefore, the correct answer is:

$cos \beta = -\sqrt\dfrac{2}{3}$