# Easy way of memorizing values of sine, cosine, and tangent

My math professor recently told us that she wanted us to be able to answer $\sin\left(\frac{\pi }{2}\right)$ in our head on the snap. I know I can simply memorize the table for the test by this Friday, but I may likely forget them after the test. So is there a trick or pattern you guys usually use to remember it? For example, SOHCAHTOA tells us what sine, cosine, and tangent really mean. If not, I will just memorize the table. But just wanted to know what memorization techniques you guys use. I feel this is the perfect place to ask, because I bet a bunch of people in the math stackexchange, also had to go through the same thing freshman year of college.

Oh here is a picture of the unit circle: • Well, you don't really need to remember $\tan$ equals what, because it's just $\sin/\cos$. Also, notice the symmetry in the values of $\sin$ and $\cos$ when the angle is less than $\pi/2$. Finally, remember that $\sin(\pi - x) = \sin x$ and $\sin(\pi/2 - x) = \cos x$, and $\cos(\pi/2 - x) = \sin x$, and that you can add as much $2 \pi$'s as you'd like (and I almost forgot: that $\sin$ is an odd function, and $\cos$ is even).
– user258700
Nov 30, 2015 at 23:58
• To be tasked with memorising a handful of values of the trigonometric functions seems to me to be a cruel and unusual punishment with no pedagogical purpose. Dec 1, 2015 at 0:16
• @JamesSmith: my sympathies to you. If it helps at all: all I have ever committed to memory is that $\pi/2$ is a right-angle, that $t = s/c$ and that $s$ and $c$ can be read as short for "sideways" and "central" (where $s$, $c$ and $t$ stand for $\sin$, $\cos$ and $\tan$). Dec 1, 2015 at 0:27
• $cos\ 60$ is $\dfrac{1}{2}$ for me. That's all Dec 1, 2015 at 3:57
• Real world applications tend to use angles in degrees. Dec 2, 2015 at 2:56

Note the pattern:

$$\sin 0^{\circ} = \frac{\sqrt{0}}{2}$$ $$\sin 30^{\circ} = \frac{\sqrt{1}}{2}$$ $$\sin 45^{\circ} = \frac{\sqrt{2}}{2}$$ $$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$$ $$\sin 90^{\circ} = \frac{\sqrt{4}}{2}$$

This is something of a mathematical coincidence as far as I know, so don't try to extend this to other angles; and it goes backwards for $\cos$. Once you have these, you can find the other angles you want by drawing them on the unit circle and figuring out whether the values should be positive or negative, whether they should be $0$ or $1$, or otherwise whether they are "small" ($\frac{1}{2}$), "medium" ($\frac{\sqrt{2}}{2}$) or "large" ($\frac{\sqrt{3}}{2}$).

• +1! It was only after I was a grad student that I saw this. Dec 1, 2015 at 2:12
• This is so AMAZING! Dec 1, 2015 at 2:26
• In so far as there's a pattern here, it does extend - if you want a bunch of angles with sines $\frac{\sqrt{k}}{(2n)^2}$, the sequence of angles is always going to have that the reverse of the sequence is the complement of each angle, the first angle is $0$. The middle angle will be $45$ by the symmetry. The other angles just need to be filled in. (Unfortunately, this extension kind of highlights that the angles on the left don't follow much of a pattern beyond what's necessary - $30$ is just there as the angle whose sine is $\frac{1}2$) Dec 1, 2015 at 3:47
• @MiloBrandt: Regarding patterns, this recent answer of mine provides a common form for the sines (and cosines) of multiples of $3^\circ$. There's no particularly-obvious pattern (certainly not a "counting" one), but it seems like the list is trying to tell me something.
– Blue
Dec 1, 2015 at 4:46
• Basically, this is due to a combination of the Niven's theorem (If $q$ and $\cos(q^\circ)$ are both rational, then $\cos(q^\circ)$ equals $0$, $\pm\frac12$, or $\pm1$) and the fact that $\sin^2x=\frac12(1-\cos2x)$. Dec 2, 2015 at 5:28

Cosine goes horizontally (from the y-axis), sine goes vertically (from the x-axis).

