I'm going to need to know this (the unit circle, English) for a test, but I'm so horrible with both charts and memorization, the whole thing is failing to stick into my brain. Using non-complicated English (literally, I could be the dumbest person here [at least I admitted it, so be kind Dx]), is there a mnemonic (I know, numbers and English), a list, some ridiculously easy hack for this??

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    $\begingroup$ "I'm going to need to know this for a test". What is "this", exactly? $\endgroup$ – Arturo Magidin May 12 '12 at 22:07
  • $\begingroup$ It's not clear what you need to know about the Unit Circle, from what you've said. What are you being tested on? $\endgroup$ – Ronald May 12 '12 at 22:09
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    $\begingroup$ I don't understand what you mean by "the unit circle, English". I don't understand what "English" has to do with anything, and there are a lot of things to be known about the unit circle. Presumably, you are being tested about some of them. Which? $\endgroup$ – Arturo Magidin May 12 '12 at 22:16
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    $\begingroup$ This question might help you. $\endgroup$ – Alex Becker May 12 '12 at 22:25
  • $\begingroup$ See the StackExchange thread Tips for understanding the unit circle, and note the distinction I make in my answer between what students often see as the unit circle and what teachers see as the unit circle. $\endgroup$ – Dave L. Renfro Jul 16 '12 at 16:31

I use a "ridiculously easy" trick to memorize trigonometric values of angles:

There are 5 angles primarily used from the unit circle: 0, 30, 45, 60, 90. This trick will provide the sine, cosine and tangent values for all 5.


1) Start by writing out the numbers 0 to 4 in ascending order: $0, 1, 2, 3, 4$

2) Divide all values by 4: $\frac{0}{4}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}$

3) Simplify all fractions: $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1$

4) Square root the numerators and denominators: $\left(\sqrt 0\right), \left(\frac{\sqrt(1)}{\sqrt(4)}\right), \left(\frac{\sqrt(1)}{\sqrt(2)}\right), \left(\frac{\sqrt(3)}{\sqrt(4)}\right), \left(\sqrt{1}\right)$

5) Simplify the fractions: $0, \left(\frac{1}{2}\right), \left(\frac{1}{\sqrt(2)}\right), \left(\frac{\sqrt(3)}{2}\right), 1$

And the resulting values are your sine values in ascending degree order:

$\sin 0 = 0 $

$\sin 30 = \left(\frac{1}{2}\right)$

$\sin 45 = \left(\frac{1}{\sqrt(2)}\right) $ This can be rationalized to $\left(\frac{\sqrt(2)}{2}\right)$ by multiplying by $\left(\frac{\sqrt(2)}{\sqrt(2)}\right)$

$\sin 60 =\left(\frac{\sqrt(3)}{2}\right)$

$\sin 90 = 1 $


Your cosine values are the same fractions above, but in descending degree order. (reverse order of sine)

$\cos 0 = 1 $

$\cos 30 = \left(\frac{\sqrt(3)}{2}\right)$

$\cos 45 = \left(\frac{1}{\sqrt(2)}\right) $ This can be rationalized to $\left(\frac{\sqrt(2)}{2}\right)$ by multiplying by $\left(\frac{\sqrt(2)}{\sqrt(2)}\right)$

$\cos 60 =\left(\frac{1}{2}\right)$

$\cos 90 = 0 $


$\tan\theta = \frac{\sin\theta}{\cos\theta}$

Therefore the tangent of any angle can be figured out by dividing the sine of that angle by the cosine of that angle.

Cotangent, Cosecant, Secant

These are simply the inverses of the tangent, sine and cosine of an angle.

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Here's how I know the common trig values (I'm assuming this is what you're looking for - hard to make out from your post):

Simply remember that $\sin(0) = 0$. Since $\sin(0) = 0, \cos(0)$ must equal $1.$ Also, it is easy to see that $\sin(30) = \frac{1}{2}$ by drawing a picture. At this point, I know in order to get a hypotenuse of $1$ (for the unit circle), you could deduct that $\cos(30) = \frac{\sqrt3}{2}$ by using the Pythagorean Theorem. Here, now note that $\sin$ and $\cos$ alternate or "flip" for lack of a complicated word (complement!). What I mean by this is that, $\sin(60) = \frac{\sqrt3}{2} = \cos(30)$ and $\cos(60) = \frac{1}{2} = \sin(30)$.

