# Calculus! #12: Early Transcendentals 1.6C

Okay. Let’s do it. Let’s get into inverse trig functions.

In my experience, inverse trig functions are one of those things that, for whatever reason, students dread. A lot of kids already dread regular trig, and now there’s an inverted version!

I admit, I once had a touch of that. But, in fact, inverse trig is relatively easy to work with, and opens up some cool possibilities when we get to integrals.

Remember, when confronted with new math that might make you nervous, don’t think of it as a scary impossible monster. It’s a tool you’re about to master.

So, let’s do it.

First off, inverse trig functions have to be modified slightly to make sense. As we discussed a while back, inverse functions are supposed to pass the horizontal line test. They’re supposed to be “one-to-one,” meaning that for each x value you put in, you get exactly one y value out.

But trig functions cycle up and down! So, for a given x, you’re going to see the same y infinity times. That’s too many times. So, clearly trig functions aren’t naturally one-to-one.

As we’ve said in the past, sine is the root of all trig, so we’ll start there.

Inverse Sine

So, we compel them to be one-to-one by restricting the domain to a single non-repeating cycle (half a wavelength, in physics terms). So, for example, if we want to run arcsin on y = sinx, we restrict x to be between -π/2 and π/2. This may feel a bit like cheating, but you’ve actually done this before. The common example is this: when you say $y=\sqrt{x}$, you restrict the domain to positive x values if you don’t want imaginary numbers.

Or, think of it this way: For y = sin(x), the available y values are inclusively between -1 and 1, right? So, if we convert that to arcsin(y) = x, the available values are inclusively between arcsin(-1) and arcsin(1), aka -π/2 and π/2. (Credit to incognitoman o twitter for this insight).

So, it’s really not so weird to be restricting the domain. Let’s move forth!

I wanna go over an example in the book with you because I think it gives some good insight into how working with inverse trig is not so hard.

Example 12-B: Evaluate tan(arcsin(1/3)).

Man, that looks like a bitch, right? But, step back, think, and put together everything you know.

First off, you can say arcsin(1/3) = something. We’ll call that something theta (ɵ).

1) arcsin(1/3) = ɵ.

Since we know sine is the inverse of arcsin, we can take a further step:

2) 1/3 = sin(ɵ)

Since we know sine is the ratio of the opposite to the hypotenuse (I like to say “O/H”), we can construct a triangle with a hypotenuse of length 3 and a side opposite ɵ of length 1.

Go ahead and draw that triangle.

Now, recall from step 1 that we defined ɵ as arcsin(1/3), and recall from the beginning that we’re trying to find out what in the world tan(arcsin(1/3)) is. By combining those two expressions, we get a new one:

3) tan(ɵ)

Now, look at your triangle. Since you know the opposite side is 1 and the hypotenuse is 3, you can use the Pythagorean Theorem to get the length of the adjacent side. When you solve, you should get $\sqrt{8}$.

Now you know all the sides of the triangle in question. All that’s left is to solve for the tangent. And, as you know, tangent is the ratio of the opposite side to the adjacent side. In this case, that’s $\dfrac{1}{\sqrt{8}}$.

So you see how by a little logic and a little algebra and by remembering your trig, you can turn an ugly expression into something nice and simple.

Lastly, remember that restricting the domain is a real thing you’ve done! It’s not a think you do with a wink at your textbook to allow you to solve certain problems. It puts real constraints on what you can do. The book defines these nicely as follows:

arcsin(sin(x)) = x, for [-π/2, π/2]

sin(arcsin(x)) = x, for [-1,1]

This is different from regular function inversion. If you have a function that adds 1 and you invert it by subtracting 1, that’s the whole mathematical picture. For trig, there’s a little more going on.

Cosine

Cosine obeys pretty much the same rules, except we restrict the domain slightly differently.

arccos(cos(x)) = x, for [0, π]

cos(arccos(x)) = x, for [-1,1]

Make sense? Remember the graph of cos is just the graph of sin shifted over by π/2.

Inverse Tangent

Check out the graph of tangent and you’ll readily see where we need to restrict. However, you may not notice that the x value never actually touches π/2. So, for arctan, the y values are (-π/2, π/2), as opposed to [-π/2, π/2] for sine.

Let’s do the book’s example here again. I want to show how by using what you know about trig, inverse trig functions remain fairly easy to work with.

Example 13: Simply the expression cos(arctanx)

The book wants you to know a certain trig shortcut, but I’m going to give you the complete under-the-hood version of how to answer this question. Here goes:

1) arctan(x) = ɵ

In other words, the arctan(x) equals some angle, which I’ll designate ɵ.

Now, draw a triangle with the following: An angle ɵ, the opposite side of which is labeled “O,” the adjacent side of which is labeled “A,” and the hypotenuse of which is labeled “H.” This isn’t standard nomenclature, but we’re under the hood right now, so it’ll help.

2) x = tan(ɵ)

This is obvious from (1). Really just algebra so far.

Now, from (2) and our knowledge that tangent is the ratio of the opposite to the adjacent side, we can say this:

3) O/A =x

That is, x is the ratio of opposite to adjacent. By some algebra, we can convert this into another form:

4) O = Ax

Now, from the Pythagorean Theorem, we know $O^2 + A^2 = H^2$. Combining that equation with (4), we get this:

5) $A^2 x^2 + A^2 = H^2$

Still with me? I just substituted Ax for O, which we established was legal in step (4).

Now, let’s simplify:

6) $H^2 = A^2 (1 + x^2)$

Now, let’s simplify again.

7) $\dfrac{H^2}{A^2} = 1 + x^2$

Now, you may remember that cosine is just the ratio of the adjacent and the hypotenuse, or (A/H). Knowing that, we can simplify (7) into this:

8) $\dfrac{1}{\cos^2 \theta} = 1 + x^2$

And now we can make that much prettier as:

9) $\cos \theta = \dfrac{1}{\sqrt{1+x^2}}$

OKAY, now remember at the outset we wanted cos(arctanx). Then, we redefined arctanx as ɵ. So, what we’re looking for is just cos(ɵ). And we have it in (9).

$\dfrac{1}{\sqrt{1+x^2}}$

WOOH! Now, that might seem like a lot of steps, but you could actually skip a lot if you know the trig identity that states:

$\sec^2 x = 1 + \tan^2 x$

That said, it’s always nice to know that everything under the hood makes sense.

The Other Inverse Trig Functions

The chapter closes by going through the seldom-used arccosecant, arcsecant, and arccotangent. If you want’em, they’re in a tidy list on page 70. But, I suspect this is about the last you’ll ever see of them.

And that’s the end of Chapter 1! I’ll do a quick review post, then we’re on to REAL LIVE CALCULUS!

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### 2 Responses to Calculus! #12: Early Transcendentals 1.6C

1. Jason Dick says:

Actually, you have to restrict it to just half a cycle, or half a wavelength. With the sine, for instance, you have the half of the cycle where the sine increases, and the half where it decreases. You can only include one of these.

• ZachWeiner says:

Fixed!