Trig is the math of angles, so before any sine or cosine shows up, we need to agree on what an angle even is, where it starts, which way it turns, and how to measure it. Nail this page and the unit circle becomes a lot less scary: every trick there starts with drawing an angle, peeling off extra laps, and finding its reference angle.
The practice is mixed into the lesson: each question sits right after the idea it needs, and the Mixed Practice at the bottom takes that hint away. Flip the switch on a question to type your answer instead: degrees as a plain number like 225, radians like 5pi/4, and quadrants like III.
First, we need everybody measuring angles the same way. So we agree on a starting line.
An angle is in standard position when its vertex is at the origin and its starting side (the initial side) lies along the positive $x$-axis. The side where it stops is the terminal side. Turning counterclockwise makes a positive angle; turning clockwise makes a negative angle.
So the sign of an angle tells you which way to turn, and its size tells you how far. Watch a few of them go:
We'll usually name angles with Greek letters, mostly $\theta$ ("theta"). When a second angle shows up, like the reference angle at the end of this page, it gets $\alpha$ ("alpha"). Now, the axes chop the plane into four regions, and you need to know them by name, because the region an angle lands in tells you a ton about it.
The $x$- and $y$-axes split the plane into four quadrants, numbered I, II, III, IV counterclockwise, starting from the top right.
Since positive angles turn counterclockwise, they walk through the quadrants in order: Quadrant I is everything between $0^\circ$ and $90^\circ$, Quadrant II is between $90^\circ$ and $180^\circ$, and so on. But what about an angle that stops exactly on an axis? It's not in any quadrant. It gets its own name.
If an angle is in standard position and its terminal side lies along the $x$- or $y$-axis, we call it a quadrantal angle.
Here's one whole lap. Each quadrant lights up as the terminal side passes through it, and every time the side stops on an axis, that's a quadrantal angle.
To find an angle's quadrant, picture the turn: which way, and how many quarter turns it gets past.
Where does each angle land? (a) $100^\circ$ (b) $290^\circ$ (c) $-45^\circ$ (d) $270^\circ$
(a) $90^\circ < 100^\circ < 180^\circ$, so Quadrant II.
(b) $270^\circ < 290^\circ < 360^\circ$, so Quadrant IV.
(c) Negative, so turn clockwise $45^\circ$. That stops below the positive $x$-axis: Quadrant IV again.
(d) Three quarter turns lands exactly on the negative $y$-axis. It's a quadrantal angle, so it isn't in any quadrant.
Your turn. Some of these are in radians, some are negative, and some go around more than once, so turn each one in your head first.
Now one with nothing to drag. Turn it in your head, then pick the quadrant.
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You've measured angles in degrees your whole life, where one full turn is $360^\circ$. Mathematicians prefer a different unit, because it ties the angle directly to the circle itself.
One radian is the central angle that cuts off an arc exactly as long as the circle's radius.
So how many radians make a full turn? Bend copies of the radius around the circle and count how many fit.
Six radii and a little bit: a full turn is $2\pi$ radians, about $6.28$. So radians just count radius-lengths of arc, which means you can read an angle straight off an arc.
A circle has radius $3$ cm. An arc of the circle is $6$ cm long. What angle does the arc make at the center, in radians?
The arc is $6 \div 3 = 2$ radii long, so the angle is $2$ radians (a bit less than a third of a turn, about $115^\circ$).
The size of the circle never matters: a bigger circle makes a longer arc, but the same number of radii fit along it.
Read one off an arc yourself. Watch for the versions that give you the diameter instead of the radius.
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A full turn is $360^\circ$ and also $2\pi$ radians, so half a turn gives us the only conversion fact you ever need to memorize.
To go from degrees to radians, multiply by $\dfrac{\pi}{180^\circ}$. To go from radians to degrees, multiply by $\dfrac{180^\circ}{\pi}$.
Let's use it both ways.
Convert $135^\circ$ to radians.
Multiply by $\frac{\pi}{180^\circ}$ and simplify the fraction:
Your turn: degrees in, radians out. Reduce the fraction all the way.
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Going the other way is even easier.
Convert $\dfrac{7\pi}{6}$ radians to degrees.
Since $\pi$ is $180^\circ$, just swap it in:
A full turn is $2\pi$ radians, and the circumference of a circle with radius $1$ is also $2\pi$. That's no coincidence: on a radius-$1$ circle, the radian measure of an angle is exactly the distance you walk along the circle.
Notice what happened there: $\frac{7\pi}{6}$ is just seven copies of $\frac{\pi}{6}$. That works every time.
Learn the "slices." $\frac{\pi}{6} = 30^\circ$, $\frac{\pi}{4} = 45^\circ$, $\frac{\pi}{3} = 60^\circ$, $\frac{\pi}{2} = 90^\circ$. Then $\frac{5\pi}{6}$ is just five $30^\circ$ slices, or $150^\circ$. No calculator, no multiplying.
Here's the whole circle with both scales on it, degrees outside and radians inside. Drag the pointer and count the slices.
Now build a few angles out of slices. Pick the slice that matches the denominator, then count.
Try it the other way: radians in, degrees out.
