
    sgg                        d Z ddlmZ ddlmZ ddlmZ ddlmZ ddl	m
Z
 ddlmZ ddlmZ dd	lmZ dd
lmZ ddZddZddZedd       Zedd       Zedd       Zedd       ZddZy)a  A module for special angle formulas for trigonometric functions

TODO
====

This module should be developed in the future to contain direct square root
representation of

.. math
    F(\frac{n}{m} \pi)

for every

- $m \in \{ 3, 5, 17, 257, 65537 \}$
- $n \in \mathbb{N}$, $0 \le n < m$
- $F \in \{\sin, \cos, \tan, \csc, \sec, \cot\}$

Without multi-step rewrites
(e.g. $\tan \to \cos/\sin \to \cos/\sqrt \to \ sqrt$)
or using chebyshev identities
(e.g. $\cos \to \cos + \cos^2 + \cdots \to \sqrt{} + \sqrt{}^2 + \cdots $),
which are trivial to implement in sympy,
and had used to give overly complicated expressions.

The reference can be found below, if anyone may need help implementing them.

References
==========

.. [*] Gottlieb, Christian. (1999). The Simple and straightforward construction
   of the regular 257-gon. The Mathematical Intelligencer. 21. 31-37.
   10.1007/BF03024829.
.. [*] https://resources.wolframcloud.com/FunctionRepository/resources/Cos2PiOverFermatPrime
    )annotations)Callable)reduce)Expr)S)igcdex)Integersqrt)cacheitc                     | syt        |       dk(  rd| d   fS t        |       dk(  rt        | d   | d         \  }}|f|fS t        | dd  \  }}t        | d   |      \  }}|gfd|D        |fS )aN  Compute extended gcd for multiple integers.

    Explanation
    ===========

    Given the integers $x_1, \cdots, x_n$ and
    an extended gcd for multiple arguments are defined as a solution
    $(y_1, \cdots, y_n), g$ for the diophantine equation
    $x_1 y_1 + \cdots + x_n y_n = g$ such that
    $g = \gcd(x_1, \cdots, x_n)$.

    Examples
    ========

    >>> from sympy.functions.elementary._trigonometric_special import migcdex
    >>> migcdex()
    ((), 0)
    >>> migcdex(4)
    ((1,), 4)
    >>> migcdex(4, 6)
    ((-1, 1), 2)
    >>> migcdex(6, 10, 15)
    ((1, 1, -1), 1)
    ) r      )r   r      Nc              3  (   K   | ]	  }|z    y wNr   ).0ivs     d/var/www/html/venv/lib/python3.12/site-packages/sympy/functions/elementary/_trigonometric_special.py	<genexpr>zmigcdex.<locals>.<genexpr>S   s     "1Q"s   )lenr   migcdex)xuhygr   s        @r   r   r   .   s    2 
1v{QqTz
1v{1qt$1a1vqyAabE?DAqQqT1oGAq!#""#Q&&    c                 p    | sydd}t        ||       }| D cg c]  }||z  	 }}t        | \  }}|S c c}w )a  Compute the partial fraction decomposition.

    Explanation
    ===========

    Given a rational number $\frac{1}{q_1 \cdots q_n}$ where all
    $q_1, \cdots, q_n$ are pairwise coprime,

    A partial fraction decomposition is defined as

    .. math::
        \frac{1}{q_1 \cdots q_n} = \frac{p_1}{q_1} + \cdots + \frac{p_n}{q_n}

    And it can be derived from solving the following diophantine equation for
    the $p_1, \cdots, p_n$

    .. math::
        1 = p_1 \prod_{i \ne 1}q_i + \cdots + p_n \prod_{i \ne n}q_i

    Where $q_1, \cdots, q_n$ being pairwise coprime implies
    $\gcd(\prod_{i \ne 1}q_i, \cdots, \prod_{i \ne n}q_i) = 1$,
    which guarantees the existence of the solution.

    It is sufficient to compute partial fraction decomposition only
    for numerator $1$ because partial fraction decomposition for any
    $\frac{n}{q_1 \cdots q_n}$ can be easily computed by multiplying
    the result by $n$ afterwards.

    Parameters
    ==========

    denoms : int
        The pairwise coprime integer denominators $q_i$ which defines the
        rational number $\frac{1}{q_1 \cdots q_n}$

    Returns
    =======

    tuple[int, ...]
        The list of numerators which semantically corresponds to $p_i$ of the
        partial fraction decomposition
        $\frac{1}{q_1 \cdots q_n} = \frac{p_1}{q_1} + \cdots + \frac{p_n}{q_n}$

    Examples
    ========

    >>> from sympy import Rational, Mul
    >>> from sympy.functions.elementary._trigonometric_special import ipartfrac

    >>> denoms = 2, 3, 5
    >>> numers = ipartfrac(2, 3, 5)
    >>> numers
    (1, 7, -14)

