- Tue 08 November 2016
- Maths
- #permanent
The trigonometric functions \(\sin{(\theta)}\) and \(\cos{(\theta)}\) satisfy the identity
$$ \cos^{2}{(\theta)} + \sin^{2}{(\theta)} = 1. $$
Solving for either the square cosine or square sine gives
$$ \sin^{2}{(\theta)} = 1 - \cos^{2}{(\theta)}, \qquad \cos^{2}{(\theta)} = 1 - \sin^{2}{(\theta)}. $$
These can be written nicely in terms of the determinant of 2-by-2 symmetric matrices:
$$ \sin^{2}{(\theta)} = \det{ \begin{pmatrix} 1 & \cos{(\theta)} \\ \cos{(\theta)} & 1 \end{pmatrix} }, \qquad \cos^{2}{(\theta)} = \det{ \begin{pmatrix} 1 & \sin{(\theta)} \\ \sin{(\theta)} & 1 \end{pmatrix} }. $$
The hyperbolic functions \(\sinh{(\xi)}\) and \(\cosh{(\xi)}\) satisfy the identity
$$ \cosh^{2}{(\xi)} - \sinh^{2}{(\xi)} = 1. $$
Solving for either the square hyperbolic cosine or square hyperbolic sine gives
$$ \sinh^{2}{(\xi)} = \cosh^{2}{(\xi)} - 1, \qquad \cosh^{2}{(\xi)} = \sinh^{2}{(\xi)} + 1. $$
The first can be written as the determinant of a symmetric matrix:
$$ \sinh^{2}{(\xi)} = \det{ \begin{pmatrix} \cosh{(\xi)} & 1 \\ 1 & \cosh{(\xi)} \end{pmatrix} } = - \det{ \begin{pmatrix} 1 & \cosh{(\xi)} \\ \cosh{(\xi)} & 1 \end{pmatrix} }. $$
The second identity can be written in terms of the permanent of a symmetric matrix:
$$ \cosh^{2}{(\xi)} = \operatorname{per}{ \begin{pmatrix} \sinh{(\xi)} & 1 \\ 1 & \sinh{(\xi)} \end{pmatrix} } = +\operatorname{per}{ \begin{pmatrix} 1 & \sinh{(\xi)} \\ \sinh{(\xi)} & 1 \end{pmatrix} }. $$