One might (arbitrarily) define the "special angles" as angles $\theta \in [0, \pi)$ such that

$\sin(\theta) = \left. \left\{ \frac{\sqrt{k}}{2} \right\} \right|_0^4 = \left\{0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1 \right\}$

which are of course the angles

$\theta' = \left\{ 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} \right\}$.

(Using prime to denote "special", i.e. belonging to this set.)

I'm wondering if there is a pattern underlying the relationship between the "special angles" and the sine-outputs; i.e. if a discrete index-based formula could be constructed that yields the value of sine for any of the special angles. I'm thinking of the form

$f(\theta ') = \sin(\theta ')$

which I only expect to be valid for the special angles $\theta '$ but might possibly be a very simple function (if we are lucky).

I believe constructing such a formula requires writing the sequence

$ \left\{ \infty, 6, 4, 3, 2 \right\} \left(=\frac{\pi}{\left\{ \theta' \right\} }\right)$

in standard nth-term form $ \left\{ a_n \right\}$ with a discrete index $n$, but I'm not seeing the pattern. If $\frac{\pi}{5}$ were a special angle, the pattern would nearly trivial. This makes me wonder how such an apparently "unnatural" pattern (which excludes 5) arises in the natural mathematics of trigonometry.

Any insight into constructing $f(\theta ')$ or explaining the anomalous nature of $\frac{\pi}{5}$ would be appreciated.

  • 1
    $\begingroup$ The "regularity" of sines and cosines for these angles is mere chance. It relies on no important hidden concept. Thus we have not to think to $\pi/5$ as a kind of "paria". $\endgroup$ – Jean Marie Dec 24 '16 at 19:52

The angles $\pi/2$ and $\pi/3$ are special because $\pi/2$ is a square angle and $\pi/3$ is the interior angle of an equilateral triangle; because of that, they have particularly easy sines and cosines.

The next two, $\pi/4$ and $\pi/6$, are their halves, and because of that they also have easy formulas (when $x=\pi/2$ of $\pi/3$): $$ \sin\frac x2=\sqrt{\frac{1-\cos x}2} $$ (in the first half of the circle). The angle $\pi/5$ does not fit in this, and so I see no reason to expect a particularly easy formula.


Overly extensive information is given in Wolfram's page on $\sin(\pi/5)$. It even provides the closed form solution:



For every integer $n\ge1$, $\sin \pi/n$ is a root of a polynomial $p_n(x)$ with integer coefficients; these are closely related to the Chebyshev polynomials, which express $\cos nx$ as a polynomial in $\cos x$. Here are the minimal polynomials of $\sin \pi/n$ for $n\le 10$:

$$ \begin{array}[c|l] n&p_n(x)\\ \hline\\ 1&x\\ 2&x-1\\ 3&4x^2-3\\ 4&2x^2-1\\ 5&4x^2-20x+5\\ 6&2x-1\\ 7&64x^6-112x^4+56x^2-7\\ 8&8x^4-8x^2+1 \end{array} $$

The only thing special about $n=1$, $2$, $3$, $4$ and $6$ is that these are the only values of $n$ such that $p_n(x)$ has degree at most two. $\sin\pi/5$ is still pretty special, as the roots of $p_5(x)$ can be expressed in terms of radicals. This will be true for infinitely many, but not all $n$; the smallest $n$ for which $\sin \pi/n$ cannot be expressed in terms of radicals is $n=11$.

A nice way to see these polynomials is via deMoivre's formula. Write $\sin x = (u-1/u)/2i$ where $u=e^{ix}$. For $n=3$ the binomial theorem gives $$\begin{align*} (2i)^3\sin^3 x &=(u-1/u)^3\\ &=(u^3-3u+3/u-1/u^3)\\ &=2i(\sin 3x -3\sin x) \end{align*}$$ giving $\sin 3x=3\sin x- 4\sin^3x$. Taking $x=\pi/3$ and $z=\sin \pi/3$, we get $$0= 3z-4z^3,$$ giving $p_3(z)$. You can get $p_5$ similarly.

  • $\begingroup$ The Chebyshev polynomials are fascinating, thank you for introducing me to them. $\endgroup$ – electronpusher Dec 26 '16 at 1:06

You can use the addition and subtraction formula together with a table of sine and cosine values to calculate exponentially more values.

$$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)$$ together with $$\sin\left(\sum_{k=1}^n \frac{d_k\pi}{2^k}\right), \text{where } d_k\in\{0,1\}$$

Factoring out $\pi$, and knowing the binary number system can represent all numbers ( with sufficient number of digits provided ), we can accomplish what you ask for, using a small table of preset values to calculate any sine or cosine up to that resolution (in bits).

  • $\begingroup$ I am not fluent enough in binary to make good use of this but I appreciate the alternative perspective! $\endgroup$ – electronpusher Dec 26 '16 at 1:05

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