# Integers in cosine

## Integers in cosine

From trigonometry we know that

$$\\sin(a) =-\cos(a + \frac{(4*m + 1)\pi}{2})\$$

where $$\a\$$ is an angle and $$\m\in\mathbb{Z}\$$ (integer).

For an input of a positive integer $$\n\$$ calculate the value of the integer $$\k\$$ with $$\n\$$ digits that minimizes the difference between $$\\sin(a)\$$ and $$\-\cos(a + k)\$$. Or in a simpler way, generate the sequence A308879.

1 digit : 8
2 digits: 33
3 digits: 699
4 digits: 9929
5 digits: 51819
6 digits: 573204

### Output

You can create a function or print the result on the standard output. Your program needs to be able to calculate at least the 6 cases in the section above.

### Scoring

This is , shortest code in bytes wins.

### Reference

https://www.iquilezles.org/blog/?p=4760

• This is A308879. You know, it's fairly obvious. – user92069 May 11 at 14:34
• I tried searching for it and didn't find anything, thanks for finding me this :) – Gábor Fekete May 11 at 15:01
• Can the C code in OEIS really compute the answer for 12 digit – user9207 Jul 18 at 14:58

# Wolfram Language (Mathematica), 25 bytes

This uses a trick: Mathematica doesn't expand Sin[51819] into a proper number even when Max is called on a list of such things, and because of how Last works, it doesn't care that Sin[51819] is not a list and simply returns 51819.

Last@Max@Sin@Range[10^#]&


Try it online!

• -1 replacing Last@ with #&@@. – att Jul 18 at 19:22

# Ruby, 37 bytes

->n{(1..10**n).max_by{|k|Math.sin k}}


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Minimising the difference between $$\\sin(a)\$$ and $$\-\cos(a+k)\$$ amounts to minimising $$\|k-(4m+1)\pi/2|\$$ for some $$\m\$$, or equivalently maximising $$\\sin(k)\$$.

Iterating over (1..10**n) works at least for $$\n\le6\$$, as required. More generally, according to the OEIS (reference courtesy of @Λ̸̸'s 05AB1E answer):

It is not guaranteed that each term in the sequence produces a better approximation than the previous one, although numerical evidence suggests so.

Iterating over (10**~-n...10**n) is guaranteed to work for any $$\n\$$, at a cost of 7 extra bytes.

• This seems to work for n <= 8 for me. – the default. May 11 at 14:14
• @mypronounismonicareinstate Hence why I said 'at least for $n\le6$'... admittedly I was too lazy to keep checking beyond that. Thanks for improving the upper bound :) – Dingus May 11 at 14:20
• Nice work, and +1. I think it would break for an n where the answer for n+1 is 10**n though (due to the inclusive range) - no idea if such an n exists, but I imagine it does. – Jonathan Allan May 11 at 18:00
• @JonathanAllan Yes, you're right. One of the 7 extra bytes in the last sentence of my answer is taken up by switching to an exclusive range. – Dingus May 11 at 22:08

# 05AB1E, 7 bytes

Port of Dingus's answer. (It's A308879. From that page, it says that the sequence is also "the n-digit integer m that maximizes sin(m)"...)

°LΣÅ½}θ


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## Explanation

°       10 ** n
L      1 range
Σ     Sort by:
Å½}  Sine
θ Maximum under that mapping

• Nice approach with the sort-by with last item! :) An alternative could have been this 7-byter: °LÅ½Zk>, with an implicit map for the Sine builtin; maximum without popping; and getting the index+1 of this maximum in the list. Too bad 05AB1E doesn't use 1-based indexing in this case. ;p – Kevin Cruijssen May 11 at 15:06
• Including 10**n could be problematic (as well as inluding numbers less than 10**(n-1) - as noted on the OEIS page). – Jonathan Allan May 11 at 18:17

# R30 27 bytes

Following the same thread as @Dingus' Ruby answer, as indicated in the entry to A308879, i.e. looking at the maximum value of $$\\sin(k)\$$ over $$\1,\ldots,10^n\$$:

which.max(sin(1:10^scan()))


Try it online!

• an alternative (which comes out to the same length, I think) is to use which.max. – JDL May 12 at 7:42
• @JDL: thanks. This was my first guess (see TIO) but it adds one byte. I am wondering why which.max(sin(1:10^scan())) alone does not work... – Xi'an May 12 at 10:13
• it does work for me (though with scan you need to enter an empty line after the n) – JDL May 12 at 10:18
• Yes thank you, I just realised I had left the f= header in my tio code! I thus corrected my answer. – Xi'an May 12 at 10:21

# Java 8, 91 bytes

n->{double r=0,m=0,s,k=Math.pow(10,n);for(;k-->1;)if((s=Math.sin(k))>m){m=s;r=k;}return r;}


Port of @Dingus' Ruby answer, so make sure to upvote him!

Try it online. (It's barely able to reach up to $$\n=9\$$ on TIO before it times out.)

Explanation:

n->{                           // Method with integer parameter and double return-type
double r=0,                  //  Result-double, starting at 0
m=0,                  //  Maximum m, starting at 0
s,                    //  Integer s for the Sine calculation, uninitialized
k=Math.pow(10,n);for(;k-->1; //  Loop k in the range (10^input, 1]:
if((s=Math.sin(k))         //   Set s to the Sine of k
>m){                    //   If this s is larger than the current maximum m:
m=s;                     //    Replace the maximum with this s
r=k;}                    //    And replace the result with k
return r;}                   //  Return the result after the loop

• @JonathanAllan I am? (Although I must admit it's coincidentally to be completely honest.) It currently loops in the range (10^n, 1] (exclusive on the left; inclusive on the right) or as notation that's inclusive on both side: [10^n-1, 1]. Dunno if the same applies to the other answers though, since I'm not too familiar with those other languages. To clarify: it starts k at $10^n$, but in the loop-check it uses k-->1 (check if $k>1$, and decrease it by 1 right after that with k--. So in the body of the very first iteration, k will be $10^n-1$. – Kevin Cruijssen May 11 at 18:09
• Oh, so you are, my bad! – Jonathan Allan May 11 at 18:15

# JavaScript (Node.js), 64 63 bytes

n=>[...Array(t=10**n)].map(_=>[9-Math.sin(--t),t]).sort()[0][1]


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# Japt-h, 10 bytes

ApU õ ñ!sM


Try it

# APL (Dyalog Extended), 9 bytes

⊃⍒1○⍳10*⎕


-7 bytes from ngn and Adám.

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# APL (Dyalog Extended), 20 16 bytes

{i[⊃⍒1○i←⍳10*⍵]}


-4 bytes from Bubbler.

Try it online!

Implements Dingus's algorithm.

## Explanation

{i[⊃⍒1○i←⍳10*⍵]}
{              } function body
i     i←⍳10*⍵   assign to i the range of 1 to 10^n
1○          find the sine of each value in i
⊃⍒          Grade(sort) by descending value, select index of first(maximum) value
i[           ]  return k in i where sin k is maximum


# Python 3, 51 bytes

import math
lambda n:max(range(10**n),key=math.sin)


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Uses Dingus's observation about maximizing sine.

# Fortran (GFortran), 53 51 bytes

read*,n
print*,maxloc([(sin(1d0*k),k=1,10**n)])
end


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Port of my own Ruby answer. Because why not?