Here's the deal. Sine and cosine sound fancy, but on the unit circle they're just the $x$ and $y$ coordinates of a point. That's it. If you can find a point on a circle, you can find sine and cosine, and the other four trig functions are just those two flipped or divided. On the standard unit circle, every one of those points comes from just three triangles. This page assumes you can already draw an angle in standard position, peel off extra laps, and find a reference angle. If any of that sounds fuzzy, hit the Angles page first.
Remember SOH-CAH-TOA? Sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, tangent is opposite over adjacent. Look at the triangle below: $y$ is the opposite side, $x$ is the adjacent side, and $r$ is the hypotenuse. Same definitions, now living on a circle.
Now let's move trig onto a circle. A circle of radius $r$ centered at the origin has the equation $x^2 + y^2 = r^2$.
Pick a point $P(x, y)$ on the circle and draw a ray from the origin through it. Boom: an angle $\theta$ in standard position. Drop a line from $P$ to the $x$-axis and you've got a right triangle with legs $x$ and $y$ and hypotenuse $r$. Fair warning: $x$ and $y$ can be negative, depending on the quadrant. The radius $r$ never is.
The six trig functions of $\theta$ are all ratios of $x$, $y$, and $r$. The first three are the stars.
The sine of $\theta$ is $\sin(\theta) = \dfrac{y}{r}$, where $P(x, y)$ is the point where the terminal side of $\theta$ meets the circle $x^2 + y^2 = r^2$.
Cosine swaps in the other leg.
The cosine of $\theta$ is $\cos(\theta) = \dfrac{x}{r}$.
And tangent skips the hypotenuse entirely and compares the two legs.
The tangent of $\theta$ is $\tan(\theta) = \dfrac{y}{x}$, defined when $x \neq 0$.
Now here's the good news: you're basically only defining three functions. The other three are those same fractions flipped upside down.
The cosecant of $\theta$ is $\csc(\theta) = \dfrac{r}{y}$, defined when $y \neq 0$. It's sine flipped.
Hold on. Cosecant is the flip of sine, and secant is the flip of cosine? Whoever handed out the "co"s was not paying attention.
Next up, secant.
The secant of $\theta$ is $\sec(\theta) = \dfrac{r}{x}$, defined when $x \neq 0$. It's cosine flipped.
And last, the flip of tangent.
The cotangent of $\theta$ is $\cot(\theta) = \dfrac{x}{y}$, defined when $y \neq 0$. It's tangent flipped.
Notice something sneaky: the size of the circle never matters. Make the circle twice as big and $x$, $y$, and $r$ all double, so every ratio stays exactly the same. Go see for yourself.
Drag P around the circle and drag the square knob to resize it. Then try the missions.
Since the definitions are all built from the same three numbers, they're tangled up with each other. Statements that are always true like that have a name.
A trig identity is a statement that is true for all values of the variable (every angle where both sides are defined).
For example, divide the top and bottom of $\frac{y}{x}$ by $r$:
The same trick works for cotangent, and the flips come straight from the definitions. That gives the first identities you need to know cold.
Every one of these is a fraction, so every one of them breaks when its denominator is $0$. When that happens, the answer is undefined. Not $0$, not infinity, not "error." Undefined.
Since the size doesn't matter, everybody uses the easiest circle there is.
The unit circle is the circle of radius $1$ centered at the origin. Its equation is $x^2 + y^2 = 1$.
With $r = 1$, dividing by $r$ does nothing, and the definitions get way simpler.
If the terminal side of $\theta$ meets the unit circle at $P(x, y)$, then
In other words, the point is $P = (\cos\theta, \sin\theta)$: the $x$-coordinate is cosine and the $y$-coordinate is sine.
As $P$ travels around the circle, its height $y = \sin\theta$ rises to $1$, falls to $-1$, and comes back, over and over. Plot that height against $\theta$ and you get the famous sine wave.
Grab the point and drag it around. Watch the legs of the triangle: red means the coordinate is positive, blue means negative. Keep going past one full turn, or drag clockwise past $0^\circ$, and see what happens to the angle. This works no matter which quadrant $P$ lands in.
Drag the purple point around the circle. Let go near a dot to snap to an exact angle.
Since the point is on the circle, $x^2 + y^2 = 1$. That means $(\cos\theta)^2 + (\sin\theta)^2 = 1$ for every angle. It's a great way to check your work, and it comes in clutch when a problem gives you one coordinate and you need the other.
So once you know the point, you know everything. Find sine, cosine, and tangent from the point, then flip them to get the other three.
The terminal side of an angle $\theta$ meets the unit circle at $P\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)$. Find all six trig function values of $\theta$.