    >>> Rational(1, Mul(*denoms))
    1/30
    >>> out = 0
    >>> for n, d in zip(numers, denoms):
    ...    out += Rational(n, d)
    >>> out
    1/30
    r   c                    | |z  S r   r   )r   r   s     r   mulzipartfrac.<locals>.mul   s    1ur   )r   intr   r#   returnr#   )r   r   )denomsr"   denomr   ar   _s          r   	ipartfracr)   V   sL    ~  3E#$!$A$A;DAqH 	%s   3c                x    g }dD ]3  }t        | |      \  }}|dk(  s|} |j                  |       | dk(  s1|c S  y)z}If n can be factored in terms of Fermat primes with
    multiplicity of each being 1, return those primes, else
    None
    )           i  r   r   N)divmodappend)nprimespquotient	remainders        r   fermat_coordsr6      sQ    
 F# $Ql)>AMM!Av r   c                 "    t         j                  S )z-Computes $\cos \frac{\pi}{3}$ in square roots)r   Halfr   r   r   cos_3r9      s     66Mr   c                 $    t        d      dz   dz  S )z-Computes $\cos \frac{\pi}{5}$ in square rootsr,   r      r
   r   r   r   cos_5r<      s     GaK1r   c                 V   t        dt        d      z   dz  t        d      t        dt        d      z
        t        t        d      dt        dt        d      z         z  dt        d      z
  t        dt        d      z
        z  z
  z  dt        d      z  z   dz         z   z  dz  z         S )	z.Computes $\cos \frac{\pi}{17}$ in square roots   r-       r   ir      "   r
   r   r   r   cos_17rB      s     	d2h"tAw$rDH}*=T!WT"tBx-00ARL
rDH}
4  !"T"X.023 	4+4  579 : 	:; ;r   c                    dd} dd} | t         j                  t        d            \  }} | |t        d            \  }} | |t        d            \  }} | |dd|z   d|z  z   z        \  }}	 | |dd|z   d|z  z   z        \  }
} | |dd|z   d|z  z   z        \  }} | |dd|z   d|z  z   z        \  }} | |d||z   |z   d|
z  z   z        \  }} | |d||z   |z   d|z  z   z        \  }} | |d||z   |	z   d|z  z   z        \  }} | |d||z   |
z   d|z  z   z        \  }} | |	d||	z   |z   d|z  z   z        \  }} | |
d||
z   |z   d|z  z   z        \  }} | |d||z   |z   d|z  z   z        \  }} | |d||z   |z   d|	z  z   z        \  }} ||d||z   |z   |z   z        }  ||d||z   |z   |z   z        }! ||d||z   |z   |z   z        }" ||d||z   |z   |z   z        }# ||d||z   |z   |z   z        }$ ||d||z   |z   |z   z        }% ||  d|!|"z   z         }& ||# d|$|%z   z         }'d	 ||& d|'z        z  }(t        t        d      t        |(dz         z  d
z  t         j                  z         S )a  Computes $\cos \frac{\pi}{257}$ in square roots

    References
    ==========

    .. [*] https://math.stackexchange.com/questions/516142/how-does-cos2-pi-257-look-like-in-real-radicals
    .. [*] https://r-knott.surrey.ac.uk/Fibonacci/simpleTrig.html
    c                ^    | t        | dz  |z         z   dz  | t        | dz  |z         z
  dz  fS Nr   r
   r'   bs     r   f1zcos_257.<locals>.f1   s9    DAN"a'!d1a4!8n*<)AAAr   c                0    | t        | dz  |z         z
  dz  S rE   r
   rF   s     r   f2zcos_257.<locals>.f2   s    DAN"A%%r      @   r;   r,   r      )r'   r   rG   r   r$   ztuple[Expr, Expr])r'   r   rG   r   r$   r   )r   NegativeOner	   r   r8   ))rH   rJ   t1t2z1z3z2z4y1y5y6y2y3y7y8y4x1x9x2x10x3x11x4x12x5x13x6x14x15x7x8x16v1v2v3v4v5v6u1u2w1s)                                            r   cos_257rx      s)   B& ws|,FBGBK FBGBK FBAq2v"}%&FBAq2v"}%&FBAq2v"}%&FBAq2v"}%&FBBR"qt+,-FBRb2",-.GBRb2",-.GBRb2",-.GBRb2",-.GBRb2",-.GBRb2",-.GCRb2",-.GB	BBGbL2%&	'B	BBGbL2%&	'B	BBGcMC'(	)B	BBHsNS()	*B	CS3Y_s*+	,B	CS2X]R'(	)B
bS"b2g,
	B
bS"b2g,
	B	BsBrEN	BQR!V$Q&/00r   c                 0    t         t        t        t        dS )ag  Lazily evaluated table for $\cos \frac{\pi}{n}$ in square roots for
    $n \in \{3, 5, 17, 257, 65537\}$.

    Notes
    =====

    65537 is the only other known Fermat prime and it is nearly impossible to
    build in the current SymPy due to performance issues.

    References
    ==========

    https://r-knott.surrey.ac.uk/Fibonacci/simpleTrig.html
    )r+   r,   r-   r.   )r9   r<   rB   rx   r   r   r   	cos_tablerz      s      	 r   N)r   r#   r$   ztuple[tuple[int, ...], int])r%   r#   r$   ztuple[int, ...])r1   r#   r$   zlist[int] | None)r$   r   )r$   zdict[int, Callable[[], Expr]])__doc__
__future__r   typingr   	functoolsr   sympy.core.exprr   sympy.core.singletonr   sympy.core.intfuncr   sympy.core.numbersr	   (sympy.functions.elementary.miscellaneousr   sympy.core.cacher   r   r)   r6   r9   r<   rB   rx   rz   r   r   r   <module>r      s   !D #     " % & 9 $%'PHV  	 	
 	 	
 	; 	; 	'1 	'1Tr   