Considering the three "main points" on the unit circle, $30^\circ,45^\circ,60^\circ$ (or $\frac \pi 6, \frac \pi 4, \frac \pi 3$ rads)...

From each axis,

• the "long" distance is $\frac {\sqrt3} 2$
• the "medium" distance is $\frac {\sqrt2} 2$
• the "short" distance is $\frac {1} 2$

When you go horizontally left, cosine value will be negative, and similarly for the sine value when going vertically down.

Green is cosine, red is sine: • +1 This is the way I remember it, and probably this is the only way which helps in understanding the values. Dec 1, 2015 at 4:08
• This is the way. Also, knowing the Pythagorean theorem means you just have to remember the \frac{1}{2} and can calculate the rest. Dec 1, 2015 at 8:40

I would not expect a student to memorize trig functions of easy angles. (Never memorized them myself.) I would expect a student to have enough understanding to be able to figure them out in seconds.

The trig functions for $30^\circ,45^\circ,60^\circ$ are based on two simple geometric figures: the square and the equilateral triangle.

The square has four sides of equal length, which we take to be $1$. It has four equal angles of $90^\circ.$

Next, cut the square along the diagonal, making two triangles. Either one of these triangles has angles of $90^\circ,45^\circ,45^\circ$; no need to memorize $45$, just divide $90$ by $2$. The triangle has two sides of length $1$; if you have memorized the theorem of Pythagoras, you can figure out that the length of the third side is $\sqrt2.$

Unfortunately, you have to memorize the definitions of the sine and tangent: $\sin=\text{opp }/\text{ hyp}$ and $\tan=\sin/\cos.$ The cosine is easier: cosine = complement's sine, so $\cos\theta=\sin(90^\circ-\theta).$

The point is that you can just read off the trig functions of $45^\circ$ from the $45^\circ$-$45^\circ$-$90^\circ$ triangle: $\sin45^\circ=1/\sqrt2,\$ $\cos45^\circ=\sin(90^\circ-45^\circ)=\sin45^\circ=1/\sqrt2,$ and $\tan45^\circ=\sin45^\circ/\cos45^\circ=(1/\sqrt2)/(1/\sqrt2)=1.$

Next, take an equilateral triangle; each of the three sides has length $1,$ and each of the three angles is $60^\circ.$ (No need to memorize $60,$ just divide $180$ by $3.$) Cut the equilateral triangle in half by bisecting an angle and look at one of the resulting triangles. The angles are $30^\circ,\ 60^\circ,$ and $90^\circ$; the sides are $1$ and $1/2$ and (if you still have Pythagoras memorized) $\sqrt3/2.$ From this triangle you can read off the trig functions of $30^\circ$ and $60^\circ.$

Executive summary. Take the two simplest polygons, the equilateral triangle and the square. An angle bisector divides each of those figures into two congruent right triangles. The trigonometric functions of $30^\circ,60^\circ,$ and $45^\circ$ can be read off of those triangles.

• Why did you bold all the usages of "memorize" in this answer? Some are "do memorize" and others are "don't memorize" so it comes off as a bit random or meaningless. Don't overuse special formatting. Dec 1, 2015 at 14:26
• It's not a low quality answer. It just has some questionable style choices which distract from the text. Of course it is entirely your choice what to do with my suggestions, but it's certainly not deserving of a downvote. Dec 1, 2015 at 21:31
• @bof ok, noted. Dec 2, 2015 at 3:58

Memorizing patterns for $\sin$ and $\tan$: $$\begin{array}{c|c|c|c|c|c} &0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}\\[1.5ex]\hline \sin{}&\frac{\sqrt{\color{#a00}0}}{2}&\frac{\sqrt{\color{#a00}1}}{2}&\frac{\sqrt{\color{#a00}2}}{2}&\frac{\sqrt{\color{#a00}3}}{2}&\frac{\sqrt{\color{#a00}4}}{2}\\[1.5ex]\hline \tan{}&\frac{\color{#090}0}{\sqrt{3}}&\frac{\color{#090}1}{\sqrt{3}}&\frac{\sqrt{3}}{\sqrt{3}}&\frac{\sqrt{3}}{\color{#090}1}&\frac{\sqrt{3}}{\color{#090}0}\\ \end{array}$$