Read sine and cosine straight off the point: $\sin(\theta) = y = \frac{\sqrt{3}}{2}$ and $\cos(\theta) = x = \frac{1}{2}$. Then
Check: $\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$. The point really is on the unit circle.
For most angles, sine and cosine are ugly decimals. But you need to find the trig functions of $30^\circ$, $45^\circ$, and $60^\circ$ without a calculator. For those angles, the triangle is one of the two special right triangles from geometry, and the answers come out exact.
In a $45^\circ$-$45^\circ$-$90^\circ$ triangle, the sides are in the ratio $1 : 1 : \sqrt{2}$.
We need the hypotenuse to be $1$, not $\sqrt{2}$. So divide every side by $\sqrt{2}$. Each leg becomes $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
Find $\cos\left(\dfrac{\pi}{4}\right)$ and $\sin\left(\dfrac{\pi}{4}\right)$.
$\frac{\pi}{4} = 45^\circ$, so the triangle is the $45$-$45$-$90$ one, scaled down to hypotenuse $1$. Both legs are $\frac{\sqrt{2}}{2}$, so the point is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ and
In a $30^\circ$-$60^\circ$-$90^\circ$ triangle, the sides are in the ratio $1 : \sqrt{3} : 2$. The short leg is across from $30^\circ$, and the long leg is across from $60^\circ$.
Same trick: divide every side by $2$ so the hypotenuse is $1$. The short leg becomes $\frac{1}{2}$ and the long leg becomes $\frac{\sqrt{3}}{2}$. (Short leg times $\sqrt{3}$ gives the long leg; short leg times $2$ gives the hypotenuse.) The only question is which way the triangle is lying.
Find $\cos\left(\dfrac{\pi}{6}\right)$ and $\sin\left(\dfrac{\pi}{3}\right)$.
At $\frac{\pi}{6} = 30^\circ$ the triangle lies flat, so the long leg $\frac{\sqrt{3}}{2}$ runs along the $x$-axis: $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.
At $\frac{\pi}{3} = 60^\circ$ the same triangle stands up tall, so the long leg is the height: $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.
Mixing up $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ is the number one mistake on this whole topic. $\frac{\sqrt{3}}{2} \approx 0.87$ is the bigger one, so it always goes with the longer leg.
Now flip and divide to get all six values for each angle. Here's the whole set for Quadrant I. Look for the patterns instead of memorizing 18 numbers: the $30^\circ$ and $60^\circ$ columns are the same numbers shuffled around.
| $30^\circ$ $\frac{\pi}{6}$ |
$45^\circ$ $\frac{\pi}{4}$ |
$60^\circ$ $\frac{\pi}{3}$ |
|
|---|---|---|---|
| $\sin(\theta)$ | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ |
| $\cos(\theta)$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ |
| $\tan(\theta)$ | $\frac{\sqrt{3}}{3}$ | $1$ | $\sqrt{3}$ |
| $\csc(\theta)$ | $2$ | $\sqrt{2}$ | $\frac{2\sqrt{3}}{3}$ |
| $\sec(\theta)$ | $\frac{2\sqrt{3}}{3}$ | $\sqrt{2}$ | $2$ |
| $\cot(\theta)$ | $\sqrt{3}$ | $1$ | $\frac{\sqrt{3}}{3}$ |
Flipping $\frac{\sqrt{3}}{2}$ gives $\frac{2}{\sqrt{3}}$, and you're not allowed to leave a square root in the bottom. Multiply top and bottom by $\sqrt{3}$: $\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$. Handy shortcut: $\frac{3}{\sqrt{3}} = \sqrt{3}$, because in general $\frac{x}{\sqrt{x}} = \sqrt{x}$.
The quadrantal angles are the easiest points on the whole circle. There's no triangle at all: the point is sitting right on an axis.
Think about $\sin(0^\circ)$ for a second. As the angle shrinks toward $0^\circ$, the opposite side of the triangle gets super small, and at $0^\circ$ it's gone. So $\sin(0^\circ) = 0$.
Find all six trig function values of $270^\circ$.
At $270^\circ$ the point is $P(0, -1)$, so $\sin(270^\circ) = -1$ and $\cos(270^\circ) = 0$. Then
Notice cotangent: $\cot = \frac{1}{\tan}$ is useless here, since $\tan(270^\circ)$ doesn't exist. Use $\cot = \frac{\cos}{\sin}$ instead and you get a perfectly good $0$.
$\cos(90^\circ) = 0$ is sitting right there in the bottom of $\sec(90^\circ) = \frac{1}{0}$. Just once, I want to divide by it and see what happens.
