Describing sine in terms of cosine

Question

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 1

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 $\sin x$

Answer 2

Because, if $a\geqslant0$, $\sqrt a$ is the non-negative square root of $a$. Therefore, and since $\sin^2\theta=1-\cos^2\theta$, if $\sin\theta\geqslant0$, then $\sin\theta=\sqrt{1-\cos^2\theta}$, and, if $\sin\theta<0$, then $\sin\theta=-\sqrt{1-\cos^2\theta}$.

Answer 3

You answer yourself.

From $$\cos^2\theta+\sin^2\theta=1,$$

$$\sin\theta=\pm\sqrt{1-\cos^2\theta}$$

and if $\sin\theta$ is negative, then...