Chapter 2, Selected problems and solutions to NFCM, 2nd Ed.

 

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Section 5:

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Click on this link to find Problem (5.6a) on page 78 of NFCM. Problem (5.6a) on page 78 of NFCM.

The Problem

On page 78 of NFCM (see References), we find problem (5.6a): Given that $I^2=-1$ and that $I$, $A$, and $B$ mutually commute, show that \begin{equation} \cos\, (A+IB) = \cos A\, \cosh B - I\sinh B\, \sin A \,.\label{eq:First} \end{equation}

A few identities we will need are \begin{align} e^X &= \cosh X + \sinh X \,,\label{eq:Hyperbolic1}\\ 2\sinh X &= e^X - e^{-X} \,,\label{eq:Hyperbolic2}\\ 2\cos X &= e^{IX} + e^{-IX} \,.\label{eq:Hyperbolic3} \end{align} Now, \begin{align} 2\cos\, (A+IB) &= e^{I(A+IB)} + e^{-I(A+IB)} \notag\\ &= e^{IA}e^{-B} + e^{-IA}e^{B} \notag\\ &= e^{IA}e^{-B} +e^{-IA}e^{-B} - e^{-IA}e^{-B} + e^{-IA}e^{B}\quad\text{(add in and subtract out)} \notag\\ &= (e^{IA} +e^{-IA})e^{-B} + e^{-IA}(e^{B} - e^{-B})\notag\\ &= 2\cos A\,( \cosh B - \sinh B)+ 2(\cos A - I\sin A)\sinh B\notag\\ &= 2(\cos A\, \cosh B - I\sinh B\, \sin A)\,.\label{eq:NearDone} \end{align} On cancelling the factors of 2, we get \begin{equation} \cos\, (A+IB) = \cos A\, \cosh B - I\sinh B\, \sin A \,.\label{eq:Done} \end{equation}

References

D. Hestenes, New Foundations in Classical Mechanics, 2nd Ed., Kluwer Academic Publishers, 1999.

NFCM Problems and Solutions