Question

Cos (theta) = sqrt2/2, and 3pi/2

Answer 1

Answer:

sin theta= -sqrt2/2

tan theta=-1

Step-by-step explanation:

I just did it.

Answer 2

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

sin²x + cos²x = 1 ⇒sinx = ± $\sqrt{1-cos^2x}$ and

tanx = $\frac{sinx}{cosx}$

Given

cosΘ = $\frac{\sqrt{2} }{2}$, then

sinΘ = ± $\sqrt{1-(\frac{\sqrt{2} }{2})^2 }$

Since $\frac{3\pi }{2}$ < Θ < 2π ← fourth quadrant

Then sinΘ and tanΘ are both negative

sinΘ = - $\sqrt{1-\frac{1}{2} }$

       = - $\frac{1}{\sqrt{2} }$ = - $\frac{\sqrt{2} }{2}$

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tanΘ = $\frac{-\sqrt{2} }{2}$ ÷ $\frac{\sqrt{2} }{2}$

       = - $\frac{\sqrt{2} }{2}$ × $\frac{2}{\sqrt{2} }$ = - 1