Secant and Arcsecant

Recall from elementary trigonometry that secant is one of the reciprocal functions and so sec(θ) = 1/cos(θ). Also, cosine is the cofunction of sine, so cos(θ) = sin(90° - θ). Put these facts together to arrive at the secret ingredient required for slide rules: sec(θ) = 1/sin(90° - θ).

The S scale, of course, gives the sine of an angle. Then merely find the reciprocal of the sine of the complement, and we arrive at the secant. The CI scale helps make this pretty automatic.

You can always compute the complement of an angle mentally, but on many slide rules these are shown in red on the S scale, easing your labor. On my favorite Pickett 1006-ES, that's not the case, but instead the backward going complements are denoted by a "<" sign as a reminder to read right to left.

Note: Click any of the photos below to zoom in for a closer inspection.

I've already covered the sine, cosine, tangent and cosecant in such detail previously, that I'm not going to waste your time repeating all that. Just refer to those sections if you need a bit of review. Let's get going with some problems at once.

Exercise 1.1: Find sec(32.5°).
  1. Mentally form the complement of the angle: 57.5°.
  2. Set the cursor at 57.5° on the S scale.
  3. Read the result under the hairline on the CI scale: 1.186.
On my Pickett, it looks like the following.

Note: I'm using the DI scale here instead of CI, since it appears on the same side as S of this duplex rule. Use whichever is most convenient for you. (But if employing DI, be sure to close the rule.)

Cursor to 57.5°, result at DI:1.186

Exercise 1.2: Find sec(87°).

That argument, being so close to 90°. should immediately suggest that the complement is "small," i.e., less than 5.74°. Thus, we switch over to the ST scale for this one.
  1. Mentally form the complement of the angle: 3°.
  2. Set the cursor at 3° on the ST scale.
  3. Read the result under the hairline on the CI (or DI) scale: 19.1.
Here's how I did it:

Cursor to ST:3°, result at DI:19.1
You might be curious about how to place the decimal point properly. Well, we know sin(5.74°) is 0.1, therefore cos(84.26°) is also 0.1. This implies that sec(84.26°) is 10. But 87° is even closer to 90°, so we must be in the next order magnitude, i.e., something greater than 10, hence 19.1.

Exercise 2.1: Find arcsec(2.5), in degrees.
  1. Set the cursor at 2.5 on the CI (or DI) scale.
  2. Read the angle under the hairline on S: 23.6°.
  3. Mentally form the complement: 66.4°.
Graphically, we have:

Cursor to DI:2.5, S:23.6° under hairline, result is 66.4°

Exercise 3.1: Find sec(1.2). The argument is in radians.
  1. Set the cursor to 1.2 on the C scale.
  2. Read the number under the hairline on the ST scale: 68.75°. This is the argument in degrees. 
  3. Mentally form the complement: 21.25°.
  4. Move the cursor to 21.25° on the S scale.
  5. Read the result under the hairline on the CI (or DI) scale: 2.76.
It'll take two pics to capture all this:

Cursor to C:1.2, read ST:68.75°, form complement: 21.25°
Cursor to S:21.25°, result at DI:2.76

It's a little tricky maintaining accuracy on a problem like this (especially with a pocket size rule) what with the number-transfer, among other things. But you should be able to get at least one decimal place decently.

Example 4.1: Find arcsec(8.5), in radians.
  1. Set the cursor at 8.5 on the CI (or DI) scale.
  2. Read the number under the hairline on the S scale: 6.75°.
  3. Form the complement: 83.25°. This is almost the result, but in degrees.
  4. Move the cursor to 83.25° on the ST scale.
  5. Read the result under the hairline on the C scale: 1.45 radians.
In pictures:

Cursor to DI:8.5, read S:6.75°, form complement: 83.25°
Cursor to ST:83.25°, result at C:1.45

Well, that's a pretty good sampling of problems involving the secant. But why don't you make a few more practice ones for yourself to really fix the ideas in place. Complements combined with reciprocals seem pretty confusing at first, but with just a smattering of repetition, the steps ought to become second nature.

Next installment: Cotangent and Arccotangent