Study Guide

# The Unit Circle - Solving Trigonometric Equations

## Solving Trigonometric Equations

When hunting mathematical abstractions, your strategy is key. (Or maybe your tactics; we can never keep the two straight.)

If you want to catch a trigonometric expression, we use an angle as bait. We've done a lot of this already. "See that tangent up there? Caught that using an empty swimming pool, five rolls of duct tape, and ."

Now we're going to turn things around: put on your safari hat, because we're going hunting for wild angles. We'll use the skills we've honed so far to track them back to their lairs, hiding within trigonometric equations.

The unit circle will continue to be our #1 tool of the trade for solving problems. By now, we can find the x- and y-coordinates for every variation of we care to name. We suggest drawing your own unit circle, with all the angles between 0 and 2π filled out, to use as a reference.

We also suggest drawing Batman fighting a dinosaur, because that is seriously cool.

• ### Simple Angle Equations

What is an equation? A miserable little pile of secrets. At least, it can feel pretty miserable if you're stuck on an even-numbered problem for hours. Luckily, we here at Shmoop don't believe in withholding the answers from you.

A trigonometric equation works just like any other; we have a variable, an equal sign, and a request to find out where the two sides meet. The only difference is that our variable is holed up inside a trig function, refusing to come out.

Take, for instance, We want to know every value of t that will make this equality true. We're nosy like that. We do that by getting (sin x) by itself, maybe ask it a few questions. Now, what values of t result in sine having a value of ? Let's use the unit circle to find out. Sine is associated with the y-coordinate. Looking around the circle, we see that the angles and have positive as their y values.

We're not done yet, though. Much like history, your rambling uncle, and a school assembly, the unit circle repeats itself. If we add—or subtract—2π from an angle, we end up in the same spot on the unit circle, and that new angle will have the same co-ordinates. Thus, we have:  Where n = 0, ±1, ±2, …. There are an infinite number of angles that make . We tried counting to infinity once, but stopped after 5. We got bored.

Sometimes, we only want some of the solutions for a trigonometric equation; those that fall within some interval.

### Sample Problem

If a tree falls in the forest, and no one is around, how much wood could a woodchuck chuck from it?

No, wait, that's the wrong question.

Solve within the interval -2π to 2π.

Hey, we've mostly solved this problem already. Isn't that nice? It's much easier than that woodchuck problem, in any case.

All we need to do is take our solution from before, when we didn't specify any interval, and plug in some values for n.  When n = 0, Both of these solutions are in-between ±2π, so we're good there. For n = +1:  And for n = -1:   and fall in the interval, but and are out of bounds. Dumping them like a bag of sweaty socks at the laundromat, we have: as our solutions. If we used n = ±2, we wouldn't find anything. Now if you'll excuse us, we have some laundry to do.

• ### Complex Angle Equations

We're kicking things up a notch. Now, there will be all kinds of expressions inside of the trig functions, not just t by itself. Do you think you can handle it? If not, take a break and come back to it tomorrow. We'll be here, we can wait.

### Sample Problem

Solve sec (4t) = 2 on the interval [0, π].

The second-most important thing to know about this problem is that it doesn't matter how many t's are inside the trig function: they don't change the right-hand side of the equation. We have sec (something) = 2, and we solve it the same way as last time. By checking the unit circle. Secant is the reciprocal of cosine (not sine! never sine!), so it will equal 2 when or (plus or minus 2π).  Do not, do not, do not say that this stuff equals t. That's the most important thing to know about this problem. It is super easy to forget this; we did it ourselves when writing this problem, embarrassing as that is. Instead, our (something) should be 4t:  We now divide each side of these equations by 4, to get t by itself:  Yes, even 2πn gets divided. That's easy to forget, too. Don't do it, or the Ghost of Mathematics Future will show up and tell you how horribly you've mucked things up.

(By the way, is he someone that died in the future and traveled back in time to tell us about it, or someone who is a ghost now and can see the future? We just don't know.)

Anyway, now we plug in some values for n to find the other solutions within the interval. When n = 1:   falls in the interval [0, π], but doesn't. Checking n = 2 gives , which is also too big.

Going in the other direction, with n = -1, we have:  We add to our stable of correct answers, while tossing aside. If we subtract another out, we end up with another negative number, which means we've exhausted our supply of angles.

That means our final answer is . We keep getting the right answers, but Regis refuses to give us our million dollars.

### Sample Problem

Solve sin (π – t) = cos (π – t)

We follow our standard procedure here. Isolate the trig function with t; find what (something) equals; find all the solutions; celebrate with fireworks.

However, we're seeing double here. There are two functions with t in them. In this case, we can fix this with a little division. Knowing that sine divided by cosine equals tangent, we just divide: tan (π – t) = 1

That's better. It only worked, though, because the stuff inside each trig function, (π – t), was the same. If they weren't identical, we'd be out of luck.

To find where sine and cosine are equal (and so tangent will be 1), we check the unit circle. Will the unit circle ever stop being so helpful? We hope not. The angles we want are π/4 and /4; remember, tangent is positive in quadrants I and III.  There's a trick we can use before going any further that will make this problem a little easier. Our two base solutions are on opposite sides of the circle, π away from each other in both directions. Instead of having two sets of solutions that repeat every 2π, we can have one set of solutions that repeat every π. We can get the correct answer either way. It's like choosing between two different routes to your destination, where one route is shorter and has nicer weather.

Now we need to replace that (something) with (π – t), while keeping an eye on that negative sign. If you were watching the signs, then you've already noticed that we didn't change it for πn. That's because n can be any integer, positive or negative, so it just absorbs the negative sign. It's kind of gross, really.

Anyway, we have our solutions. We can go back to the original problem and double check we did everything right. For instance, when : And now, we lie back in our easy chair and watch the boom.

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