In one line
Putting an imaginary argument into a circular function gives a hyperbolic function and vice versa: \(\sin ix=i\sinh x\), \(\cos ix=\cosh x\), \(\sinh ix=i\sin x\), \(\cosh ix=\cos x\).
Explained simply
Circular and hyperbolic functions are two members of one family, connected by the number \(i\). Multiplying the input by \(i\) is like switching a translator on: everything you know about \(\sin\) and \(\cos\) translates into a fact about \(\sinh\) and \(\cosh\). This bridge is why we can separate \(\sin(x+iy)\) into real and imaginary parts later.
Technically, replace \(z\) by \(ix\) in \(\sin z=\dfrac{e^{iz}-e^{-iz}}{2i}\): \(\sin ix=\dfrac{e^{-x}-e^{x}}{2i}=\dfrac{-2\sinh x}{2i}=i\sinh x\) (because \(\frac{-1}{i}=i\)).
How it works
- \(\cos ix=\dfrac{e^{i\cdot ix}+e^{-i\cdot ix}}2=\dfrac{e^{-x}+e^{x}}2=\cosh x\).
- \(\sin ix=i\sinh x\) (shown above).
- Divide: \(\tan ix=\dfrac{i\sinh x}{\cosh x}=i\tanh x\).
- Reverse direction: \(\sinh ix=\dfrac{e^{ix}-e^{-ix}}2=i\sin x\), \(\cosh ix=\cos x\), \(\tanh ix=i\tan x\).
- Reciprocals follow: \(\sec ix=\mathrm{sech}\,x\), \(\mathrm{cosec}\,ix=-i\,\mathrm{cosech}\,x\), \(\cot ix=-i\coth x\).
Key formula
| Circular with \(ix\) | Hyperbolic with \(ix\) |
|---|---|
| \(\sin ix=i\sinh x\) | \(\sinh ix=i\sin x\) |
| \(\cos ix=\cosh x\) | \(\cosh ix=\cos x\) |
| \(\tan ix=i\tanh x\) | \(\tanh ix=i\tan x\) |
Here \(x\) may be real or complex; \(i^2=-1\).
Quick example
Use the bridge to get a hyperbolic identity from a trig one. Start with \(\cos^2\theta+\sin^2\theta=1\) and put \(\theta=ix\):
\(\cos^2ix+\sin^2ix=1\Rightarrow\cosh^2x+(i\sinh x)^2=1\Rightarrow\cosh^2x-\sinh^2x=1\).
Similarly from \(\sin2\theta=2\sin\theta\cos\theta\): \(i\sinh2x=2(i\sinh x)\cosh x\), so \(\sinh2x=2\sinh x\cosh x\).
Common mistakes
- Writing \(\cos ix=i\cosh x\); cosine of \(ix\) is real, no \(i\) appears.
- Forgetting that \(\dfrac1i=-i\), which gives the wrong sign in \(\sin ix\).
- Squaring \(i\sinh x\) as \(\sinh^2x\) instead of \(-\sinh^2x\).
How MU asks it
Asked as 3-mark Q.1 parts: "Prove that \(\sin ix=i\sinh x\)" or "Show that \(\cosh^2x-\sinh^2x=1\) using the relation with circular functions". It is also used silently in 6-mark separation problems. For full marks, start from the exponential definitions, substitute \(ix\), show the \(i^2=-1\) step, and conclude.