Answer
(a) $cosθ$=$\frac{e^{iθ} + e^{-iθ}}{2}$=$coshiθ$
(b) $i.sinθ$=$\frac{e^{iθ} - e^{-iθ}}{2}$=$sinhiθ$
Work Step by Step
We know that
$e^{iθ}$=$cosθ+isinθ$
and
$e^{-iθ}$=$cosθ-isinθ$
Using these formulas
(a) $e^{iθ}$ + $e^{-iθ}$=$(cosθ +isinθ)$ + $(cosθ -isinθ)$=$2cosθ$
$cosθ$=$\frac{e^{iθ} + e^{-iθ}}{2}$=$coshiθ$
(b) $e^{iθ}$ $-$ $e^{-iθ}$=$(cosθ +isinθ)$ $-$ $(cosθ -isinθ)$=$2i .sinθ$
$i.sinθ$=$\frac{e^{iθ} - e^{-iθ}}{2}$=$sinhiθ$
