Answer
$cos~θ=\frac{\sqrt 2}{2}$
$sin~θ=±\frac{\sqrt 2}{2}$
Work Step by Step
$sin^2θ+cos^2θ=1$
$cos^2θ=1-sin^2θ$
$\sqrt 2=\sqrt {4-x^2}$
$\sqrt 2=\sqrt {4-4~sin^2θ}$
$\sqrt 2=\sqrt {4(1-sin^2θ)}$
$\sqrt 2=\sqrt {4~cos^2θ}~~$ (Square both sides)
$2=4cos^2θ$
$cos^2θ=\frac{1}{2}$
$cos~θ=±\frac{1}{\sqrt 2}=±\frac{\sqrt 2}{2}$
Since $-\frac{\pi}{2}\ltθ\lt\frac{\pi}{2}$, $cos~θ=\frac{\sqrt 2}{2}$
$sin^2θ=1-cos^2θ=1-\frac{1}{2}=\frac{1}{2}$
$sin~θ=±\frac{1}{\sqrt 2}=±\frac{\sqrt 2}{2}$
