Using Inverse Trig Functions - At A Glance

Knowing how to find inverse trig functions is all well and good, but it's even better if we know how to apply them to real problems. We talked a bit about solving equations like y = sin x for x, but what about bigger, more complicated equations? Is that too much to ask?

No way. Paired with our newfound knowledge of inverse trig, this will be easier than giving candy to a baby. For those of you without a lot of baby experience: babies love candy. Almost as much as we do.

Sample Problem

Solve .

Hey, what's the deal? Where's x, and why has θ taken its place? Grumble grumble.

Let's just calm down. θ is a variable just like x. There's nothing wrong with switching things up a little bit, and θ is often used to stand in for unknown angles. Just call him "theta." It's cool.

When looking at trig equations, our first goal is to isolate the trig function. If that sine is really messing us up, we can always rewrite the equation as:

Oh hey, welcome back, x. You're making this equation look a lot easier. Dividing both sides by 2 and putting sine back in for x brings us to:

Once we have the trig function isolated, we get to use the inverse to find the value of θ.

We're not quite there yet. This is only one angle that satisfies the equation. For one thing, there will be an infinite numbers of angles coterminal with this one, and each one is just as much a solution to the equation as this one. There's no room for special little snowflakes in this problem. Each angle will be 2π above or below its neighbor; check your nearest unit circle if you don't agree. That means:

, for n = 0, ±1, ±2, etc.

These are all solutions to the equation. That's a lot of solutions.

We're still not done, though. For a second thing, using ASTC (Alien Symbiotes Totally Cackle), we know that sine is positive in Quadrants I and II. That means there is an angle in Quadrant II that gives the exact same results as . It—and all of its coterminal buddies—will be solutions as well.

So, we need an angle from Quadrant II that has a reference angle of . Since the reference angle refers to how far away from the x-axis we are, and the x-axis in Quadrant II is at π, our other angle is:

Putting it all together, our solutions to the equation are:

, for n = 0, ±1, ±2, etc.

, for n = 0, ±1, ±2, etc.

That is a lot of infinite solutions. Wowzers.

Sample Problem

Solve 

Hey now, what is that  stuff at the end there? Grumble grumble again.

Man, we are kinda grouchy today. Give us a minute; we're going to go have a glass of OJ and relax for a minute. That will make us feel better.

Okay, we're back. That helped. Now, on to that new stuff.

"For " means "for angles within the domain from -2π to 2π." This will be important at the end of the problem, so we'll come back to it. Let's focus on the equation part first.

Oh, the equation is the same as the last problem. Nice. We've already found the full set of possible solutions throughout the entire universe and beyond. Now we just need to narrow things down to just the solutions within our domain. All that means is plugging in values of n into our solutions above.

Here are all the values for n = -1, 0, and 1.

Well, using n = 1 was a mistake, because they're all larger than 2π. But the rest of these are good, and going lower, like n = -2, won't be in our domain either. So, the full set of solutions to the equation are:

That's a nice, manageable number of solutions.

Sample Problem

Solve cos (2θ) = -1.

This problem will be a little tricky. We have B = 2, which, if we remember, means that the period will be  instead of 2π. This changes things; we're going to be cramming more solutions into a smaller space. Make sure the overhead compartment locks in place when you close it, as items may shift during flight.

The trig function is already on its own, so we can go straight to extracting 2θ:

2θ = arccos -1 = π

We have 2θ = π. We are sorely tempted to divide by 2, but we can't. Well, we could, but then we'd get the wrong answer, which is the same as saying we can't do it. Let's use x = 2θ for a second, and soon all will become clear.

Within 0 to 2π there are no other angles that will equal our π. For cosine, there's only one point where the graph dips down to -1, and π is it.

x = π + n2π, for n = 0, ±1, ±2, etc.

That's great and all, but we don't really care about x. Sorry, x, but this isn't your show. It's θ's.

2θ = π + n2π

Now we can finally divide that 2 out.

, for n = 0, ±1, ±2, etc.

Yep, we divide the 2 out of our angle and n2π. That's why we couldn't do it before now. Where once there was one solution between 0 and 2π, there are now two. That's what we would expect from a trig function with a smaller period.

What we wouldn't expect is the Spanish Inquisition. But nobody expects that.