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Your goal is to get a gut feeling for how "big" each of these angles is, in both units, so you can sketch one without thinking. Once you can draw a positive angle, the negative ones are easy: cover the same amount, just clockwise.
Sketch each angle in standard position: (a) $240^\circ$ (b) $-150^\circ$
(a) $180^\circ < 240^\circ < 270^\circ$, so this one lands in Quadrant III. Turn counterclockwise past the negative $x$-axis, then $60^\circ$ more.
(b) Negative, so turn clockwise. $150^\circ$ is $30^\circ$ short of a half turn, so the terminal side stops $30^\circ$ below the negative $x$-axis. Also Quadrant III!
Radians look scarier, but here's the trick: split the fraction into a chunk of $\pi$ (a half turn) plus whatever is left over.
Sketch each angle in standard position: (c) $\dfrac{4\pi}{3}$ (d) $-\dfrac{5\pi}{4}$
A half turn, then one more $60^\circ$ slice. That's the same angle as (a).
A half turn clockwise, then $45^\circ$ more, which lands in Quadrant II.
Here are all four sketches, each one a half turn and then a slice. Notice (a) and (c) end up as the same picture.
Now draw them yourself. The drag remembers how far you've turned, so a negative angle means dragging clockwise, and an extra lap means going all the way around.
And the reverse: here's the angle, so which sketch is it?
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Nobody says you have to stop after one trip around. Spin $60^\circ$, and you land somewhere. Spin a full extra lap first ($60^\circ + 360^\circ = 420^\circ$), and you land in the exact same spot. Spin backwards (clockwise) $300^\circ$, and... same spot again.
Angles in standard position that share the same terminal side are called coterminal angles.
So how do you get from one coterminal angle to another? Add or subtract full turns: $360^\circ$, or $2\pi$ radians.
Wait. So $361^\circ$ is just $1^\circ$ in a trench coat? And $721^\circ$ is $1^\circ$ in two trench coats? I could spin forever and never actually get anywhere.
Here are those three spins, one after another. All of them end on the same point.
Every extra lap, either way, is one more full turn. So we can write down the whole family at once.
The angles coterminal with $\theta$ are exactly the angles $\theta + k \cdot 360^\circ$ (or $\theta + k \cdot 2\pi$ in radians), where $k$ is any integer.
That $k$ just counts laps: positive $k$ adds laps counterclockwise, negative $k$ takes them off. Pile on as many as you like. The terminal side won't budge.
One move comes up so much it deserves its own spotlight: going from a positive angle to a negative one. That's $k = -1$.
A positive angle $\theta$ with $0^\circ < \theta < 360^\circ$ is coterminal with the negative angle $\theta - 360^\circ$ (in radians: $0 < \theta < 2\pi$ and $\theta - 2\pi$).
And $\theta - 360^\circ$ is the negative coterminal angle closest to $0^\circ$: one lap fewer and you'd overshoot. Here's why it works. The counterclockwise turn and the clockwise turn split one full turn between them.
Here it is with numbers.
Find the negative angle closest to $0^\circ$ that is coterminal with $140^\circ$.
Subtract a full turn: $140^\circ - 360^\circ = -220^\circ$. Turning $140^\circ$ counterclockwise and turning $220^\circ$ clockwise end in the same spot.
That subtraction is where people slip, so here's the shortcut.
Doing $140 - 360$ in your head is annoying, because the big number is second. So flip it: $360 - 140 = 220$ is easy, and since $a - b = -(b - a)$, the real answer is just $-220$. The trap is forgetting to put the minus sign back.
Try a few. Some are in radians, where the full turn to subtract is $2\pi$.
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Here's the kind of question you'll see on quizzes all the time.
Find three angles, two positive and one negative, that are coterminal with each angle: (a) $120^\circ$ (b) $\dfrac{5\pi}{6}$
(a) Two positive: $120^\circ + 360^\circ = 480^\circ$, and nothing stops you from adding a bunch of laps at once: $120^\circ + 3 \cdot 360^\circ$ $= 120^\circ + 1080^\circ = 1200^\circ$. One negative: $120^\circ - 360^\circ = -240^\circ$.
(b) First write a full turn with the same denominator: $2\pi = \frac{12\pi}{6}$. Then
In radians, the hard part is writing $2\pi$ over the right denominator.
Keep this list handy: $2\pi = \frac{12\pi}{6} = \frac{8\pi}{4} = \frac{6\pi}{3} = \frac{4\pi}{2}$. Pick the one whose denominator matches your angle and the fractions add themselves.
Now aim for a target: add or subtract laps until you land where the question asks.
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You can also run it backwards, and peel laps off to find the friendly angle hiding inside a big one.
Find the coterminal angle on one trip around the circle: (a) $-135^\circ$ (b) $\dfrac{17\pi}{6}$
(a) Add one full turn: $-135^\circ + 360^\circ = 225^\circ$.
(b) Subtract one: $\frac{17\pi}{6} - \frac{12\pi}{6} = \frac{5\pi}{6}$.
Now unmask a few yourself. Watch out: one card on the right is a decoy.
One more, with no cards to lean on. Peel off every extra lap.
