acos¶
- acos(x: array, /) array ¶
Calculates an implementation-dependent approximation of the principal value of the inverse cosine for each element
x_i
of the input arrayx
.Each element-wise result is expressed in radians.
Note
The principal value of the arc cosine of a complex number \(z\) is
\[\operatorname{acos}(z) = \frac{1}{2}\pi + j\ \ln(zj + \sqrt{1-z^2})\]For any \(z\),
\[\operatorname{acos}(z) = \pi - \operatorname{acos}(-z)\]Note
For complex floating-point operands,
acos(conj(x))
must equalconj(acos(x))
.Note
The inverse cosine (or arc cosine) is a multi-valued function and requires a branch cut on the complex plane. By convention, a branch cut is placed at the line segments \((-\infty, -1)\) and \((1, \infty)\) of the real axis.
Accordingly, for complex arguments, the function returns the inverse cosine in the range of a strip unbounded along the imaginary axis and in the interval \([0, \pi]\) along the real axis.
Note: branch cuts follow C99 and have provisional status (see Branch Cuts).
- Parameters:
x (array) – input array. Should have a floating-point data type.
- Returns:
out (array) – an array containing the inverse cosine of each element in
x
. The returned array must have a floating-point data type determined by Type Promotion Rules.
Notes
Special cases
For real-valued floating-point operands,
If
x_i
isNaN
, the result isNaN
.If
x_i
is greater than1
, the result isNaN
.If
x_i
is less than-1
, the result isNaN
.If
x_i
is1
, the result is+0
.
For complex floating-point operands, let
a = real(x_i)
,b = imag(x_i)
, andIf
a
is either+0
or-0
andb
is+0
, the result isπ/2 - 0j
.If
a
is either+0
or-0
andb
isNaN
, the result isπ/2 + NaN j
.If
a
is a finite number andb
is+infinity
, the result isπ/2 - infinity j
.If
a
is a nonzero finite number andb
isNaN
, the result isNaN + NaN j
.If
a
is-infinity
andb
is a positive (i.e., greater than0
) finite number, the result isπ - infinity j
.If
a
is+infinity
andb
is a positive (i.e., greater than0
) finite number, the result is+0 - infinity j
.If
a
is-infinity
andb
is+infinity
, the result is3π/4 - infinity j
.If
a
is+infinity
andb
is+infinity
, the result isπ/4 - infinity j
.If
a
is either+infinity
or-infinity
andb
isNaN
, the result isNaN ± infinity j
(sign of the imaginary component is unspecified).If
a
isNaN
andb
is a finite number, the result isNaN + NaN j
.If
a
isNaN
andb
is+infinity
, the result isNaN - infinity j
.If
a
isNaN
andb
isNaN
, the result isNaN + NaN j
.
Changed in version 2022.12: Added complex data type support.