Given input \$x \in \left\{0,3,6,...,90\right\}\$, output \$\sin\left(x°\right)\$ using integer and \$+ - \times \div ( ) \sqrt{\cdot}\$(square root), e.g. \$\sin(45°)=\sqrt{1\div 2}\$.

Flexible output format. You needn't do any simplification to output, aka. it's fine to output \$\sin(0°)=\sqrt{42}-\sqrt{6\sqrt{-9+58}}\$. Complex intermediate value is fine but please mention.

Shortest code wins.

  • \$\begingroup\$ What exactly is the output format? Especially for square root, how would that be represented, as I understand we're supposed to output a string of some sort. \$\endgroup\$ Commented Nov 3, 2023 at 21:39
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    \$\begingroup\$ May we use square roots to produce complex numbers in intermediate steps? \$\endgroup\$
    – xnor
    Commented Nov 4, 2023 at 1:14
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    \$\begingroup\$ May we use - as both subtraction and as negation? Your example shows it as subtraction. (Ideally we shouldn't have to guess the pattern from examples). \$\endgroup\$
    – Wheat Wizard
    Commented Nov 4, 2023 at 2:27

6 Answers 6


Python, 89 bytes

f=lambda d:d and'+*'+y+(x:=f(d-3))+'*'+c+x*2+c+y*2or'0'

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It times out when trying to run it in ATO when trying to run it for large degrees, but it should work in theory. It outputs in prefix notation, where s is the unary square root operator; all other operators are binary. Furthermore, in this notation, there are only single digits (e.g. 20 represents a 2 then a 0).

It works by repeatedly applying the formula $$ \sin(a+3^{\circ}) = \sin(a) \cos(3^{\circ}) + \cos(a) \sin(3^{\circ}) \\ = \sin(a) \cos(3^{\circ}) + \sqrt{1-\sin(a)\sin(a)} \sqrt{1-\cos(3^{\circ})\cos(3^{\circ})}$$

where \$\cos(3^\circ) = \sin(87^\circ)\$ equals $$\frac12 \sqrt{2+\frac12 \sqrt{7+\sqrt{5}+\sqrt{6(5+\sqrt{5})}}}$$

(according to Wolfram-Alpha).

  • 1
    \$\begingroup\$ You could golf your notation slightly by only allowing single digits, which would save thee bytes \$\endgroup\$
    – emanresu A
    Commented Nov 3, 2023 at 23:01
  • \$\begingroup\$ @emanresuA Is that definitely valid? \$\endgroup\$ Commented Nov 3, 2023 at 23:04
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    \$\begingroup\$ It's a perfectly valid and unambiguous representation, and I don't think it's particularly overoptimised for this challenge (which would probably fall under a loophole) \$\endgroup\$
    – emanresu A
    Commented Nov 3, 2023 at 23:12
  • \$\begingroup\$ What's the ^ in output? \$\endgroup\$
    – l4m2
    Commented Nov 3, 2023 at 23:23
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    \$\begingroup\$ y+y or'0' => y*2or'0' \$\endgroup\$
    – l4m2
    Commented Nov 3, 2023 at 23:29

Python 3, 75 bytes

lambda x:f"(1-1{(s:=x//3*'*(1+S(5)+S(S(20)-10))/(S(12)+S(-4))')})/S(-4{s})"

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This produces an expression with complex numbers as intermediates that can be evaluated in Python with S=cmath.sqrt. For instance, x=3 gives:


The main idea is to work with complex numbers, in particular roots of unity. Since \$e^{i \theta}\ = \cos(\theta) + i\sin(\theta)\$ where \$i = \sqrt{-1}\$, we can work with multiples of \$\theta\$ as

$$\ \cos(k\theta) + i\sin(k\theta)= e^{i k\theta}\ = \left(e^{i \theta}\right)^k = \left( \cos(\theta) + i\sin(\theta)\right)^k$$

From there, we can extract just the sine with:

$$\sin(k \theta) = \frac{\left(e^{i \theta}\right)^k-1/\left(e^{i \theta}\right)^k}{2i}$$

We can take the \$k\$ power by multiplying \$k\$ times, that is concatenating \$k\$ copies of *stuff in the expression, since the challenge doesn't allow exponents in the formula.

So we express \$e^{i \theta}\$ for \$\theta = 3^{\circ}\$ (three degrees), and take its \$k\$-th power by multiplying it \$k\$ times, using string repetition.

