# Memorizing the identities $\cos {\pi \over 3}=\sin {\pi \over 6} = {1 \over 2}$

I memorized $\sin {\pi \over 4} = \cos {\pi \over 4}= {1\over \sqrt{2}}$ easily by using the diagonal inside the unit square.

I am having great trouble memorizing the identities $\cos {\pi \over 3}=\sin {\pi \over 6} = {1 \over 2}$ because I keep confusing whether it is $\cos {\pi \over 3}$ or $\cos {\pi \over 6}$ that equals ${1\over 2}$.

Is there a picture similar to the unit square picture or something like it to memorize this identity?

• Why would you want to memorize these identities?! Feb 13, 2015 at 5:30
• Why wouldn't you want to memorize these? I certainly don't want to consult a reference or calculator every time I need to know what $\cos\dfrac{\pi}{3}$ is. Feb 13, 2015 at 5:32
• @Anna, See geogebratube.org/student/m2446 Feb 13, 2015 at 5:36
• @BaronVT, it is a rather useless piece of information. I for one have never used it as a mathematician, ever. If you are in a situation in which you end up having to consult a reference often for a particular value (for example, in doing exercises about values of trigonometric funcions...) then the very act of consulting repetitively the reference will help you memorize it! This applies to everything, from values of trigonometric functions to useful things. Feb 13, 2015 at 6:04
• In (mathematical and real) life, rotations of order 3, 4 and 6 are special: they are the ones popping up in crystallography, in symmetries of elliptic curves, in the torsion of $GL_2(\mathbb Z)$... Basically, all these more or less equivalent facts reduce to the fact that there are very few unit complex numbers whose real part is an integer or half-integer. I think that's enough of a reason to remember that $\cos(\pi/3) = 1/2, \cos(\pi/2) = 0, \cos(2\pi/3) = -1/2$. And while I must accept you never use them "as a mathematician", I know loads of mathematicians who treat them with less disdain. Feb 13, 2015 at 9:16

There is an old trick, just memorize that $$\sin 0 =\frac{\sqrt{0}}{2}\qquad \sin \frac{\pi}{6}=\frac{\sqrt{1}}{2} \qquad \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2} \qquad \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} \qquad \sin \frac{\pi}{2}=\frac{\sqrt{4}}{2} \qquad$$ and $\cos x$ goes the other way.

For a 30-60-90 triangle, the sides follow the pattern $x, x \sqrt{3}, 2x$. You can see this by drawing a 30-60-90 triangle and noticing that it is half of an equilateral triangle. If you know that $\sin\frac\pi4=\frac1{\sqrt2}$, it is easy to remember that $\frac12$ is the value of $\sin\frac\pi6$, not $\cos\frac\pi6$, because $\sin$ is increasing on $[0,\pi/2]$.

• don't know why but i confuse pi/6 and pi/3 with 30 and 60, yes putting pi as 180 works but in quick application the denominators do confuse. :( Feb 13, 2015 at 5:52
• You are right, I don't know why I didn't think of that! Than you for your answer.
– Anna
Feb 13, 2015 at 8:11

Work with them enough, and they will become second nature. In the meantime, here is a mnemonic that might help:

For the "important" angles $0, \dfrac\pi6, \dfrac\pi4, \dfrac\pi3, \dfrac\pi2$, the sines of the angles are:

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

(of course most of these can be reduced, but the "square root of $0,1,2,3,4$" pattern is what is easy to remember)

Yes, there is a quick mental picture, the inscribed hexagon:

$\hskip 1in$ (Or really just the top right triangle, as others have noted.)

\begin{array}{|c|c|c|c|c|c|} \hline {}&{0^\circ}&{30^\circ}&{45^\circ}&{60^\circ}&{90^\circ} \\ \hline {\rm sin (\alpha)} &\sqrt 0 \over 2&{\sqrt 1 \over 2}&{\sqrt 2 \over 2}&{\sqrt 3 \over 2}&{\sqrt 4 \over 2}\\ \hline {\rm cos (\alpha)} &{\sqrt 4 \over 2}&{\sqrt 3 \over 2}&{\sqrt 2 \over 2}&{\sqrt 1 \over 2}&{\sqrt 0 \over 2}\\ \hline {\rm tan (\alpha)} &0&{{1} \over \sqrt{3}}&{1}&{\sqrt 3}&\infty\\ \hline {\rm ctan (\alpha)} &\infty&{\sqrt{3}}&1&{1 \over \sqrt 3}&0 \\ \hline \end{array}

I memorize "an equilateral triangle has equal sides and equal angles." Since I previously memorized "the angles in a triangle add up to $180^\circ$", that means it has $60^\circ$ angles. By cutting it in half I get a triangle with angles of $30^\circ$, $60^\circ$, and $90^\circ$. If I call the side length of the equilateral triangle $1$, then the right triangle has a hypotenuse of length $1$ and the side opposite the $30^\circ$ angle has length $\frac12$. I use Pythagoras (which I also have memorized) to get the third side. Um, I also need to have memorized that the size of an acute angle in a right triangle is the ratio of the opposite side to the hypotenuse.

• Thank you, I like this answer because it is like what I did with the square. I don't even need to find the third side: I use the cosine and the two sides that I have. Thank you very much!
– Anna
Feb 13, 2015 at 8:13

A few possibilities:

• Draw a right triangle with angle $\pi/3$ (or so). It'll be clear that $\sin \frac{\pi}{3} > \cos \frac{\pi}{3}$.

• Remember that $\cos \theta$ is decreasing on $[0, \frac{\pi}{2}]$ (a diagram should make that clear if you don't remember it offhand), and note thta $\cos \frac{\pi}{2} = \frac{1}{\sqrt{2}} < \frac{1}{2}$.

• Use the double-angle formula: $$\cos \frac{\pi}{3} = 2\cos^2 \frac{\pi}{6} - 1 = 2\sin^2 \frac{\pi}{3} - 1 = 1 - 2\cos^2 \frac{\pi}{3}.$$ Solve the quadratic equation to get $\cos \frac{\pi}{3} = \frac{1}{2}$. (This is a terrible mnemonic, but it's useful to rederive the value of $\cos \frac{\pi}{3}$ if you forget it.)

In any case, after working with it for a while, you won't need to make an effort to remember it; it'll just become second nature.

$$\large \text{Practise and Practice: Do more questions!}$$

I remembered: "sin 30 is half", dont know why but that fat(past form of fit) into my mind.

Another aid for "sin 30=cos60" is that the ratio of sides you ust be knowing as $1,\sqrt3,2$ and this ratio must be one of $1/2,\sqrt3/2$. Imagine the triangle, the side opposite to 30 will be smaller than side opposite to 60 so can you know assign the ratios, with of course the largest side, the hypotenuse as 2?

• @Anna edited... Feb 13, 2015 at 5:50
• As for your postscript: American English only uses the 'practice' spelling. British English uses 'practice' for the noun and 'practise' for the verb. Canadian English follows the same setup, although it allows but disprefers 'practice' as a verb. Feb 13, 2015 at 5:57
• @anomaly What kind of English uses 'disprefer' as a verb?
– bof
Feb 13, 2015 at 6:13
• @bof: The kind of English used by people who study linguistics. Feb 13, 2015 at 7:14