## Algebra and Trigonometry 10th Edition

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