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Introduction

The two most common trigonometric functions, sine and cosine (or sin and cos for short), can be extended to be matrix-valued functions. One way to compute the matrix-valued analogs is as follows:

Consider these two important trigonometric identities: $$ e^{i x} = \cos(x) + i \sin(x) \\ \sin^2(x) + \cos^2(x) = 1 $$ Using these identities, we can derive the following equations for sin and cos: $$ \sin(x) = \frac{e^{i x} - e^{-i x}}{2 i} \\ \cos(x) = \frac{e^{i x} + e^{-i x}}{2} $$ The matrix exponential exists for all square matrices and is given by: $$ e^A = \sum_{k = 0}^\infty \frac{1}{k!} A^k $$ where A0 is the identity matrix I with the same dimensions as A. Using the matrix exponential, these two trigonometric functions (and thus all the other trigonometric functions) can be evaluated as functions of matrices.

The Challenge

Given a square matrix A, output the values of sin(A) and cos(A).

Rules

  • Input and output may be in any convenient, reasonable format (2D array, your language's matrix format, etc.).
  • You may write a single program, two independent programs, a single function, or two functions. If you choose to write two functions, code may be shared between them (such as imports and helper functions).
  • The input matrix's values will always be integers.
  • Your solution may have accuracy issues as the result of floating-point imprecision. If your language had magical infinite-precision values, then your solution should work perfectly (ignoring the fact that it would require infinite time and/or memory). However, since those magical infinite-precision values don't exist, inaccuracies caused by limited precision are acceptable. This rule is in place to avoid complications resulting from requiring a specific amount of precision in the output.
  • Builtins which compute trigonometric functions for matrix arguments (including hyperbolic trig functions) are not allowed. Other matrix builtins (such as multiplication, exponentiation, diagonalization, decomposition, and the matrix exponential) are allowed.

Test Cases

Format: A -> sin(A), cos(A)

[[0]] -> [[0]], [[1]]
[[0, 2], [3, 5]] -> [[-0.761177343863758, 0.160587281888277], [0.240880922832416, -0.359709139143065]], [[0.600283445979886, 0.119962280223493], [0.179943420335240, 0.900189146538619]]
[[1, 0, 1], [0, 0, 0], [0, 1, 0]] -> [[0.841470984807897, -0.158529015192103, 0.841470984807897], [0, 0, 0], [0, 1, 0]], [[0.540302305868140, -0.459697694131860, -0.459697694131860], [0, 1, 0], [0, 0, 1]]
[[1, 0, 0, 0, 0], [0, 1, 0, 0, 0], [0, 0, 1, 0, 0], [0, 0, 0, 1, 0], [0, 0, 0, 0, 1]] -> [[0.841470984807897, 0, 0, 0, 0], [0, 0.841470984807897, 0, 0, 0], [0, 0, 0.841470984807897, 0, 0], [0, 0, 0, 0.841470984807897, 0], [0, 0, 0, 0, 0.841470984807897]], [[0.540302305868140, 0, 0, 0, 0], [0, 0.540302305868140, 0, 0, 0], [0, 0, 0.540302305868140, 0, 0], [0, 0, 0, 0.540302305868140, 0], [0, 0, 0, 0, 0.540302305868140]]
[[-3, 2, -6], [3, 0, 4], [4, -2, 7]] -> [[-0.374786510963954, 0.135652884035570, -1.35191037980742], [1.14843105375406, 0.773644542790111, 1.21625749577185], [1.21625749577185, -0.135652884035570, 2.19338136461532]], [[4.13614256031450, -1.91289828483056, 5.50873853927692], [-2.63939111203107, 1.49675144828342, -3.59584025444636], [-3.59584025444636, 1.91289828483056, -4.96843623340878]]

Further Reading

This excellent question over at Math.SE includes some alternate derivations of the matrix-valued analogs of trigonometric functions.

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  • \$\begingroup\$ I got sin([[1, 0, 1], [0, 0, 0], [0, 1, 0]]) = {{0.841, -0.158, 0.841}, {0, 0, 0}, {0, 1, 0}} with Mathematica, can you check? \$\endgroup\$
    – kennytm
    May 5, 2016 at 17:37
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    \$\begingroup\$ @kennytm That is what the test case shows. \$\endgroup\$
    – user45941
    May 5, 2016 at 17:38
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    \$\begingroup\$ @Mego Apparently all of the existing answers should be deleted then. \$\endgroup\$
    – feersum
    May 5, 2016 at 19:30
  • 3
    \$\begingroup\$ @Mego It's completely unreasonable to think that all of the floating-point based builtins use an exact algorithm (or one that would be exact if floating-point operations were replaced with "real number" operations). \$\endgroup\$
    – feersum
    May 5, 2016 at 19:35
  • 1
    \$\begingroup\$ @feersum I've addressed that in my latest edit: (ignoring the fact that it would require infinite time and/or memory) \$\endgroup\$
    – user45941
    May 5, 2016 at 19:52

9 Answers 9

7
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Julia, 33 19 bytes

A->reim(expm(A*im))

This is a function that accepts a 2-dimensional array of floats and returns a tuple of such arrays correponding to the cosine and sine, respectively. Note that this is the reverse of the order given in the test cases, in which sine is listed first.

For a real-valued matrix A, we have

sine

and

cosine

That is, the sine and cosine of A correspond to the imaginary and real parts of the matrix exponential eiA. See Functions of Matrices (Higham, 2008).

