Tangent and Arctangent

A couple sections back, you learned about the sine and arcsine functions which used the S, ST, and C scales. Now we turn our attention to the tangent and arctangent functions. The slide rule manipulations are quite similar, with a trivial adjustment beyond 45°. Now, however, we'll be using the T scale (instead of S), and half of the time will need CI instead of C.

My recommendation is to review the earlier installment just mentioned before proceeding. Otherwise, put on your seatbelt and let's get going.

How Tangents Behave on a Slide Rule


Pause for a moment to examine your slide rule and compare the S and T scales. They come close to corresponding early on, but then you'll notice they diverge as you move to the right.

Here's why. The sine ranges from 0 to 1 as its argument goes from 0° to 90°. On the other hand, the tangent ranges from 0 to 1 as its argument goes from 0° to 45°, but then grows without bound as the angle gets closer and closer to 90°. Clearly 45° is an important point.

Let's investigate tangent's range more carefully. On the low end, for small angles, the tangent is closely approximated by the arc length corresponding to the argument. This is because the limit of tan(θ)/θ = 1, as θ (in radians) goes to 0. In everyday language, tan(θ) ≈ θ if θ is tiny.

And you'll also remember from Sine and Arcsine that sin(θ) ≈ θ as well when θ is small. Thus the ST scale can be used to approximate both sine and tangent for little angles.

How little, I hear you ask? For the sine, it's angles less than 5.74°. For the tangent, it's angles less than 5.71°. And for the arc length function (on the ST scale), it's angles less than 5.73°. Close enough for State work!

In short, we'll simply use ST to mimic either sine or tangent when the argument is less than about 5.73°. You might want to remember that this corresponds to a range of 0.0 to 0.1, for either sine or tangent.

Now, what happens if stay above 5.73°? Well, things go according to plan all the way up to 45°, where the tangent takes on the value of 1.0. But that's all the farther we can move on the T scale! How do accommodate larger angles?

Answer: by recalling that tan(θ) = 1/tan(90° - θ), now with θ in degrees. (This is easy to prove using some basic properties; try it!)

While that may look messy, just think how it sounds in words: to find the tangent of an angle, take the reciprocal of the tangent of the complement. And, of course, we're old hands at computing reciprocals using the CI scale.

We now have everything we need to compute the tangent for any angle:
  1. When θ < 5.71°, use the ST scale, and read the value on C; it will be a number less than 0.1.
  2. When 5.71° < θ < 45°, use the T scale, and read the result on C; it will be a number between 0.1 and 1.0.
  3. When 45° < θ < 84.29°, use the T scale (with the complement of θ), and read the result on CI; it will be a number between 1.0 and 10.0.
  4. When 84.29° < θ < 90°, use the ST scale (with the complement of  θ), and read the result on CI; it will be a number greater than 10.0.
You might also want to remember that angles greater than 89.43° yield tangents beyond 100.

This sounds much worse than it really is! After working a couple problems, it all becomes very second-nature. I'll tell you something else, too: you'll understand the behavior of the tangent function more fully than you ever would if stuck with a scientific calculator only.

Obviously, for negative angles just remember that tangent is an odd function. And for angles beyond 90°, simply work with a reference angle, bearing in mind which quadrant it lies in.

Tangent of an Angle Measured in Degrees


Enough theory; let's work some problems!

Example 1.1: Find tan(37.5°).

We recognize that the argument resides in the plain-Jane portion of the domain, and hence expect a result between 0.1 and 1.0. So, proceed directly:
  1. Set the cursor at 37.5° on the T scale.
  2. Read the result under the hairline on the C scale: 0.767.
Graphically, we have the following. (Be sure to click the photo to magnify it).

Cursor to T:37.5°, result at C:0.767

Example 1.2: Find tan(72.2°).

This angle is larger than 45° (but not too large), so we'll be working the complement, finding the reciprocal result on the CI scale. We expect something greater than 1.0
  1. Set the cursor at 17.8° (which is the complement of 72.2°) on the T scale.
  2. Read the result on the CI scale: 3.11
Now a couple notes are in order. First, many slide rules explicitly call out the complements on the T scale, typically in red. On my Pickett 1006-ES, they're denoted with a "<" symbol to remind the user the complements read right to left. Second, on my Pickett it's simpler to read the result on the DI scale (with the slide rule closed). So, my photos here all go that route. But feel free to use either the CI or DI, whichever is more convenient on your slide rule.