Do the same thing for every quadrantal angle and you get these. Watch where the "undefined"s land: always exactly where a $0$ ends up on the bottom of a fraction.
The $0^\circ$ and $360^\circ$ cards match, because they're coterminal: same point, same values.
Notice the sine and cosine lines are the same numbers, shifted over by one card ($90^\circ$)? That's no accident. It's the cofunction identity, $\sin(A) = \cos(90^\circ - A)$, and it's the reason the "co" in cosine is there at all. More on that when we get to trig identities.
Here's the best news on the whole page: the other quadrants need no new triangles. By symmetry, every point in Quadrants II, III, and IV is a mirror image of a point in Quadrant I. For example, $135^\circ$ mirrors $45^\circ$ over the $y$-axis: $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$. Same numbers, one new sign.
Remember reference angles? Here's why they matter so much: the reference angle is always one of the angles inside the triangle for the point $P$. So the reference angle tells you exactly which triangle you're dealing with.
Same $\frac{\pi}{6}$ triangle, four times. The numbers never change, only the signs.
To find the point for any key angle:
1. Find the reference angle. It tells you which triangle (Tall, Mid, or Wide, coming up next), so it tells you the numbers.
2. Find the quadrant. It tells you the signs: $x$ is negative on the left, $y$ is negative on the bottom.
3. Put them together.
Find $\cos(-240^\circ)$.
Negative angle? Add $360^\circ$ to find a friendlier coterminal angle: $-240^\circ + 360^\circ = 120^\circ$. The reference angle is $60^\circ$, so the numbers are $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$. At $60^\circ$ the triangle is tall, so the $x$-value is the small one, $\frac{1}{2}$. Quadrant II is on the left, so $x$ is negative:
If we keep going all the way around, we get the full unit circle. This is the picture to know. The spokes are color-coded by triangle shape.
Every key angle that isn't on an axis gives you one of exactly three triangle shapes (the three colors on that chart). Learn to spot the shape, and the coordinates come for free.
The quadrant picks the signs. Here's a shortcut: in radians, a denominator of $3$ means Tall, $4$ means Mid, and $6$ means Wide.
Even giant angles fall into one of the three bins once you peel off the extra laps. Drag each card into its bin and watch its whole "butterfly" of four matching triangles light up.
Drag the angle card into the bin for its triangle. Sort six to finish the mission.
Peeling off laps is just coterminal angles: add or subtract $2\pi$ until you land between $0$ and $2\pi$. Write $2\pi$ with your angle's denominator ($\frac{6\pi}{3}$, $\frac{8\pi}{4}$, $\frac{12\pi}{6}$) and it's just adding numerators.
Evaluate: (a) $\sin\left(\dfrac{4\pi}{3}\right)$ (b) $\sec\left(-\dfrac{19\pi}{3}\right)$ (c) $\cot\left(-\dfrac{13\pi}{6}\right)$ (d) $\cos\left(-\dfrac{11\pi}{4}\right)$
(a) Denominator $3$: a Tall triangle. $\frac{4\pi}{3} = \pi + \frac{\pi}{3}$ is in Quadrant III, so the point is $\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$ and $\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$.
(b) Add full turns until it's friendly: $-\frac{19\pi}{3} + 4 \cdot 2\pi$ $= -\frac{19\pi}{3} + \frac{24\pi}{3}$ $= \frac{5\pi}{3}$. Tall triangle in Quadrant IV: $\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$. So
(c) $-\frac{13\pi}{6} + \frac{24\pi}{6}$ $= \frac{11\pi}{6}$. Wide triangle in Quadrant IV: $\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$. So
(d) Denominator $4$: Mid. $-\frac{11\pi}{4} + \frac{16\pi}{4} = \frac{5\pi}{4}$, in Quadrant III, so $\cos\left(-\frac{11\pi}{4}\right) = -\frac{\sqrt{2}}{2}$.
Cosine is $x$ and sine is $y$, and everything else is built from those two. So the quadrant alone tells you the sign of every trig function.
Cosecant has the same sign as sine, secant the same as cosine, and cotangent the same as tangent. (A number and its flip always have the same sign.)
Remember it with All Students Take Calculus. Going counterclockwise from Quadrant I: All are positive, then only Sine, then only Tangent, then only Cosine. (Plus their flips.)
Name the quadrant in which both conditions are true: (a) $\sin(\theta) < 0$ and $\tan(\theta) > 0$ (b) $\sec(\theta) > 0$ and $\cot(\theta) < 0$
(a) Sine is negative in Quadrants III and IV. Tangent is positive in Quadrants I and III. The only overlap is Quadrant III.
