Answer
$-1\leq sin~θ\leq1$
$0\leq\sqrt {1-cos^2θ}\leq1$
Example:
$θ=225°$:
$sin~225°=-\frac{\sqrt 2}{2}\ne\sqrt {1-cos^2225°}=\sqrt {1-(\frac{\sqrt 2}{2})^2}=\frac{\sqrt 2}{2}$
Work Step by Step
$sin^2θ+cos^2θ=1$
$sin^2θ=1-cos^2θ$
$sin~θ=±\sqrt {1-cos^2θ}$
