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.
$\endgroup$ 23 Answers
$\begingroup$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 $\sin\theta=-\sqrt{1-\cos^2\theta}$.
$\endgroup$ 2 $\begingroup$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 $\sin x$
$\endgroup$ $\begingroup$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...
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