Answer:
[tex]\huge\boxed{\sin\theta=-\dfrac{\sqrt2}{2};\ \tan\theta=-1}[/tex]
Step-by-step explanation:
We have:
[tex]\\cos\theta=\dfrac{\sqrt2}{2},\ \dfrac{3\pi}{2}<\theta<2\pi[/tex]
For sine use:
[tex]\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x[/tex]
Substitute:
[tex]\sin^2\theta=1-\left(\dfrac{\sqrt2}{2}\right)^2\\\\\sin^2\theta=1-\dfrac{(\sqrt2)^2}{2^2}\\\\\sin^2\theta=1-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4}{4}-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4-2}{4}\\\\\sin^2\theta=\dfrac{2}{4}\to\sin\theta=\pm\sqrt{\dfrac{2}{4}}\\\\\sin\theta=\pm\dfrac{\sqrt2}{\sqrt4}\\\\\sin\theta=\pm\dfrac{\sqrt2}{2}[/tex]
θ in IV quadrant, therefore sine is negative.
[tex]\sin\theta=-\dfrac{\sqrt2}{2}[/tex]
For tangent use:
[tex]\tan x=\dfrac{\sin x}{\cos x}[/tex]
Substitute:
[tex]\tan\theta=\dfrac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=-\dfrac{\sqrt2}{2}\cdot\dfrac{2}{\sqrt2}=-1[/tex]