Cursor at T:17.8, result under hairline at DI:3.11

Example 1.3: Find tan(1.5°).

Here we have a small angle, and so will be using the ST scale, anticipating a result lying between 0 and 0.1.
  1. Set the cursor at 1.5° on the ST scale.
  2. Read the result under the hairline on the C scale: 0.0262
On my unit, I see:

Cursor at ST:1.5°, result at 0.0262

Example 1.4: Find tan(85.2°).

This argument is fairly close to 90°, so we'll be working with the complement along the ST scale, then finding the reciprocal.
  1. Set the cursor at 4.8° on the ST scale.
  2. Read the result under the hairline on the CI scale: 11.9.

Cursor to ST:4.8°, result at CI:11.9

Arctangent in Degrees


All of the comments made in previous sections concerning arcsine and arccosine apply here in a general sort of way. Really, all that's new is we need to remember that "fold-over" point at 45°.

Exercise 2.1: Find arctan(0.7), in degrees.

Since the argument is less than 1.0, we expect an angle less than 45°.
  1. Align the cursor with 0.7 (7.0) on the C scale.
  2. Read the result under the hairline on the T scale: 35°
Here's what she looks like:

Cursor to C:0.7, result at T:35°

Exercise 2.2: Find arctan(0.05), in degrees.

We've definitely got a small argument here, and so proceed accordingly with the ST scale.
  1. Set the cursor at 0.05 (5.0) on the C scale.
  2. Read the result under the hairline on the ST scale: 2.86°.
In summary:

Cursor to C:0.05, result at ST:2.86°

Exercise 2.3: Find arctan(8.5), in degrees.

You should be pretty skilled at recognizing what to do now upon seeing the size of the argument. The result has got to be more than 45°, right? That calls for a reciprocal.
  1. Set the cursor at 8.5 on the CI scale.
  2. Read the angle under the hairline on the T scale: 6.7°
  3. Interpret result as the complement: 83.3°
Here's how it goes:

Cursor to CI:8.5, hairline at T:6.7°, complement is 83.3°

Tangent of an Angle Measured in Radians


The process here has been covered extensively in Sine and Arcsine, so let's just jump in.

Exercise 3.1: Find tan(0.6).

A moment's reflection ought to convince you that the "fold-over" point for T lies at .79 radians (equivalent to 45°). So we'll be working with the angle directly here, not its complement.
  1. Set the cursor at 0.6 on the C scale. This is the angle in radians.
  2. Read the number under the hairline on the ST scale: 34.3° This is the angle in degrees.
  3. Leaving the cursor intact, move the slide so that 34.3° on the T scale aligns with the hairline.
  4. Read the result under the hairline on the C scale: 0.684.
In pictures:

Cursor at C:0.6, read ST:34.3°, leave cursor untouched
Align T:34.3° with hairline, result at C:0.684

Arctangent in Radians


The next problem takes a little bit of muggling, but if you keep your wits about you, makes perfect sense. Bear in mind that 7.5 suggests the angle must lie beyond the "fold-over" point. So, we'll be working with the complement.

Exercise 4.1: Find arctan(7.5), in radians.

That 7.5 ought to suggest where you're heading.
  1. Set the cursor at 7.5 on the CI scale.
  2. Read the number under the hairline on the T scale: 7.6°.
  3. Form the complement: 82.4°.
  4. Leave the cursor alone, but move the slide so that 82.4° on the ST scale aligns with the hairline.
  5. Read the result on the C scale: 1.44 radians.
In the first picture below, you'll note in step 1 that I'm actually using DI, simply because it happens to appear on the same side of the slide rule as T and ST. I could have just as easily used CI, but that would require some duplex-flipping.

Cursor at CI:7.5, read T:7.6°, complement: 82.4°, leave cursor untouched
Align ST:82.4° with hairline, result at C:1.44

And with that we notch up the tangent and arctangent as "done." Hence it must be time to consider the three reciprocal trig functions. Come back next time as we take on the scary cosecant and arccosecant functions.

Next installment: Cosecant and Arccosecant