Try it online! (includes all test cases)

Saved 14 bytes thanks to Dennis!

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Mathematica, 27 bytes

{Im@#,Re@#}&@MatrixExp[I#]&

Based on @Rainer P.'s solution.

Takes the square matrix A as an argument and outputs a list containing {sin(A), cos(A)}.

The input is formatted with N to get a numerical value instead of a long exact formula and Column to display the results of sin(A) and cos(A) as separate matrices instead of a nested list.

Example

Calculating the values separately requires 38 bytes

{(#2-#)I,+##}/2&@@MatrixExp/@{I#,-I#}&
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6
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Jelly, 23 22 bytes

³æ*÷!
®Ḥ‘©r0Ç€s2_@/µÐL

Try it online!

Background

This approach directly computes the Taylor series for sine and cosine, i.e.,

formula

It increases the number of initial terms of both series until the result no longer changes, so its accuracy is only limited by the precision of the floating point type.

How it works

®Ḥ‘©r0Ç€s2_@/µÐL  Main link, Argument: A (matrix)

             µÐL  Loop; apply the chain until the results are no longer unique.
                  Return the last unique result.
®                   Yield the value of the register (initially zero).
 Ḥ                  Unhalve/double it.
  ‘©                Increment and copy the result (n) to the register.
    r0              Range; yield [n, ..., 0].
      ǀ            Apply the helper link to each k in the range.
        s2          Split the results into chunks of length 2. Since n is always
                    odd, this yields [[Ç(n), Ç(n-1)], ..., [Ç(1), Ç(0)]].
          _@/       Reduce the columns of the result by swapped subtraction,
                    yielding [Ç(1) - Ç(3) + ... Ç(n), Ç(0) - Ç(2) + ... Ç(n - 1)].


³æ*÷!             Helper link. Argument: k (integer)

³                 Yield the first command-line argument (A).
 æ*               Elevate A to the k-th power.
    !             Yield the factorial of k.
   ÷              Divide the left result by the right one.
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4
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C++, 305 bytes

#include<cmath>
#include<iostream>
#include<vector>
int x,i=0, j;void p(std::vector<double> v){int x=sqrt(v.size());for(i=0;i<x;i++){for(j=0;j<x;j++) std::cout << v[x] << " ";std::cout << "\n";}}int main(){std::vector<double> s, c;while(std::cin >> x){s.push_back(sin(x));c.push_back(cos(x));}p(s);p(c);}

Input is a list of numbers that are a perfect square on stdin. Output is a pretty printed 2d array on stdout

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Matlab, 37 bytes

@(A){imag(expm(i*A));real(expm(i*A))}
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1
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Matlab, 138 121 52 50 bytes

Since matrix exponentiation is allowed, (what I didnt notice first, d'oh) I do not need to define my helper funciton anymore, and the whole thing can be trivially solved:

A=input('')*i;a=expm(A);b=expm(-A);[(b-a)*i,a+b]/2

The input should be a matrix e.g. [1,2;4,5] or alternatively [[1,2];[3,4]]

An unexpected thing (in hindsight not so unexpected) is that the cosine and sine matrix still satisfy

I = sin(A)^2+cos(A)^2
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  • \$\begingroup\$ Isn't A^0 the same as eye(size(A))? \$\endgroup\$ May 5, 2016 at 17:15
  • \$\begingroup\$ Oh, you're right, thank you! \$\endgroup\$
    – flawr
    May 5, 2016 at 17:17
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    \$\begingroup\$ Why not use expm? \$\endgroup\$
    – Luis Mendo
    May 5, 2016 at 17:18
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    \$\begingroup\$ As per the identity: I should hope they satisfy that identity, considering that the scalar form was used to extend the functions to matrices! \$\endgroup\$
    – user45941
    May 5, 2016 at 17:27
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    \$\begingroup\$ Well then the whole thing becomes almost trivial. \$\endgroup\$
    – flawr
    May 5, 2016 at 18:27
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Julia 0.4, 28 bytes

A->imag({E=expm(im*A),im*E})

Input is a matrix of floats, output is an array of matrices. Try it online!

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Sage, 44 bytes

lambda A:map(exp(I*A).apply_map,(imag,real))

Try it online.

This anonymous function returns a list of 2 matrices corresponding to sin(A) and cos(A), respectively. exp(I*A) computes the matrix exponential for I*A (A with all elements multiplied by the imaginary unit), and matrix.apply_map(f) returns a matrix where f has been applied to all of its elements. By applying imag and real (the functions for getting the imaginary and real parts of a scalar value) to the matrices, we get the values of sin(A) and cos(A), thanks to Euler's famous identity (referenced in the challenge text).

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0
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Itr, 8+9=17 bytes

cosine

iäeà~e+½

sine:

iä~eàe-½i

Takes a matrix as input, also works for complex inputs

Explanation

implements the equations

\$ \sin(x)= i*\frac{e^{-ix}-e^{ix}}{2}\$

\$ \cos(x)= \frac{e^{ix}+e^{-ix}}{2}\$

i         ; i times the implicit input
 ä        ; duplicate
  ~       ; negate
   e      ; exponential
    à     ; swap
     e    ; exponential
      -   ; subtract
       ½  ; half
        i ; times i
          ; implicit output
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