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What is the difference between these functions? $$f(x)= \frac {\sqrt{\frac12(1-\cos(2x))}}{x}\qquad g(x)=\sqrt{2} \cdot\sqrt{ \frac{1-\cos(2x)}{4x^{2} } }$$

Manipulating $g(x)$ to get $f(x)$:

$$\begin{align} g(x)&=\sqrt{2} \cdot\sqrt{ \frac{1-\cos(2x)}{4x^{2}}} \tag1\\[4pt] &=\sqrt{ \frac{(1-\cos(2x))}{2\cdot x^{2}}} \tag2\\[4pt] &=\sqrt{ \frac{\frac12(1-\cos(2x))}{x^{2}}} \tag3\\[4pt] &=f(x) \tag4 \end{align}$$

Yet, $f(x)$ and $g(x)$ have different graphs. I don't think I messed with the domain, so what is the mistake here? Since they have different graphs, they have different limits at $x=0$.

For graphs: https://www.desmos.com/calculator/t8cga1ayzw

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$f(x)$ has $x$ in the denominator. So it includes both positive and negative values of $x$.

$g(x)$ has $\sqrt{x^2} = |x|$ (correction to be made here). This involves only positive values.

So, $g(x)$ is identical to $f(x)$ for $x>0$ and is the reflection of $f(x)$ about $x$ axis for $x<0$ as, $|x| = \begin{cases}x &\text{for } x\ge0 \\ -x &\text{for } x<0 \end{cases}$

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They will have different signs at $x<0$.

Watch out their denominators. Note that $\sqrt{x^2}=\left|x\right|$ instead of merely $x$.

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Hint:

For real $y,$

$$\sqrt{y^2}=|y|=\begin{cases} y &\mbox{if } y\ge0 \\ -y & \mbox{if } y<0 \end{cases} $$

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It's $$\frac{|\sin{x}|}{x}-\frac{|\sin{x}|}{|x|},$$ which is $0$ for $x>0$ and $\frac{2|\sin{x}|}{x}$ for $x<0$.

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