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For example:

  1. $\sin(15^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$
  2. $\sin(18^\circ) = \frac{\sqrt{5}}{4} - \frac{1}{4}$
  3. $\sin(30^\circ) = \frac{1}{2}$
  4. $\sin(45^\circ) = \frac{1}{\sqrt{2}}$
  5. $\sin(67 \frac{1}{2}^\circ) = \sqrt{ \frac{\sqrt{2}}{4} + \frac{1}{2} }$
  6. $\sin(72^\circ) = \sqrt{ \frac{\sqrt{5}}{8} + \frac{5}{8} }$
  7. $\sin(75^\circ) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$
  8. ?

Is there is a list of known exact values of $\boldsymbol \sin$ somewhere?

Found a related post here.

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4  
See this. –  J. M. Jul 30 '12 at 16:51
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Since we have the half angle formula, the list must be infinite... And I am guessing you are asking which values can be expressed as an algebraic number, since all values are known analytically using the formula $\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(1n+1)!}$. –  copper.hat Jul 30 '12 at 16:53
    
@copper.hat: Yes, algebraically is what I am looking for, not approximately or a series. –  ja72 Jul 30 '12 at 16:59
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@ja72 While $\sin(m\pi/n)$ is algebraic for all integer $m$ and $n$, that doesn't mean it's expressable in terms of radicals for all those values. You may find the notion of constructibility ( en.wikipedia.org/wiki/Constructible_polygon )interesting, though that doesn't cover the case of other radicals; for instance, the value of $\sin(\pi/7)$ can be expressed with cube roots (but not with cube roots or real numbers). Have a look at en.wikipedia.org/wiki/Root_of_unity for more information on that case... –  Steven Stadnicki Jul 30 '12 at 17:41
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In what sense do we know the value of $\sqrt{2}$ exactly? –  Qiaochu Yuan Jul 30 '12 at 20:07

4 Answers 4

up vote 7 down vote accepted

We use radian notation. Every rational multiple of $\pi$ has trigonometric functions that can be expressed using the ordinary arithmetic operations, plus $n$-th roots for suitable $n$.

This is almost immediate if we use complex numbers, since $(\cos(2\pi/n)+i\sin(2\pi/n)^n=1$.

But it is known, for example, that there is no expression for $\sin(\pi/9)$ that starts from the integers, and uses only the ordinary operations of arithmetic and roots in which every component is real.

The following more restricted problem has a long history because of its close connection with the problem of which angles are constructible by straightedge and compass.

Let $\theta=\frac{m}{n}\pi$, where $m$ and $n$ are relatively prime. Restrict our algebraic operations to the ordinary operations of arithmetic, plus square roots only, The trigonometric functions of $\theta$ are so expressible iff $n$ has the form $$n=2^k p_1p_2\cdots p_s,$$ where the $p_i$ are distinct Fermat primes.

A Fermat prime is a prime of the form $2^{\left(2^t\right)}+1$. There are only five Fermat primes known: $3$, $5$, $17$, $257$, and $65537$. It is not known whether or not there are more than five.

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@tomasz: Thanks for the suggestion. Done. –  André Nicolas Jul 30 '12 at 19:39

$\sin 3^\circ=\frac{(\sqrt{3}+1) (\sqrt{5}-1)}{8 \sqrt{2}}-\frac{(\sqrt{3}-1) \sqrt{5+\sqrt{5}}}{8}$.

Solving a cubic equation you can get a huge expression for $\sin 1^\circ$ in radicals, and therefore, for any $\sin n^\circ$.

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Can you show your work for $\sin 3^\circ$? –  ja72 Jul 30 '12 at 17:46
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@ja72: Note that $3^\circ=18^\circ-15\circ$. Now use the formula $\sin(a-b)=\sin a\cos b-\cos a\sin b$. In your post you mentioned expressions for the sines of $18^\circ$ and $15^\circ$. From those and the Pythagorean Theorem you get their cosines. –  André Nicolas Jul 30 '12 at 17:55
    
But solving the cubic equation for $\sin 1^\circ$ will require taking the cube root of a complex number, won't it? –  Rahul Jul 30 '12 at 19:12
    
@RahulNarain: Yes, you are right. –  Vladimir Reshetnikov Jul 30 '12 at 20:42
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It's really not so bad if you don't try to put it in one expression. Here's how Maple can optimize it. $$\eqalign{t_1 &= 5^{1/2}\cr t_4 &= (30+6 t_1)^{1/2}\cr t_9 &= (2 t_4(1-t_1)+28-4 t_1)^{1/2}\cr t_{11} &= (-1+t_1+t_4+i t_9)^{1/3}\cr t_{13} &= t_{11}^2\cr t_{22} &= (128+32 t_{11}+2 t_4 t_{13}+2 t_1 t_{13}-2 t_{13}- 2 i t_9 t_{13})^{1/2}\cr t_{25} &= (32-2 t_{22})^{1/2}\cr \text{result} &= t_{25}/8 \cr}$$ –  Robert Israel Jul 31 '12 at 7:20

In January 2008 I posted several references published in the 1800s of tables that give exact values for the sine and cosine of $3$, $6$, $9$, …, $90$ degree angles. (Among the integer degree angles, only those that are multiples of $3$ can be expressed in real-radical form.) See Math Forum archive for 1st post and Math Forum archive for 2nd post.

The best table I know of was prepared by the Belgium mathematician E. Gelin in the 1880s. His table gives a list of values, with rationalized denominators, for all six trig. functions evaluated at $3$, $6$, $9$, …, $90$ degree angles. I know of three places where his table has been published:

Mathesis Recueil Mathematique (1) 8 (1888), Supplement 3. [See pp. 327-333 of the downloaded .pdf file.]

Mathesis Recueil Mathematique (3) 6 (1906), Supplement 3. [See pp. 338-348 of the downloaded .pdf file.]

E. Gelin, Éléments de Trigonométrie Plane et Sphérique (1888). [See pp. 59-62, which is equivalent to pp. 66-69 of the downloaded .pdf file.]

I believe Johann Heinrich Lambert was the first person who published exact radical values for the sine of $3$, $6$, $9$, etc. degree angles. A table of values is in Volume 1 of his Collected Works. The table is in an item that was published in 1770. Lambert’s table was reprinted two or three times in the first half of the 1800s (e.g. one was in Crelle’s Journal [= Journal für die reine und angewandte Mathematik]), but I don’t have the exact references with me now.

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Starting with $\tan(\pi/3)=\sqrt{3}$ and $\tan(\pi/4)=1$ and using $$ \tan(x/2)=\frac{\sqrt{1+\tan^2(x)}-1}{\tan(x)}\tag{1} $$ and $$ \tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}\tag{2} $$ and $$ (\cos(x),\sin(x))=\frac{(1,\tan(x))}{\sqrt{1+\tan^2(x)}}\tag{3} $$ we can construct the sine and cosine of all rational multiples of $\pi$ where the denominator is a power of $2$, or $3$ times a power of $2$.

For example, $x=\pi/4$ with $(1)$ gives $$ \tan(\pi/8)=\sqrt{2}-1\tag{4} $$ then $(2)$ and $(4)$ yields $$ \begin{align} \tan(3\pi/8) &=\tan(\pi/4+\pi/8)\\ &=\frac{1+\tan(\pi/8)}{1-\tan(\pi/8)}\\ &=\sqrt{2}+1\tag{5} \end{align} $$ Then $(3)$ and $(5)$ give $$ (\cos(3\pi/8),\sin(3\pi/8))=\frac{(1,\sqrt{2}+1)}{\sqrt{4+2\sqrt{2}}}\tag{6} $$

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