Also, for $45$ degrees, it should be easy to see that both $\sin$ and $\cos$ need to be $\frac{\sqrt2}{2}$ since our hypotenuse is $1$ for a unit circle.

Alternative way:

$\sin (\theta)$ for $0, 30, 45, 60, 90$ degrees follows the order of:

$$\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$$

For $\cos(\theta)$, it is simply the reverse.

Also, if you're having troubles with what function is positive in what quadrant, I suggest the simple tool I learned in school: ASTC (all students take calculus). All functions are positive in the first quadrant, only $\sin$ is positive in the second quadrant, only $\tan$ is positive in the third, and only $\cos$ is positive in the fourth quadrant.

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  • $\begingroup$ What picture tells you that $\sin(30)=\frac{1}{2}$? $\endgroup$ – Chris Eagle May 12 '12 at 22:26
  • $\begingroup$ Since the hypotenuse of a $30-60-90$ triangle is two times the side across from the $30$ degrees, it follows that our side across from the $30 degrees$ is half the hypotenuse of our triangle (which is $1$ for a unit circle). $\frac{1}{2} \cdot 1 = \frac{1}{2} = \sin(30)$ $\endgroup$ – Joe May 12 '12 at 22:31
  • $\begingroup$ @JayElectronica yes that is what i was asking, and also the answer I was looking for. @ everyone else i'll be more specific in the future. $\endgroup$ – Caroline May 12 '12 at 22:45
  • $\begingroup$ @Caroline You can still be specific in this particular question. You can edit it to address Arturo's question. $\endgroup$ – user23211 May 12 '12 at 23:04
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    $\begingroup$ @Caroline Glad I could help. Others, any reason for the downvote? $\endgroup$ – Joe May 12 '12 at 23:18

$$\begin{array}{c|ccccc} & \;\;0^\circ & \;\;30^\circ\;\; & \;\;45^\circ\;\; &\;\; 60^\circ\;\; & \;\;90^\circ\\[0.1in]\hline \sin & \frac{\sqrt{0}}{2} & \frac{\sqrt{1}}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{3}}{2} & \frac{\sqrt{4}}{2}\\[0.1in]\hline \cos & \frac{\sqrt{4}}{2} & \frac{\sqrt{3}}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{1}}{2} & \frac{\sqrt{0}}{2} \end{array}$$

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    $\begingroup$ Then maybe you want to write $0$ as $\frac{\sqrt{0}}{2}$ and $\frac{1}{2}$ as $\frac{\sqrt{1}}{2}$? $\endgroup$ – Arturo Magidin May 12 '12 at 22:34
  • $\begingroup$ Yes ;) but it was a simple case and i think that it can see.. $\endgroup$ – diofanto May 13 '12 at 8:15

Assuming you are talking about values of the trig functions at typical "special" angles (where the angle measure and the values of the trig functions are relatively simple), I tend to keep track of the values based on three facts:

  • $(\cos\theta,\sin\theta)$ is the rotation image of $(1,0)$ by $\theta$ about the origin—that is, the cosine and sine of an angle are, respectively, the $x$- and $y$-coordinates of the point on the unit circle.
  • The sides of a 45°-45°-90° triangle are in the ratio $1:1:\sqrt{2}$. (diagram from wikipedia)
    45°-45°-90° triangle
  • The sides of a 30°-60°-90° triangle (outlined in blue, half of an equilateral triangle) are in the ratio $1:\sqrt{3}:2$. (diagram from wikipedia)
    30°-60°-90° triangle

To figure out the cosine and sine of a special angle:

  1. find the corresponding point on the unit circle
  2. if the point is on the $x$- or $y$-axis, no triangles needed
  3. otherwise, place a triangle similar to one of the two triangle from above inside the unit circle so that the hypotenuse is a radius of the circle and one leg lies on the $x$-axis (not the $y$-axis); since we're working in the unit circle, the sides will be $\frac{1}{2}$, $\frac{\sqrt{3}}{2}$, and 1 or $\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{2}}$, and 1.
  4. if the point is not in the first quadrant, add negative signs to the legs as needed (if the leg/legs is/are in the negative $x$- or $y$-direction)
  5. read off the cosine and sine as the $x$- and $y$-coordinates, which are the legs of the triangle with negative signs inserted as needed.

If you need one of the other trig functions, find the sine and cosine and use those to calculate the other trig function.

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