# Proof of Osborne's Rule

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.

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Your substitution attempt is supposed to work. Can you give an example of where it fails? – J. M. Apr 30 '12 at 11:57
Also, $i^{4n+2}=-1$, so... – J. M. Apr 30 '12 at 11:58
This is somewhat related also to Wick rotation. – Willie Wong Apr 30 '12 at 12:01
Generally better to keep questions self-contained. Since Osbourne's rule is relatively short, no reason not to include it in the problem: $$\cosh(x-y) = \cosh x \cosh y + \sinh h \sinh y$$ – Thomas Andrews Apr 30 '12 at 12:24
@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. – J. M. Apr 30 '12 at 12:53

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.

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