Question

Obtain an equation for cosh(3x) in terms of cosh(x) and sinh(x).

Answer 1

`\cosh ax+\sinh ax=e^(ax)=(e^x)^a=(\cosh x+\sinh x)^a`


`\cosh x=\cos ix`

`\sinh x=-i\sin ix`

By DeMoivre's theorem,


`\cos3ix-\sin3ix=(\cos ix-i\sin ix)^3=\cos^3ix-3i\cos^2ix\sin ix-3\cos ix\sin^2ix+i\sin^3ix`

You then have


`\cos3ix=\mathrm{Re}\left[(\cos ix-i\sin ix)^3\right]`

`\cos3ix=\cos^3ix-3\cos ix\sin^2ix`

`\cos3ix=\cos^3ix+3\cos ix(-i\sin ix)^2`

Converting back to hyperbolic functions, you get


`\cosh3x=\cosh^3x+3\cosh x\sinh^2x`