Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identities
sin²x + cos²x = 1 ⇒sinx = ± [tex]\sqrt{1-cos^2x}[/tex] and
tanx = [tex]\frac{sinx}{cosx}[/tex]
Given
cosΘ = [tex]\frac{\sqrt{2} }{2}[/tex], then
sinΘ = ± [tex]\sqrt{1-(\frac{\sqrt{2} }{2})^2 }[/tex]
Since [tex]\frac{3\pi }{2}[/tex] < Θ < 2π ← fourth quadrant
Then sinΘ and tanΘ are both negative
sinΘ = - [tex]\sqrt{1-\frac{1}{2} }[/tex]
= - [tex]\frac{1}{\sqrt{2} }[/tex] = - [tex]\frac{\sqrt{2} }{2}[/tex]
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tanΘ = [tex]\frac{-\sqrt{2} }{2}[/tex] ÷ [tex]\frac{\sqrt{2} }{2}[/tex]
= - [tex]\frac{\sqrt{2} }{2}[/tex] × [tex]\frac{2}{\sqrt{2} }[/tex] = - 1
