Cotangent and Arccotangent

And here we are at the last of the six trig functions (and its inverse). By this point, I've beaten how to use the S, T and ST scales into the ground, emphasized the magical angles to look for, explained the workings with either radians or degrees, and more. So, I'll keep this installment brief and to the point. If you need extra detail, then I'll refer you back to the preceding five sections.

Here's what you need to know, in a nutshell. You'll recall that when finding the tangent of angles from 0° to 45° you could work directly with the angle, then read the result on C. On the other hand, if the argument lies between 45° and 90°, then you work with the complement and read the result off of CI.

Well, since cot(θ) = 1/tan(θ), then it should be clear that all we have to do is reverse the roles of C and CI. And again, the value 1 (corresponding to 45°) is the "turning point." Lastly, you'll want to keep in mind that cotangent is a decreasing function. With that, let's work some problems.

Exercise 1.1: Find cot(33.5°).
  1. Set the cursor at 33.5° on the T scale.
  2. Read the result under the hairline on the CI scale: 1.51.
Here's how it looked when I did it. Note that I'm using DI (equivalent to CI with the rule closed), to keep everything on the same side of my duplex rule, the Pickett 1006-ES.

Cursor to T:33.5°, result at DI:1.51

Exercise 1.2: Find cot(75°).

As with the tangent, here you'll be looking for the complement, 15° since the argument exceeds 45°. Some slide rules show the complements in red (running in reverse) on the T scale.
  1. Mentally form the complement: 15°.
  2. Set the cursor to 15° on the T scale.
  3. Read the result under the hairline on the C scale: 0.268.
For emphasis, if this was the tangent, we'd be using CI in step 3; with the cotangent, this is flipped with C.

Form complement 15°, cursor to T:15°, result at C:0.268

Example 1.3: Find cot(2.5°).

This is a small argument, so you'll remember to use the ST scale, right?
  1. Set the cursor at 2.5° on the ST scale.
  2. Read the result under the hairline on the CI (or DI) scale: 23.
Cursor to ST:2.5, result at DI:23

We know the cotangent function decreases, so given that 2.5° is fairly close to 0°, the result must be 23, and not 2.3.

Example 1.4: Find cot(87°).

Once more we should use the the ST scale. This time the complement, 3°, is quite small.
  1. Mentally form the complement of the argument: 3°.
  2. Set the cursor to 3° on the ST scale.
  3. Read the result under the hairline on the C scale: 0.0522.
Form complement: 3°, cursor to ST:3°, result at C:0.0522

In this case, we're very near 90°, so the result must be small (i.e., less than 0.1).

Let's try some problems involving arccotangents in degrees.

Example 2.1: Find arccot(2.5), in degrees.

Hmm...an argument greater than 1, so the result must be an angle less than 45°.
  1. Set the cursor to 2.5 on the CI scale.
  2. Read the result under the hairline on the T scale: 21.8°
Here's how it goes on my Pickett:

Cursor to CI:2.5, result at T:21.8°

Example 2.2: Find arccot(0.7), in degrees.

This time the argument is smaller than one, so we'll be looking for an angle exceeding 45°. (Remember, the cotangent function is decreasing).
  1. Set the cursor at 0.7 on the C scale.
  2. Read the number under the hairline on the T scale: 35°.
  3. Mentally form the complement: 55°.
In pictures:

Cursor to C:0.7, read T:35°, form complement: 55°

We probably ought to see some examples in radians.

Example 3.1: Find cot(1.3).

First order of business is to convert 1.3 radians to degrees, then carry on as usual. Whatever the degree equivalent is, we know it exceeds 45°
  1. Set the cursor at 1.3 on the C scale.
  2. Read the number under the hairline on the ST scale: 74.5°.
  3. Mentally form the complement: 15.5°.
  4. Move the cursor to 15.5° on the T scale.
  5. Read the result under the hairline on C: 0.28.
Of course, the argument being larger than 45° implies the result must be less than 1.0.

Graphically, we have:

Cursor to C:1.3, read ST:74.5°, form complement: 15.5°
Cursor to T:15.5°, result at C:0.28.

These damn radians! Not only is reading the angle conversion off of ST a bit problematic, but then having to do a number-transfer to T introduces a bit of imprecision. Still, you should be able to get a workable number for most applications.

Example 4.1: Find arccot(0.95), in radians.

Note that the argument is less than (but close to) 1.0. So we expect an angle close to and greater than π/4. The C scale must be needed, then.
  1. Set the cursor to 0.95 on the C scale.
  2. Read the number under the hairline on T: 43.5°.
  3. Mentally form the complement: 46.5°. This is the desired angle in degrees.
  4. Move the cursor to 46.5° on the ST scale.
  5. Read the radian equivalent on the C scale: 0.81.
Here's a summary:

Cursor to C:0.95, read T:43.5°, form complement: 46.5°
Cursor to 46.5°, result at C:0.81

And with that we've now seen how to compute all six trig functions and their inverses, in either radians or degrees, just with a simple type slide rule. When I get caught up, I'll add an installment explaining how to use the T1 and T2 scales (available on more advanced slide rules) to expedite computation of tangents and cotangents, as well as dipping into the P scale for speedy cosines.

Next installment: Common Logs with the L Scale