It remains to express \$e^{i \theta}\$ for \$\theta = 3^{\circ}\$. We could directly write out its sine and cosine, but it's a bit shorter to express it in terms of 36 and 30 degrees:

$$e^{i (3^\circ)} = \sqrt{e^{i (6^\circ)}} = \sqrt{e^{i (36^\circ)}e^{i (-30^\circ)}} = \sqrt{(\cos(36^\circ)+i \sin (36^\circ))(\cos(30^\circ)-i \sin (30^\circ))}$$

These angles have relatively nice sines and cosines. A bit of tweaking and combining terms gives this in the code's notation as:


If we call S(c) the value above, the value we want to output is:

(S(c) - 1/S(c))/S(-4)

using that \$\sqrt{-4} = 2i\$. Finally, an improvement from @loopwalt that saved 6 bytes is to simplify this to

  • 1
    \$\begingroup\$ Bit of a different answer, but you can save quite a few bytes by using a prefix notation similar to's answer \$\endgroup\$
    – emanresu A
    Commented Nov 4, 2023 at 9:55
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    \$\begingroup\$ Can't you write S(20) for 2*S(5)? \$\endgroup\$
    – loopy walt
    Commented Nov 4, 2023 at 15:53
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    \$\begingroup\$ -6 I believe lambda x:f"(1-1{(s:=x//3*'*(1+S(5)+S(S(20)-10))/(S(12)+S(-4))')})/S(-4{s})" \$\endgroup\$
    – loopy walt
    Commented Nov 5, 2023 at 11:58

Vyxal, 39 bitsv2, 4.875 bytes


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Outputs square root as sqrt(...), and division as /. Uses sympy to calculate the exact sine value, and then casts to string.


∆R     # ‎⁡Convert the input to radians
  ∆s   # ‎⁢find sine of that
    S  # ‎⁣cast to string, which stops conversion to float. Instead, it uses the exact representation, which is expressed in terms of integers.

Created with the help of Luminespire.

  • \$\begingroup\$ I always feel a little bit silly when I write a solution in the triple-digit byte counts, then see a solution in a golfing language in under a dozen bytes :P \$\endgroup\$ Commented Nov 3, 2023 at 23:13
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    \$\begingroup\$ @ I feel, if anything, the opposite. Meaning I'll sometimes feel the builtin is just a cheat. Not that I am against golfing languages by a longshot. They certainly have their own beauty and finding the exact right tool or builtin for the job is undoubtedly a skill, can require broad knowledge of many languages in some cases, and is its own sort of satisfaction. But I also often feel the more work you are doing as a programmer, from a pure problem-solving perspective, the more impressive. TLDR: don't feel silly, it's apples and oranges. \$\endgroup\$
    – Jonah
    Commented Nov 4, 2023 at 1:21

Python 3 + SymPy, 41 39 bytes

lambda n:sin(rad(n))
from sympy import*

Try it online! Link outputs all results from to 90°. Edit: Saved 2 bytes thanks to @loopywalt.

  • \$\begingroup\$ You can use sympy.rad to save two bytes. \$\endgroup\$
    – loopy walt
    Commented Nov 4, 2023 at 9:20

Ruby, 132 110 bytes

f=->x{(x%60-24).abs==6?%w{(s5-1)/4 1/2 (s(30+s180)+s5-1)/8 1}[x/30]:"s(1/2+#{q=x<=>45}*(#{f[q*(x*2-90)]})/2)"}

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While existing answers are very clever (and shorter than this!), the output expressions are rather long. This recursive function uses the half-angle formulas to produce expressions that fit on a single line in all cases.

Half angle formulas (rules for determining correct sign in all quadrants are unnecessary and omitted):

cos(a/2)=sqrt(1/2+(cos a)/2) or cos x=sqrt(1/2+(cos x*2)/2)

sin(a/2)=sqrt(1/2-(cos a)/2) or sin x=sqrt(1/2-(cos x*2)/2)

Below is a development version of the code, showing how the half angle formula is used to recursively derive expressions for all inputs. There are four infinite cycles as follows: 30>30, 90>90, 18>54>18, 78>66>42>6>78. These are broken with explicit values for 30,90,18 and 78 (note even the fairly complex base expression for 78 is much simpler than the base expression for 3 used in other answers.)

Development code with table showing how each value is derived by recursion


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Mathematica/Wolfram, 27 bytes


Try it online (thx to @noodle man)

This is a function solution - # introduces a placeholder for argument and & says whatever comes before is a pure function. ° is builtin for number of radians in a degree, @ is function application without brackets. Beyond that just relying on the underlying symbolic computation engine.

The full list of answers can be generated via:

Table[f[i * 3], {i, 0, 30}]

enter image description here

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    \$\begingroup\$ Welcome to Code Golf, and nice first submission! Mathematica is actually on TIO: Try it online! \$\endgroup\$ Commented Nov 6, 2023 at 18:20
  • \$\begingroup\$ @noodleman thanks! I was kind of assuming Wolfram would never license it, so I didn't bother to actually check - apparently my assumptions were incorrect. \$\endgroup\$ Commented Nov 6, 2023 at 20:35

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