trigonometry - Describing sine in terms of cosine

The problem says:

Show if $\sin{\theta} = \sqrt{1 - \cos^2{\theta}}$ if $\sin{\theta}$ is positive and $\sin{\theta} = -\sqrt{1 - \cos^2{\theta}}$ if $\sin{\theta}$ is negative.

I know that $\sin^2{\theta} + \cos^2{\theta}=1$ which can be proven with the Pythagorean Theorem. I also understand we can rearrange the above formula to get $\sin{\theta} = \sqrt{1 - \cos^2{\theta}}$, but how to show that if $\sin{\theta}$ is negative, then $\sin{\theta} = -\sqrt{1 - \cos^2{\theta}}$?

I can't bring that together with the Pythagorean Theorem (this was my first thought to use) to proof it.

Answer

Note that if $x^2=16$ we get $x=\pm 4$ where $4$ is positive and $-4$ is negative.

Similarly $$\sin ^2x = 1- \cos ^2 x \implies \sin x = \pm \sqrt {1-\cos ^2 x}$$

where the positive sign is for the positive $\sin x$ and the negative sign for the negative $\sin x$