Chapter A, Problem 24P

Problems

Show that

$\cos bx=\frac{1}{2}\left(e^{ibx}+e^{-ibx}\right)$

and

$\sin bx=\frac{1}{2i}\left(e^{ibx}-e^{-ibx}\right)$.

(A comparison of these formulas with the corresponding formulas

$\cosh bx=\frac{1}{2}\left(e^{bx}+e^{-bx}\right)$

and

$\sinh bx=\frac{1}{2}\left(e^{bx}-e^{-bx}\right)$

indicates why the trigonometric and hyperbolic functions satisfy similar identities.)