• This is how I learned it, +1 Dec 2, 2015 at 19:25
• I have a question unrelated to math, how did you make that table and different color text in LaTeX? Feb 15, 2017 at 21:58
• @KKZiomek See \color command and array environment for tables. Feb 17, 2017 at 13:50

If you are so ahead in math, then you may have encountered the complex plane. When combining the two, you see that cos refers to the a and sin refers to the b in a+bi. That is:$$\cos(\theta)+i\sin(\theta)$$Oddly, this is how I remember my values. In particular, I've spent many hours researching the roots of unity, if you are so inclined. You will find that roots of unity lie on this circle and follow the above equation.

For example, the second root of $$1$$ is $$1,-1$$. $$\theta=0,2\pi, 4\pi, 6\pi,\dots$$ Taking the second root is square rooting or dividing theta by 2. If done so, then we see $$\theta=0,\pi$$. Plugging it back into the above equation gives $$1+0i$$ and $$-1+0i$$. This reminds me that $$\sin(0,\pi)=0$$ (writing $$\sin(a,b)=c$$ for shorthand here for $$\sin(a)=\sin(b)=c$$) Doing, say, the twelfth root of one is the equivalent to most of the positions of your circle of unity.

This is somewhat advanced trigonometry that my teacher never taught me (thank god, prayed my friends), so I assume you will never learn this. If you do, it is a nice... reminder.

Also, it is one of the only ways you can check if you got your answer right. For example, if:$$\sqrt{1}=\cos\left(0,\frac{2\pi}{3}\right)+i\sin\left(0,\frac{2\pi}{3}\right)$$then:$$1=\left(\cos\left(0,\frac{2\pi}{3}\right)+i\sin\left(0,\frac{2\pi}{3}\right)\right)^3$$ If you can use the binomial expansion theorem from there, your guesses for cos and sin should work out and indeed equal 1.

• Did you want to have $i$ somewhere in the last two equations? Dec 1, 2015 at 6:35

Other people are using a hell of a lot of formulas. It's easier than that.

Going around the unit circle, the cosine is the x-coordinate and the sine is the y-coordinate. So for the multiples of 90° ($\pi/2$), these are easy: at 0, the x-coordinate is 1 and the y-coordinate is 0. So sin(0) = 0 and cos(0) = 1. We also might just remember the graphs of sin and cos, and remember that sin goes through the origin and cos has a peak at (0, 1). It should be easy to go around the circle by 90° and get the other 3.

For the "45s", remember that a 45-45-90 triangle has side lengths that go $1$ - $1$ - $\sqrt2$. We need the hypotenuse to be 1, so divide by $\sqrt2$ (and rationalize the denominator) to get $\frac{\sqrt2}{2}$ - $\frac{\sqrt2}{2}$ - $1$. This is nice and symmetrical, so all sines and cosines of these angles will be $\pm\frac{\sqrt2}{2}$, just check the quadrant for the signs.

For the 30s/60s, it's the same trick with a 30-60-90 triangle. Remember that these side lengths go $1$ - ${\sqrt3}$ - $2$, or for a hypotenuse of 1, $\frac{1}{2}$ - $\frac{\sqrt3}{2}$ - $1$. Visualize the angle, and the smaller coordinate will be $\pm \frac{1}{2}$ and the larger one will be $\pm \frac{\sqrt 3}{2}$

The only trick is if you're not used to thinking of angles in radians. Remember that $\pi$ is half of the circle. So for example the angle $\frac{2\pi}{3}$ is 2/3 of a half circle, or 120°. This is one of those 30-60-90 cases, and the x-coordinate is the smaller value in the negative, or -1/2, and the y-coordinate is the larger value in the positive, or $\frac{\sqrt 3}{2}$.

Recall that $\sin^2(\theta)+\cos^2(\theta)=1$. For example, if you ONLY remember your $\sin$ values, then you see that $$\cos(\theta)=\sqrt{1-\sin^2(\theta)}$$For example, $\sin(\frac{\pi}{2})=1$

so... $$\cos(\frac{\pi}{2})=\sqrt{1-\sin^2\left(\frac{\pi}{2}\right)}=\sqrt{1-1^2}=\sqrt{0}=0$$

So you can find $cos$ by remembering $sin$ or vice versa.

