In one line
Every trigonometric identity has a hyperbolic twin, obtained by changing \(\cos\to\cosh\), \(\sin\to\sinh\) and flipping the sign of every term that contains a product of two sines (Osborn's rule).
Explained simply
You already spent years learning trig identities. Osborn's rule lets you reuse all of them. The only change is a sign flip wherever two sines multiply, because \((i\sinh x)(i\sinh y)=-\sinh x\sinh y\). It is like a recipe that works in both kitchens if you swap one spice.
The rule is not a separate theorem; it is a shortcut for substituting \(\sin\theta=-i\sinh(i\theta)\) style relations. For proofs in exams, you should derive identities from the exponential definitions.
How it works
- Take a trig identity, e.g. \(\cos2\theta=1-2\sin^2\theta\).
- Replace each function by its hyperbolic version.
- Look for products of two sines (including \(\sin^2\), and \(\tan^2=\frac{\sin^2}{\cos^2}\)). Change their sign.
- Result: \(\cosh2x=1+2\sinh^2x\).
- To prove it properly: \(1+2\sinh^2x=1+2\cdot\dfrac{e^{2x}-2+e^{-2x}}4=\dfrac{e^{2x}+e^{-2x}}2=\cosh2x\).
Key formula
- \(\cosh^2x-\sinh^2x=1\); \(\ 1-\tanh^2x=\mathrm{sech}^2x\); \(\ \coth^2x-1=\mathrm{cosech}^2x\)
- \(\sinh2x=2\sinh x\cosh x\); \(\ \cosh2x=\cosh^2x+\sinh^2x=2\cosh^2x-1=1+2\sinh^2x\)
- \(\tanh2x=\dfrac{2\tanh x}{1+\tanh^2x}\)
- \(\sinh(x\pm y)=\sinh x\cosh y\pm\cosh x\sinh y\); \(\ \cosh(x\pm y)=\cosh x\cosh y\pm\sinh x\sinh y\)
- \(\sinh3x=3\sinh x+4\sinh^3x\); \(\ \cosh3x=4\cosh^3x-3\cosh x\)
Quick example
Prove \(\tanh2x=\dfrac{2\tanh x}{1+\tanh^2x}\). Trig version: \(\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}\). The \(\tan^2\) term hides \(\sin^2\), so its sign flips, giving \(1+\tanh^2x\). Proof: \(\dfrac{2\tanh x}{1+\tanh^2x}=\dfrac{2\sinh x/\cosh x}{(\cosh^2x+\sinh^2x)/\cosh^2x}=\dfrac{2\sinh x\cosh x}{\cosh^2x+\sinh^2x}=\dfrac{\sinh2x}{\cosh2x}\).
Check \(\sinh3x\) at \(x=\log2\): \(\sinh x=\tfrac34\), so \(3(\tfrac34)+4(\tfrac{27}{64})=\tfrac94+\tfrac{27}{16}=\tfrac{63}{16}\); directly, \(\sinh(\log8)=\dfrac{8-\frac18}{2}=\dfrac{63}{16}\). Correct.
Common mistakes
- Flipping the sign of \(\cosh(x+y)=\cosh x\cosh y+\sinh x\sinh y\) twice; trig has a minus there, hyperbolic has a plus.
- Writing Osborn's rule as the proof; examiners want the exponential working.
- Forgetting that \(\tan^2\) and \(\cot^2\) also count as "product of two sines".
How MU asks it
3-mark Q.1 parts: "Prove \(\cosh^2x=\dfrac{1+\cosh2x}2\)" or "Prove \(\sinh3x=3\sinh x+4\sinh^3x\)". They also appear as steps in 6-mark problems such as "If \(\cosh x=\sec\theta\), prove \(\tanh^2\frac x2=\tan^2\frac\theta2\)". Start from definitions or from an identity already proved, and show every substitution.