## Trigonometry (11th Edition) Clone

$$\sin x=\pm\sqrt{1-\cos^2 x}$$
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 $A=\pm B$)