# Sine using square root [closed]

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.

• 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. Commented Nov 3, 2023 at 21:39
• May we use square roots to produce complex numbers in intermediate steps?
– xnor
Commented Nov 4, 2023 at 1:14
• 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). Commented Nov 4, 2023 at 2:27

# Python, 89 bytes

c='s-1*'
y='/s+2/s++7s5s*6+5s522'
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).

• You could golf your notation slightly by only allowing single digits, which would save thee bytes Commented Nov 3, 2023 at 23:01
• @emanresuA Is that definitely valid? Commented Nov 3, 2023 at 23:04
• 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) Commented Nov 3, 2023 at 23:12
• What's the ^ in output?
– l4m2
Commented Nov 3, 2023 at 23:23
• y+y or'0' => y*2or'0'
– 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:

(1*S(1+S(5)+S(2*S(5)-10))/S(S(12)+S(-4))-1/(1*S(1+S(5)+S(2*S(5)-10))/S(S(12)+S(-4))))/S(-4)


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:

S((1+S(5)+S(2*S(5)-10))/(S(12)+S(-4)))


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-c)/S(-4c)

• Bit of a different answer, but you can save quite a few bytes by using a prefix notation similar to 97.100.67.109's answer Commented Nov 4, 2023 at 9:55
• Can't you write S(20) for 2*S(5)? Commented Nov 4, 2023 at 15:53
• -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})" Commented Nov 5, 2023 at 11:58

# Vyxal, 39 bitsv2, 4.875 bytes

∆R∆sS


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Bitstring:

000100111011111001011001001010111111101


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

## Explained

∆R∆sS­⁡​‎‎⁪⁡⁪⁠⁪⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁪‏‏​⁡⁠⁡‌⁢​‎‎⁪⁡⁪⁠⁪⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁤⁪‏‏​⁡⁠⁡‌⁣​‎‎⁪⁡⁪⁠⁪⁢⁡⁪‏‏​⁡⁠⁡‌­
∆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.

• 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 Commented Nov 3, 2023 at 23:13
• @97.100.97.109 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. 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 0° to 90°. Edit: Saved 2 bytes thanks to @loopywalt.

• You can use sympy.rad to save two bytes. 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.

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

f=->x{m=x%60
m==30?"1/2"[0,90/x]:m==18?"(sqrt5-1)/4/-2+(sqrt(30+sqrt180)/8"[0,x/13*11]:"sqrt(1/2+#{s=x<=>45}sin(#{s*(x*2-90)})/2)"}


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# Mathematica/Wolfram, 27 bytes

f=FunctionExpand@Sin[#*°]&


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}]


• Welcome to Code Golf, and nice first submission! Mathematica is actually on TIO: Try it online! Commented Nov 6, 2023 at 18:20
• @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. Commented Nov 6, 2023 at 20:35