How does the Sin and Cos scale on a slide rule work and what is the formula for it? As I described in this question, I am trying to make a printable slide rule (similar to the slide rule provided in this SciAm article). I have made most of the parts of it but I can't make the Sin and the Cos scales. Although it looks very easy when I use a slide rule, but for some reason now I can't make them work. I tried various formulas but it didn't work.
From the original answer, I got the formula for the main log scale:
$$f(x,y)=(\log x)\cdot y$$
(where $f$ is the distance where the ticks have to be made, $x,y$ are integers and $y$ is fixed.)
But what is the formula for the $\sin$ and $\cos$ scales?
I searched on google but I didn't get any result.
If I diectly apply $\sin$, then I get this:

 A: Usually the $S$ scale is exactly the sine of the $C$ scale , where the argument is taken in degrees.  Say the length of the rule is $\ell $. Each point on the $C$ scale at distance $x$ from the index is marked with the number  $10^{2x/\ell }$.
Then each point on $S$ is marked with $$\sin\left (\frac\pi{180} 10^{2x/\ell}\right).$$
To use it, you have some argument on the $C$ scale of which you need to find the sine. You zero the scale (aligning the index marks) and match the cursor to the argument on the $C$ scale.
 Then you read the sine off of the corresponding value on the $S$ scale.
(To solve the reverse problem, "where on the $S$ scale should I place the label $s$?, you must invert this function: place label $s$ at distance $\frac\ell2\log\left (\frac {180}\pi\sin^{-1} s \right)$.)

Addendum: I think the foregoing would work well, but when I inspected an actual slide rule, I found that it was constructed differently: The $S$ scale, not the $A$ scale, is the argument.  I don't know if my memory was completely at fault or if different slide rules worked in different ways.   On the example slide rule I looked at, the argument (in degrees) is marked on the $S$ scale, for arguments between $\sin^{-1}\frac1{10}\approx 5.73^\circ$ and $\sin^{-1} 1 = 90^\circ$, and then the sines of these arguments are in the corresponding positions on the $D$ scale, which runs from $\frac1{10}$ to $1$.  For this slide rule, the rule is that each point at position $x$ is marked with the two labels  $\frac{180}\pi\sin^{-1}10^{(x/\ell)-1}$ and $90-\frac{180}\pi\sin^{-1}10^{(x/\ell)-1}$. 
The picture below shows that the sine of $10^\circ10' $ (on the $S$ scale) is $0.1765 $ (on the $D$ scale).

