We all are familiar with certain special values of sin and cos, e.g. $\sin(30^\circ)=0.5$, $\sin(45^\circ)=\frac{1}{\sqrt{2}}$ etc. In high school I found these values difficult to remember until I noticed they could be reformulated as this pattern:






This has been bugging me for years ever since. I know to expect there to be some deep cause of patterns when I see them, but I have no idea what causes these common special angle values to be half-roots-of-integers, or whether this pattern is just a special case of a more general notion of special trig angles.

What's the explanation? Why should this be so?

  • $\begingroup$ Lol...and you can derive more "special angles" along these lines by the way. $\endgroup$ – Simply Beautiful Art Jan 14 '17 at 17:09
  • $\begingroup$ How? arcsin((sqrt(5)/2) is transcendental, for example. $\endgroup$ – spraff Jan 14 '17 at 17:12
  • $\begingroup$ :-) I'll try to find it. $\endgroup$ – Simply Beautiful Art Jan 14 '17 at 17:17
  • $\begingroup$ To derive $\sin(45°)$, just consider a triangle with angles $90°$ and twice $45°$. It is easy to see that the lengths of the sides are $x,x,\sqrt{2}\cdot x$. Now, basic properties of triangles with an angle of $90°$ are enough to calculate $\sin(45°)$. $\endgroup$ – Peter Jan 14 '17 at 17:43
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    $\begingroup$ Yes, I can see the arguments for specific angles, I'm looking for an argument for the pattern. Or is my way of writing it an arbitrary numerical fluke with no intrinsic meaning? $\endgroup$ – spraff Jan 14 '17 at 18:08

Unsatisfying explanation: $\sin: [0,\pi/2]\to [0,1]$ is a monotonically increasing homeomorphism, so you'll always be able to find values which solve $\sin(a)=b$ for $b\in[0,1]$ and if you pick $b$ to be a sequence like $\frac{\sqrt{k}}{n}$ where $k$ runs from $0$ to $n^2$, you'll see the same sort of mirrored progression in the angles, as $\sin^2(x)+\sin^2(\pi/2-x)=1$.

To me, the real gem here is that rational multiples of $\pi$ give algebraic numbers under $\sin$ and $\cos$ (and therefore under $\tan,\sec,\csc,\cot$ where appropriately defined). Here's why:

Suppose $\theta$ is a rational multiple of $\pi$. We can write it as a rational multiple of $2\pi$: $\theta=\frac{2p\pi}{q}$. By De Moivre's formula, we have that $(\cos\theta+i\sin\theta)^q=\cos(q\theta)+i\sin(q\theta)=1$, so $\cos\theta+i\sin\theta$ is algebraic, and so each of $\cos\theta$ and $\sin\theta$ are as well (if $a+bi$ is algebraic, then $a-bi$ is also algebraic, as algebraicness is preserved under field automorphisms, and so $(a+bi)+(a-bi)=2a$ is algebraic as well, and therefore since algebraic numbers form a field, we have that $a=\frac{2a}{2}$ and $b=\frac{a+bi-a}{i}$ are algebraic).

  • $\begingroup$ I actually find that more satisfying an explanation than the others :-) $\endgroup$ – spraff Jan 17 '17 at 22:12

The intervals between the angles are not uniform. Somehow it's mathematical coincidence:

Since $\sin (90^{\circ}-\theta)=\sqrt{1-\sin^{2} \theta}$,

$$\sin 0^{\circ}=\frac{\sqrt{0}}{2} \iff \sin 90^{\circ}=\frac{\sqrt{4}}{2}$$

$$\sin 30^{\circ}=\frac{\sqrt{1}}{2} \iff \sin 60^{\circ}=\frac{\sqrt{3}}{2}$$

We might invent a mnemonic in this way:

$$\sin 35.26^{\circ}=\dfrac{\sqrt{1}}{\sqrt{3}}$$

$$\sin 54.74^{\circ}=\dfrac{\sqrt{2}}{\sqrt{3}}$$

These angles relate to tetrahedron and are special angles in another way.

Comparing the $\color{green}{\textbf{fitting curve}}$ with $\color{red}{\boldsymbol{\sin \dfrac{\pi x}{12}}}$ below:

enter image description here

  • $\begingroup$ What is the fitting line and what is it's motivation or relevance? $\endgroup$ – mathreadler Jan 17 '17 at 21:03
  • $\begingroup$ Just interpolating the special angles mentioned and comparing with the true sine curve. When I was a student, I used to sketch a sine curve with every $30^{\circ}$-interval as benchmarks - a very good fit. $\endgroup$ – Ng Chung Tak Jan 17 '17 at 21:18

Let's make a table.

$$\begin{array}{ccccc|l}\frac{\sqrt 0}{2}&\frac{\sqrt 1}{2}&\frac{\sqrt 2}{2}&\frac{\sqrt 3}{2}&\frac{\sqrt 4}{2}&\text{sin}\\\hline0&30&45&60&90&\text{angle (deg)} \\0&2&3&4&6&\text{multiple of 15}\\ 1&0&1&0&1&\text{divisible by 45?}\\ 1&1&0&1&1&\text{divisible by 30?}\\\end{array}$$

By periodicity this is a combination of the cyclic groups $\frac{360}{45} = 8$ and $\frac{360}{30} = 12$ considering rotation as the generator. This means we will stay in the set $\pm\left\{\frac{\sqrt 0}{2},\frac{\sqrt 1}{2},\frac{\sqrt 2}{2},\frac{\sqrt 3}{2},\frac{\sqrt 4}{2}\right\}$ if starting from 0, subset $\pm\left\{\frac{\sqrt 0}{2},\frac{\sqrt 1}{2},\frac{\sqrt 3}{2},\frac{\sqrt 4}{2}\right\}$ if 30 degrees and $\pm\left\{\frac{\sqrt 0}{2},\frac{\sqrt 2}{2},\frac{\sqrt 4}{2}\right\}$ if 45 degrees. Number theoretic every integer times 15 deg which is a multiple of 2 or 3. And with a rotation of 15 degrees we can get a rather peculiar overlap.


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