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

Question

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?

Answer 1

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.

Answer 2

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 $[0,\pi/2]$.

Answer 3

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)

Answer 4

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.

Answer 5

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

$\hskip 1in$

(Or really just the top right triangle, as others have noted.)

Answer 6

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

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

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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?

Answer 7

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.

Answer 8

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.

Answer 9

\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}