(b) Secant has cosine's sign, so it's positive in Quadrants I and IV. Cotangent has tangent's sign, so it's negative in Quadrants II and IV. The overlap is Quadrant IV.
Your turn. Think of it like a detective game: each clue rules out two quadrants, and θ is hiding in whatever's left.
Click a clue to cross out quadrants, then click the quadrant where θ is hiding. Crack all four.
Here is a typical problem that uses everything at once: the signs, the "type of right triangle," and the Pythagorean theorem.
Given that $\sin(\theta) = \dfrac{1}{4}$ and $\cot(\theta) < 0$, find $\cos(\theta)$.
Quadrant: $\sin(\theta) > 0$ means Quadrant I or II, and $\cot(\theta) < 0$ means Quadrant II or IV. So $\theta$ is in Quadrant II.
Triangle: sine is opposite over hypotenuse, so the opposite side is $1$ and the hypotenuse is $4$. The other leg is $\sqrt{4^2 - 1^2} = \sqrt{15}$.
Signs: in Quadrant II, $x$ is negative, so
The hypotenuse is always positive. It's the radius $r$, a distance. Every negative sign in these problems comes from $x$ or $y$, never from the hypotenuse.
Draw the triangle with plain positive lengths, then go back and attach the signs from the quadrant at the very end. If you skip that last step, you'll get the right number with the wrong sign, and that's a wrong answer.
Given that $\cos(\theta) = \dfrac{5}{13}$ and $\csc(\theta) < 0$, find $\tan(\theta)$.
Quadrant: $\csc(\theta) = \frac{1}{\sin(\theta)}$, so it has sine's sign: $\sin(\theta) < 0$, which means Quadrant III or IV. And $\cos(\theta) > 0$ means Quadrant I or IV. So Quadrant IV.
Triangle: cosine is adjacent over hypotenuse: adjacent $5$, hypotenuse $13$, so the opposite side is $\sqrt{13^2 - 5^2} = \sqrt{144} = 12$.
Signs: in Quadrant IV, $y$ is negative, so the opposite side is really $-12$:
Same idea when the problem hands you a coordinate instead. On the unit circle, $x^2 + y^2 = 1$ does the work of the Pythagorean theorem.
Let $P(x, y)$ be the point where the terminal side of $\theta$ meets the unit circle. If $P$ is in Quadrant II and $y = \dfrac{5}{13}$, find the six trig functions of $\theta$.
Plug into $x^2 + y^2 = 1$:
$P$ is in Quadrant II, where $x$ is negative, so $x = -\frac{12}{13}$. Now read everything off $P\left(-\frac{12}{13}, \frac{5}{13}\right)$:
That's every tool. Let's run the whole routine from start to finish: sketch it, find the point, then read off all six.
Sketch an angle measuring $210^\circ$. Give the coordinates of the point where its terminal side meets the unit circle. Then state the six trig functions of the angle.
$210^\circ = 180^\circ + 30^\circ$: Quadrant III, reference angle $30^\circ$, a Wide triangle. The point is $\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right) = (\cos\theta, \sin\theta)$.
In the tangent, the two negatives cancel. That's Quadrant III's tangent being positive, just like the sign rules promised.
And a lightning round. For each one: find the point, then build the function you need out of $x$ and $y$.
Evaluate each using the unit circle: (a) $\tan\left(\frac{7\pi}{4}\right)$ (b) $\csc\left(\frac{2\pi}{3}\right)$ (c) $\tan(300^\circ)$ (d) $\cot\left(-\frac{5\pi}{6}\right)$
(a) $\frac{7\pi}{4}$ is at $\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$, so $\tan\left(\frac{7\pi}{4}\right)$ $= \frac{-\sqrt{2}/2}{\sqrt{2}/2} = -1$.
(b) $\csc\left(\frac{2\pi}{3}\right) = \frac{1}{\sin\left(\frac{2\pi}{3}\right)}$ $= \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$.
(c) $300^\circ = \frac{5\pi}{3}$ is at $\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$, so $\tan(300^\circ) = \frac{-\sqrt{3}/2}{1/2} = -\sqrt{3}$.
(d) $-\frac{5\pi}{6} + 2\pi = \frac{7\pi}{6}$, at $\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$, so $\cot\left(-\frac{5\pi}{6}\right)$ $= \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$.
Time to put the whole circle together. Some blanks want the coordinates of a point and some want its angle. Wrong tiles bounce right back with a hint, and the refresh button deals a fresh set.
Drag each tile onto its blank. Hit ↻ for five new blanks.
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Alright, if you've made it this far, it's time for something with a little heat. These chain everything together: big angles, all six functions, and fractions you have to combine at the end. Evaluate each piece on its own before you add anything.
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