Secondly, recall $$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$$

You could even further simplify this by using the formula up above to get: $$\tan(\theta)=\frac{\sqrt{1-\cos^2(\theta)}}{\sqrt{\cos^2(\theta)}}=\sqrt{\frac 1{\cos^2(\theta)}-1}=\sqrt{\frac 1{1-\sin^2(\theta)}-1}$$

Use one of the above formulas to find $\tan(\theta)$.

Then you'll need to remember the formulas and one set of values, say, for $\cos$.

Then remember the following rules: sine is odd, cosine is even, and tangent is odd. This means $\sin(-\theta)=-\sin(\theta)$, $\cos(-\theta)=\cos(\theta)$, $\tan(-\theta)=-\tan(\theta)$.

You could try to remember that on your unit circle, $\sin$ is positive from 0 to $\pi$ and negative from $\pi$ to 2$\pi$. $\cos$ is positive from $-\frac{3\pi}2$ to $\frac{\pi} 2$ and negative from $\frac{\pi} 2$ to $\frac{3\pi} 2$

• You might prefix LaTeX 'sin', 'cos' and 'tan' with a backslash to make them function symbols, displayed in upright font: \sin, \tan → $\sin, \tan$. Dec 2, 2015 at 9:34
• Thanks. Never noticed that. Dec 3, 2015 at 23:51
• Yet another $\LaTeX$ trick: use \left and \right modifier to brackets to make them automatically fit to the contents height. Compare \sin (\frac 12) → $\sin (\frac 12)$, \sin \left(\frac 12\right) → $\sin \left(\frac 12\right)$ Dec 4, 2015 at 6:39
• And your just the best! Dec 6, 2015 at 1:36

Consider the simple example first and then you can derive further values from it. Lets begin with the following table. Sin Cos Tan Table

All you need to remember about this table is the arrangement of rows and columns i.e. angles are placed in columns and trigonometric functions at the rows. So draw a table and start with the sine values.

Step 1 (red): In each of the empty cells for sine of 0,30,45,60 and 90 write (0,1,2,3,4) respectively.

Step 2 (green): Then divide each of them with 4 (i.e. 0/4, 1/4, 2/4, 3/4, 4/4).

Step 3(blue): The next step is to take square root of each (i.e. sqrt(0/4), sqrt(1/4), sqrt(2/4), sqrt(3/4), sqrt (4/4)).

Step 4(brown): Now cancel any terms from numerator and denominator if applicable and simplify the results. You end up with the values of sine 0,30,45,60 and 90.

Drawing Sin Cos Tan Table

Step 5(orange):Once you have values for sine function, invert them for cosine i.e( sin 90 = cos 0, sin 60 = cos 30, sin 45 = cos 45 and so on) and you get values for cosine function.

Step 6: For tangent, put sin/cos values and simplify.

Step 7: You can extend the table for further angles by using formulas such as

sin(π−x)=sinx sin⁡(π−x)=sin⁡x and sin(π/2−x)=cosx sin⁡(π/2−x)=cos⁡x, and cos(π/2−x)=sinx cos⁡(π/2−x)=sin⁡x,

1) Know "Sine of Sixty is Square root of 3 over 2" - notice the alliteration

2) Know Sqrt 3/2 is always paired with 1/2

3) Know Sin and Cos of 45 (Pi/4) is Sqrt 2/2

Two other things that shouldn't have to be memorized:

4) Know The ordered pair (Cos,Sin), like (x, y) is in alphabetical order

5) Know Where x and y are positive and negative

I just came up with this, it's not too bad!

This method worked good for me, you need to remember two triangles.

Let the first be a equilateral triangle with all sides of length 2, you divide the triangle in two by joining the top vertex (easy to see that way), with the centre of side of the opposite one, now you have a triangle with sides as of length 1 and $\sqrt3$ and hypotenuse of 2, one angle is 30 and other is 60.

