Answer:
a. [tex]cos \ x=\frac{\sqrt{5} }{3}[/tex].
Step-by-step explanation:
Given sin x= -2/3.
Squaring both sides, we get
[tex]sin^2x=(-\frac{2}{3})^2[/tex]
[tex]sin^2x=\frac{4}{9}[/tex].
We know,
[tex]cos^2x= 1-sin^2x[/tex]
Plugging the value of [tex]sin^2x[/tex]
[tex]cos^2x= 1-\frac{4}{9}[/tex]
[tex]cos^2x= \frac{1}{1} -\frac{4}{9}[/tex]
[tex]cos^2x= \frac{9}{9} -\frac{4}{9}[/tex]
[tex]cos^2x= \frac{9-4}{9}[/tex]
[tex]cos^2x= \frac{5}{9}[/tex]
Taking square root on both sides, we get
[tex]\sqrt{cos^2x} = \sqrt{\frac{5}{9}}[/tex]
[tex]cos \ x=\frac{\sqrt{5} }{3}[/tex].
Note: In 3rd qadrant cos is positive.
Therefore, [tex]cos \ x=\frac{\sqrt{5} }{3}[/tex].
Answer:
[tex]cos x = \frac{\sqrt{5} }{3}[/tex]
Step-by-step explanation:
We are given sin = -2/3 which falls in the 3rd quadrant and we are to find the value of cos with the help of this given information.
We know that, [tex]cos^2x= 1- sin^2x[/tex]. therefore we will square the given value of sin:
[tex]sin = -\frac{2}{3}[/tex]
[tex]sin^2x = (sin = -\frac{2}{3} )^2[/tex]
[tex]sin^2x= \frac{4}{9}[/tex]
Now substituting this value of [tex]sin^2x[/tex] in the above mentioned formula to get:
[tex]cos^2x= 1- \frac{4}{9}[/tex]
[tex]cos^2x=\frac{5}{9}[/tex]
Taking square root on both the sides to get:
[tex]\sqrt{cos^2x} = \sqrt{\frac{5}{9} }[/tex]
[tex]cos x = \frac{\sqrt{5} }{3}[/tex]
Therefore, [tex]cos x = \frac{\sqrt{5} }{3}[/tex] since it is positive in the 3rd quadrant.
