5
$\begingroup$

Osborne's rule is described here.

Firstly, am I right that only signs of terms in the form $\sin^{4n+2} \theta$, $n \in \mathbb{Z^+}$ have their signs switched (i.e. terms like $\sin^4 \theta$ simply become $\sinh^4 \phi$)?

Secondly, I haven't been able to find a proof of Osborne's Rule anywhere - does anyone know of one?

My attempt was to have $\theta=i\phi$ in the trigonometric equation, so that $\cos \theta=\cosh \phi$ and $\sin \theta=i \sinh \phi$. However, the presence of $i$ in the latter seems problematic e.g. if the original trigonometric equation contains both even and odd powers of $\sin \theta$ then we will end up with both real and imaginary terms.

  • 0
    Your substitution attempt is *supposed* to work. Can you give an example of where it fails?2012-04-30
  • 0
    Also, $i^{4n+2}=-1$, so...2012-04-30
  • 1
    This is somewhat related also to [Wick rotation](http://en.wikipedia.org/wiki/Wick_rotation).2012-04-30
  • 1
    Generally better to keep questions self-contained. Since Osbourne's rule is relatively short, no reason not to include it in the problem: $$\cosh(x-y) = \cosh x \cosh y + \sinh h \sinh y$$2012-04-30
  • 0
    @Thomas: in fact OP made no mention of where a description of Osborne's rule could be found, and I had to add it in... :o Also, that isn't Osborne's rule itself, but an example where it was applied in the MathWorld link.2012-04-30
  • 0
    Ah, yes, read the article too quickly.2012-04-30

1 Answers 1

10

Observe the following facts; $\theta$ will be a real parameter throughout. From the definition $\cos z = \frac12 (e^{iz} + e^{-iz})$ and $\sin z = \frac1{2i} (e^{iz} - e^{-iz})$ for $z\in\mathbb{C}$. So

$$ \cos i\theta = \frac12 (e^{-\theta} + e^{\theta}) = \cosh \theta $$

and

$$ \sin i\theta = \frac1{2i} ( e^{-\theta} - e^{\theta}) = i\sinh\theta $$

Now, notice that $\cos$ and $\sin$ are holomorphic functions when looked at as functions on the complex plane. And a trigonometric identity can be expressed as

$$ R(\cos \theta,\sin\theta, \cos 2\theta, \sin 2\theta, \ldots, \cos k\theta, \sin k\theta) = Q(\cos\theta,\sin\theta) $$

where $R$ and $Q$ are rational functions (functions expressible as a polynomial divided by another polynomial) of suitable number of variables, this implies that a trigonometric identity is asserting that two meromorphic functions on the complex plane take the same values when restricted to the real axis. Now by a fundamental property of meromorphic and holomorphic functions, this implies that the two meromorphic functions are in fact everywhere equal: in particular they are equal on the imaginary axis. That is, we have

$$ R(\cos i\theta, \sin i\theta, \ldots) = Q(\cos i\theta,\sin i \theta) $$

which by the identities above can be written as

$$ R(\cosh \theta, i\sinh\theta, \ldots) = Q(\cosh \theta, i\sinh\theta) $$

thus showing Osborne's rule.