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This will matter a lot on the unit circle: coterminal angles land on the same point, so they share every trig value. Big ugly angles like $\frac{17\pi}{6}$ or $-300^\circ$ are just regular angles wearing a disguise.
You will need to be able to work with reference angles, and I mean need. They're the reason you only have to memorize one quadrant instead of four.
Suppose $\theta$ is an angle in standard position that is not a quadrantal angle. The reference angle for $\theta$ is the acute angle of positive measure formed by the terminal side of $\theta$ and the $x$-axis.
In plain English: it's the smallest angle from the terminal side back to the $x$-axis. Not the $y$-axis. Ever. So it's always strictly between $0^\circ$ and $90^\circ$. Here's the trap that warning is about:
So in each quadrant, you measure back to whichever side of the $x$-axis is closer. That gives one rule per quadrant.
For an angle $\theta$ between $0^\circ$ and $360^\circ$, measure back to the nearest side of the $x$-axis:
Big or negative angle? Find a coterminal angle between $0^\circ$ and $360^\circ$ first.
Those four rules aren't magic. Each one moves the wedge into Quadrant I: a flip, a spin, or a flip the other way.
One more thing before the examples.
The reference angle isn't always different! A Quadrant I angle like $60^\circ$ is already touching the positive $x$-axis, so it is its own reference angle. Don't go subtracting it from $180^\circ$ just because you're in the habit.
Let's run one angle through each quadrant.
Find the reference angle for each angle: (a) $150^\circ$ (b) $225^\circ$ (c) $310^\circ$ (d) $-45^\circ$
(a) Quadrant II: $180^\circ - 150^\circ = 30^\circ$.
(b) $225^\circ = 180^\circ + 45^\circ$, so we went $45^\circ$ past the negative $x$-axis. The reference angle is $45^\circ$.
(c) $310^\circ = 270^\circ + 40^\circ$, which puts it in Quadrant IV. Careful: that $40^\circ$ is measured from the $y$-axis, so it's not the answer. Measure to the $x$-axis instead: $360^\circ - 310^\circ = 50^\circ$.
(d) Turn $45^\circ$ clockwise and you're $45^\circ$ below the positive $x$-axis. The reference angle is $45^\circ$, because a reference angle is always positive.
Same moves in radians: just write $\pi$ or $2\pi$ over the angle's denominator first.
Find the reference angle for each angle: (e) $\dfrac{2\pi}{3}$ (f) $\dfrac{11\pi}{6}$
(e) Quadrant II, so subtract from $\pi = \frac{3\pi}{3}$:
(f) $\frac{11\pi}{6}$ is just shy of $2\pi = \frac{12\pi}{6}$, so it's in Quadrant IV:
Your turn. Find the quadrant first, then pick the rule.
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Now go hunting. Drag the terminal side anywhere, even around a few extra times or backwards, and watch the reference angle follow it back to the $x$-axis.
The hunter didn't care how many laps you took, because coterminal angles share a terminal side. On paper, that's the last line of the Finding It box: peel the laps off first.
Find the reference angle for each angle: (g) $840^\circ$ (h) $-\dfrac{17\pi}{6}$
(g) Take off two full turns: $840^\circ - 720^\circ = 120^\circ$. That's Quadrant II, so the reference angle is $180^\circ - 120^\circ = 60^\circ$.
(h) Add two full turns, $2 \cdot 2\pi = \frac{24\pi}{6}$:
That's just past $\pi = \frac{6\pi}{6}$, so Quadrant III, and the reference angle is $\frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi}{6}$.
Two steps now: unmask the angle, then find its reference angle.
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You can also run the rules backwards: knowing the quadrant and the reference angle pins down the angle.
(a) Find the angle between $0^\circ$ and $360^\circ$ in Quadrant III with reference angle $40^\circ$. (b) Find the angle between $0$ and $2\pi$ in Quadrant II with reference angle $\dfrac{\pi}{6}$.
(a) Quadrant III means $40^\circ$ past the negative $x$-axis: $180^\circ + 40^\circ = 220^\circ$.
(b) Quadrant II means $\frac{\pi}{6}$ short of the negative $x$-axis:
Run one backwards yourself.
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That's why every reference angle $\alpha$ comes with a family of four angles, one in each quadrant: $\alpha$, $180^\circ - \alpha$, $180^\circ + \alpha$, and $360^\circ - \alpha$. Drag the red side and watch the whole family move.
Find a family member on your own: one more angle with the same reference angle.
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On the unit circle, the reference angle is the boss: it tells you which triangle you're standing on, and that triangle hands you the sine and cosine for free.
That's everything you need about angles. Next stop: the unit circle, where every angle gets its own point. But first, some practice without the hints.
Every question so far came right after its idea, which is a hint a quiz never gives you. Down here the hint is gone. First, true or false: a claim about angles from anywhere on this page, and you decide whether it holds.
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Next, a mixed set: every question type on this page, shuffled. Figuring out which idea a question needs is half the work.
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Made it this far? Time for something with a little heat. These chain the whole page together: clock hands that turn the "wrong" way, angles multiplied until they lap the circle, and counting every angle in a window. Sketch first, then compute.
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