Answer:
[tex]\cos(2\theta) = 0[/tex]
Explanation:
Given
[tex]\sin(\theta) = -\frac{\sqrt 2}{2}[/tex]
[tex]270^o < \theta < 360^o[/tex]
Required
[tex]\cos(2\theta)[/tex]
Recall that:
[tex]\sin^2(\theta) + \cos^2(\theta) =1[/tex]
Substitute: [tex]\sin(\theta) = -\frac{\sqrt 2}{2}[/tex]
[tex](-\frac{\sqrt 2}{2})^2 + \cos^2(\theta) =1[/tex]
[tex]\frac{1}{2} + \cos^2(\theta) =1[/tex]
Collect like terms
[tex]\cos^2(\theta) =1-\frac{1}{2}[/tex]
[tex]\cos^2(\theta) =\frac{1}{2}[/tex]
[tex]\cos(2\theta)[/tex] is calculated as:
[tex]\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)[/tex]
So, we have:
[tex]\cos(2\theta) = \frac{1}{2} - (-\frac{\sqrt 2}{2})^2[/tex]
[tex]\cos(2\theta) = \frac{1}{2} - \frac{2}{4}[/tex]
[tex]\cos(2\theta) = \frac{1}{2} - \frac{1}{2}[/tex]
[tex]\cos(2\theta) = 0[/tex]
