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We feel like we've gotten to know angles really well lately. We've been working with trig functions a lot, using them to find all kind of errant angles. We might be the world's premier angle experts at this point.

We've been so focused on angles, though, that we might have lost the trig values themselves. Uh, oops. Don't worry, we can fix this, and we'll use the inverse trig functions to do it.

### Inverse Trig Functions

Now that we're nice and comfy cozy with finding the values of trig functions, it is time to switch things up, literally. So far, we have learned how to find the value of the trig function given the angle. For instance, when we have the function

*y*= sin*x*, we can find*y*whenever we know*x*.We'll be going in the other direction now. We know

*y*, the value of the function, but we don't know*x*, the angle that gets us that result. That's where the**inverse trig functions**come in:If sin

*x*=*y*, then sin^{-1}*y*=*x*.We write the inverse of sine as sin

^{-1}. Some people get confused about the "^{-1}" looking like an exponent (it's not), so sometimes we see the inverse written as arcsin instead. They are totes the same, though.Of course cosine, tangent, and their reciprocals can all get in on this game as well, wearing little "

^{-1}" crowns or giving themselves "arc" as a fancy title. Just like the kings and queens they are pretending to be, each inverse function has a particular domain that they rule over.### Domain-ion

Not every function can have an inverse. It's an exclusive club, and the rules are all written down here. The big one is the horizontal line test—if a horizontal line can cross the original function multiple times, then it can't have an inverse.

Uh, sine, cosine, and tangent all fail that test. Miserably. Like, they somehow got negative points on the test. That's because they are periodic functions, so they're going to repeat their

*y*values over and over again.Obviously there's a way around this problem, or else we would be talking about something else entirely here. Like, maybe

*Battlestar Galactica*, or*The Parent Trap*. What we do to get over this issue is to restrict the domain to just an area where it won't fail the horizontal line test.Restricting sine's domain (somehow) improves its test scores, so now it can have an inverse, arcsine. Notice that the domain and range of sine get swapped over in arcsine. That is, sine could have

*y*values between -1 and 1, and that is now the domain of arcsine. We've restricted the domain of sine to between and , and that's arcsine's range. And this means that sine, and arcsine, don't repeat themselves.Here are the domains, and their associated ranges, for each inverse trig function.

**Function****Domain****Range**arcsin *x*[-1,1] arccos *x*[-1,1] [0, π] arctan *x*( -∞, ∞) arccsc *x*(-∞, -1] and [1, ∞) and arcsec *x*(-∞, -1] and [1, ∞) and arccot *x*(-∞, ∞) [0, π] Don't go memorizing this table. We know, we've made you memorize a lot of stuff about trig functions already, but this is a bit too much. The only parts we'll work with on a regular basis are the ranges of arcsin, arccos, and arctan. And, like we mentioned before, they are related to the original trig functions' domains and ranges. So it is usually easier to reason out what our answer should be.

Plus, if we use a calculator on the inverse functions, it will automagically refuse stuff outside of the proper domain, and they give their results in the correct range. Forget friendship, calculators are all the magic we need.

### Sample Problem

Solve .

Let's keep our eyes on the prize here. We're looking for an angle

*x*that, if we took the sine of that angle, would give us . Sine and arcsine are inverses of each other, so if we apply arcsine to both sides of the equation, the sine will go away.Now we can try to solve for

*x*. Taking another look at this, that should look familiar to us. Think special triangles. Much like your little sister, you just can't get away from them. That's a good thing here though, because they make things easier on us. We don't even need a calculator for this one; all those who forgot their calculator, rejoice.The special triangle that has a in it is the 30-60-90 triangle. It's opposite the 60° angle, and the hypotenuse is 2. Sine is opposite side over hypotenuse, so that gets us exactly what we want.

Well, almost exactly. We work in radians nowadays, so we convert from 60° to .

### Sample Problem

Find .

A lot of the time, the inverses for cosecant, secant, and cotangent make themselves scarce. They don't even show up on most calculators. We don't really mind, though, because we have our hands full dealing with arcsine, arccos, and arctangent.

Luckily, we can still MacGyver our way into finding those inverses. It can be a little tricky, though. Let's start off by rewriting the problem.

If we take the secant of the angle

*x*, we get . Secant is the inverse of cosine, though, which is much friendlier to work with.We can rearrange this to get

*x*back by itself again.We started with , and we changed it into . So to go from an inverse function to its reciprocal, we have to take the reciprocal of the value instead. Trippy. It only takes a bit of duct tape and some elbow grease.

At this point we should recognize that

*x*must be . No other angle could be so daring.We're starting to get good at these inverse functions, but there is one more sticking point. Sometimes we have to pick our answers based on the range of the inverse function. Remember, there are an infinite number of possibilities, but we are restricted to one. We decide based on which Quadrant our angle is based in.

We're going to throw you for one more loop, because loops are fun. Since it is possible for trig functions to have a negative value, it would probably be a good idea to know how to handle this. We come back to ASTC for this one. We decide which quadrant our angle is in based on the value of the trig function. For example, inverse cosine has a range of 0 to π, which is Quadrants I and II. If we have a negative cosine, we know that it must be in quadrant II because cosine is positive in I and negative in II.

### Sample Problem

Find sin

^{-1}(-½).Let's set the negative to the side for now. We will come back to that. Once again, we have a result that looks like it came from a special triangle, this time a 30-60-90 one. We picture it in our head, or maybe on our piece of paper, and try to find the (opposite over hypotenuse) that will get us ½.

If our triangle-drawing skills are to be believed, the 30° angle fits the bill. Our angle has an absolute value of 30°, or . We're not done yet, though. Remember how we set the negative to the side earlier? Well, we're taking it back now.

Inverse sine has a range of to , which are in Quadrants I and IV. Sine is negative in quadrant IV, so our angle must be negative.

Always do a double check when working with negatives. Otherwise, we might leave things positive, and that would be a mistake this time.

### Using Inverse Trig Functions

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 (2

*θ*) = -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 2

*θ*:2

*θ*= arccos -1 = πWe have 2

*θ*= 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*= 2*θ*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 +*n*2π, 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 +

*n*2πNow we can finally divide that 2 out.

, for

*n*= 0, ±1, ±2, etc.Yep, we divide the 2 out of our angle

*and**n*2π. 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.

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