Answer
$$\sin x=\pm\sqrt{1-\cos^2 x}$$
Work Step by Step
Following the Pythagorean Identity,
$$\sin^2 x+\cos^2 x=1$$
we can rewrite as follows:
$$\sin^2x=1-\cos^2x$$
To get $\sin x$, we take the square root of both sides:
$$\sqrt{\sin^2x}=\sqrt{1-\cos^2 x}$$
$$|\sin x|=\sqrt{1-\cos^2 x}$$
$$\sin x=\pm\sqrt{1-\cos^2 x}$$ (for $|A|=B$ then $A=\pm B$)