Other is make a right angled isosceles triangle with sides 1,1 and $\sqrt2$, with this you lookup 45 and 90.

I used to make these small drawings at a place easy to lookup like last paper of answer sheet and just check them whenever needed.

Okay, here's my two cents.

There are two concepts to memorize.

A) Definition: $\cos = x$, $\sin = y$, $\tan = y/x$. $x^2 + y^2 = 1$.

B)Then for specific angles there are values. These are basically geometry and there are three cases:

1) $\theta$ ~ $0$ or $90$. These are $\theta = 0, \pi/2, \pi, 3\pi/4$ (or $-\pi/4$).

These are the points $(x,y) = (\cos \theta, \sin \theta)$ = $(0, \pm 1)$ and $\tan \theta = (0, \pm 1)$ and $x/y$ = $0$,$\pm 1/0$ (undefined).

Which is which is visually apparent by imagining the circle.

2) $\theta$ ~ 45. These are the $\theta = \text{odd}*\pi/4$ (or another way to view it $\{0, \pi, \pm \pi/2\} \pm \pi/4$; or simply $\pi/4, 3\pi/4, 5\pi/4 = -3\pi/4, 7\pi/4 = -\pi/4$).

These form 45-45-90 triangles, or an isosceles right triangle. $x = y$ and so $x^2 + y^2 = 2x^2 = 1$ so $|x| = |y| = \sqrt{2}/2$. And $|x|/|y| = 1$.

So $\sin \theta = \pm \cos \theta = \pm \sqrt{2}/2$. and $\tan \theta = \pm 1$.

Which pos/neg polarity depends upon which quandrant $(x,y) = (\cos \theta, \sin \theta)$ lies in.

3) $\theta$ ~ $30, 60$. This are $\{0,\pi, \pm \pi/2\} \pm \pi/6$. Or $0, \pi/6, \pi/3, 2\pi/3, 5\pi/6, 7\pi/6, 4\pi/3, 5\pi/3, 11\pi/6$.

These for 30-60-90 degree triangles which are equilateral triangles cut in half. As right triangles have $a^2 + b^2 = c^2$ and as $c = 1$ then $a = 1/2$ and $b = \sqrt{3} 2$.

So these are $(x,y) = (\cos \theta, \sin \theta) = (\{\pm 1/2: \pm \sqrt{3}/2\},\{\pm \sqrt{3}/2: \pm 1/2)$ and $\tan \theta = \{ \pm 1/\sqrt 3 : \pm \sqrt 3\}$

Which values depend upon whether $|x| > |y|$ or $|y| > |x|$ and which quadrant $(x,y) lie in. ====== And yes, that circular picture helps. As the circular picture is perfectly symmetric it's easy to memorize even though it does, technically, have 48 values. As it's symmetric it reduces to really only the three cases above. First of all, you know the square diagonal is$\sqrt 2$of the side, so sine and cosine of$45^\circ$is$1/\sqrt 2 = \sqrt 2/2$and tangent is$1$. Recall also the isosceles right triangle and find the described ratios in it (see https://commons.wikimedia.org/wiki/File:Reglas.svg) For the rest of the first quarter just memorize that the common 30-60-90 degree triangle has its shortest to longest edge ratio of$1:2$(see the image linked above) so sine reaches$1/2$at one-third of the right angle. And, from the other corner of the triangle, cosine of two-thirds of a right angle is$1/2$, too (since it is a ratio of the same two lengths). Now, from the Pythagorean theorem, the third side is$\sqrt{1^2-(1/2)^2} = \sqrt 3/2$of the hypotenuse, hence$\sin 60^\circ$and$\cos 30^\circ$. Finally tangent of$60^\circ$is$\frac{\sqrt 3}2 : \frac 12 = \sqrt 3$(tangent is an increasing function in the first quarter and$60^\circ > 45^\circ$, so$\tan 60^\circ > 1$) and tangent of$30^\circ\$ is a reciprocal of it.

The last three rows of your table follow directly from the first row and from trigonometric functions' main properties — being even, odd and periodic. Just familiarize with their graphs and find specific points on them. Pay attention to different symmetries and translations between them.