How do you prove Osborn's rule?
It's worth noting the origin of Osborn's Rule among a list of brief "Mathematical Notes" in The Mathematical Gazette, Vol 2 No. 34 (Jul 1902), pg 189 (JSTOR link). See this screenshot.
Transcribing in its entirety ...
- [D.6.d.] Mnemonic for hyperbolic formulae.
Hyperbolic functions are now so constantly used, that a brief mnemonic for their somewhat confusing formulae may not be unwelcome.
In any Trigonometrical formula for $\theta$, $2\theta$, $3\theta$, or $\theta$ and $\phi$, after changing sin to sinh, cos to cosh, etc., change the sign of any term that contains (or implies) a product of sinhs, eg $\tanh\theta \tanh\phi$ implies a product of sinhs,
$$\begin{align} \therefore \tanh(\theta+\phi) &= \frac{\tanh\theta+\tanh\phi}{1+\tanh\theta\tanh\phi}\; ; \\[4pt] \sinh3\theta &= 3\sinh\theta + 4\sinh3\theta\; ; \\[4pt] \cosh\theta-\cosh\phi &= +2\sinh\frac{\theta+\phi}{2}\sinh\frac{\theta-\phi}{2}\; ; \end{align}$$ and so on. This rule would likely fail for terms of the fourth degree, but it covers everything that is likely to be required, and is very convenient for teaching purposes.
G. Osborn
Despite the overly-ambitious ---one might say, hyperbolic--- claim of applicability to "any Trigonometrical formula", it seems clear that what has come to be known as "Osborn's Rule" was intended, and has ever only served, as something more like "Osborn's Rule of Thumb": it's a quick-and-easy way (especially for students) to remember when to change signs in familiar fundamental Trigonometrical formulas.
The Rule is a teaching tool, not an over-arching general principle.
You could proceed in the following way:$$ \sin x + \cos 2x \equiv \sin x + \cos^2 x - \sin^2x $$ corresponds to $$ i\sinh x + \cosh 2x \equiv i\sinh x + \cosh^2 x + \sinh^2 x \, ,$$
consequently, by equalling the imaginary parts and the real ones, we get that,
$$\begin{cases} \sinh x\equiv\sinh x\\ \cosh 2x\equiv\cosh^2 x+\sinh^2 x \end{cases}$$
and, by adding both sides,
$$\sinh x + \cosh 2x \equiv \sinh x + \cosh^2 x + \sinh^2 x\,.$$