Answer:
Option b
Step-by-step explanation:
By definition we know that:
[tex]cos ^ 2(\theta) = 1- sin ^ 2(\theta)[/tex]
We know that for this case:
[tex]sin(\theta) = \frac{1}{4}[/tex]
Then:
[tex]cos ^ 2(\theta) = 1- (\frac{1}{4})^2\\\\cos ^ 2(\theta) = 1- (\frac{1}{16})\\\\cos ^ 2(\theta) = \frac{15}{16}\\\\[/tex]
Apply square root on both sides of the equation
[tex]\sqrt{cos ^ 2(\theta)} = \sqrt{\frac{15}{16}}\\\\cos(\theta) = \sqrt{\frac{15}{16}}\\\\cos(\theta) = \frac{\sqrt{15}}{4}[/tex]
