Why is it that when we convert radians to degrees we multiply radians $\times \frac{180}π$ , but when we convert slope per radians to slope per degrees we have to multiply the inverse conversion formula slope per radians $\times \frac{π}{180}$
If we want to know an hour in terms of minutes, we multiply 1 hour $\times \frac{60}{1}$, given the result in minutes. If we want to know how convert 180 minutes, we divide $180$ minutes by $60$, i.e., multiply $180 \times \frac 1{60}$.
You'll find this phenomenon in any conversion: To convert temperature in degrees Celsius to temp in Fahrenheit, we have $F = \frac 95 C + 32$. To convert to from F to C, we need to invert this: $C = \frac 59(F-32)$
$R \text{ radians}\;\times \dfrac{180^\circ}{\pi \;\text{radians}} = \dfrac{R\times 180^\circ}{\pi}$.
"Radians" cancels as the unit, leaving a numeric value expressed in degrees.
$D \text{ degrees}\; \times \dfrac{\pi \;\text{radians}}{180\; \text{degrees}} = \dfrac{D\pi\;}{180}\;\text{ radians}$.
"Degrees" cancels as the unit, leaving the value expressed in radians.
Moved from comments:
Note that slope per radian is a ratio: $\;\dfrac{\text{slope}}{\text{radians}}.\;$ So to obtain "slope per degree", you need to have $π$ radians in the numerator, to cancel the unit "radians" from the denominator, and $180$ degrees in the denominator, to end with slope/degrees.
Slope itself is not a "unit" per se, meaning it isn't a degree, or radian, a mm, or foot. It is unit-free: even if we assign distance units (meters, say) to displacement: e.g. $Δ\,y\text{m}=y_2\,\text{m}−y_1\,\text{m}$ and $Δ\,x\,\text{m}=x_2\text{m}−x_1\,\text{m},$ then we have $$\text{slope}\,= \dfrac{Δy\,\text{ m}}{Δx\,\text{m}}=\frac{Δy}{Δx},$$ you see that the units "m: meter" attached to "change in y" and "change in x" cancel in the ratio defining slope, leaving us with a unit-free scalar which slope really is.
From Gerry Myerson's comment, we have
\begin{align*}
\frac{18\pi}{8} &= \frac{9\pi}{4} \\
&=2\pi+\frac{\pi}{4} \\
\cos \frac{18\pi}{8} &= \cos \left(2\pi+\frac{\pi}{4}\right) \\
&= \cos \frac{\pi}{4} \\
&= \frac{\sqrt{2}}{2}
\end{align*}
We then have the half-angle formula
$$
\cos \frac{1}{2}\theta = (-1)^{\lfloor (\theta + \pi) / (2\pi) \rfloor} \sqrt{\frac{1 + \cos \theta}{2}}
$$
Setting $\theta = \frac{18\pi}{8}$, we have
\begin{align*}
\cos \left(\frac{1}{2}\right)\left(\frac{18\pi}{8}\right) &= (-1)^{\lfloor ((18\pi/8) + \pi) / (2\pi) \rfloor} \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \\
\cos \frac{9\pi}{8} &= (-1)^{\lfloor(13\pi/4)/(2\pi)\rfloor}\sqrt{\frac{2 + \sqrt{2}}{4}} \\
&= (-1)^{\lfloor 13\pi / 8\rfloor} \frac{\sqrt{2 + \sqrt{2}}}{2} \\
&= (-1)^1 \frac{\sqrt{2 + \sqrt{2}}}{2} \\
\cos \frac{9\pi}{8} &= - \frac{\sqrt{2 + \sqrt{2}}}{2}
\end{align*}
Best Answer
Take a look at these ratios:
$\frac{180}{d}=\frac{\pi}{r}$
Where $d$ is the degrees and $r$ is radians. Knowing that $\pi$ radians is 180 degrees one can setup this ratio to find the values they're looking for.
Finding degrees: $d=\frac{180}{\pi}r$
Finding radians: $r=\frac{\pi}{180}d$
These equations are simply derived from the first ratio.
It doesn't only need to be $\frac{\pi}{180}$, it can also be setup as:
$\frac{360}{d}=\frac{2\pi}{r}$
Because it is also known that $2\pi$ radians is a full revolution about the circle just as 360 degrees is.