# How do I rewrite this trigonometric function?

$$\sqrt{1-\sin^{2}x}$$

I know that $1-\sin^{2}x = \cos^{2}x$ but I cannot figure out how this would become just a single $\cos x$.

According to WolframAplha, you can rewrite this as $|\cos x|$ if $x$ is positive. Could someone explain why this is?

• can you find when : $\cos x > 0 , \cos x < 0 , \cos x = 0$ – what'sup Oct 7 '13 at 22:07
• $\cos x > 0$ from $0>x>\pi$, $\cos x < 0$ from $\pi >x>2\pi$, and $\cos x = 0$ at $x=0$ and $x=\pi$. I don't really see the relation except for the fact that $1-\sin^{2}x$ has to be positive. – Kot Oct 7 '13 at 22:11
• @StevenN your inequalities don't make sense to me. – Kaster Oct 7 '13 at 22:16
• Oops, I accidentally used $>$ instead of $<$. – Kot Oct 7 '13 at 22:18
• You can rewrite it as $\lvert\cos x\rvert$ for any $x$, because $\sqrt{a^2}=\lvert a\rvert$. – egreg Oct 7 '13 at 22:20

## 1 Answer

$\sqrt{1-sin^2(x)} = \sqrt{cos^2(x)} = |cos(x)|$ or just $cos(x)$ if the corresponding value is positive. Also, $|cos(x)|$ this just talks about the actual distance of $cos(x_0)$ for some $x_0 \in R$.

• Coincidentally with trigonometry, your name is $\pi$. Must be a conspiracy. – chubakueno Oct 8 '13 at 2:20