Answer
1$-$ $(θ)^2$$/2!$ + $(θ)^4$$/4!$ $-$ $(θ)^6$$/6!$ +... =$cosθ$
θ$-$ $(θ)^3$$/3!$ + $(θ)^5$$/5!$ $-$ $(θ)^7$$/7!$ +... =$sinθ$
Work Step by Step
Using formulas of taylor series
$e^x$=1+ $x$+ $x^2$$/2!$ + $x^3$$/3!$ + $x^4$$/4!$ +...
$e^{iθ}$=1+ $iθ$+ $(iθ)^2$$/2!$ + $(iθ)^3$$/3!$ + $(iθ)^4$$/4!$ +...
and
$e^{-iθ}$=1+ $-iθ$+ $(-iθ)^2$$/2!$ + $(-iθ)^3$$/3!$ + $(-iθ)^4$$/4!$ +...
=1 $-$ $iθ$+ $(iθ)^2$$/2!$ $-$ $(iθ)^3$$/3!$ + $(iθ)^4$$/4!$ $-$...
now combing the formulas
$\frac{e^{iθ}+e^{-iθ}}{2}$ =$\frac{(1+ iθ+ (iθ)^2/2! + (iθ)^3/3!+ (iθ)^4/4! +...) + ( 1- iθ+ (iθ)^2/2! -(iθ)^3/3!+ (iθ)^4/4! -... )}{2}$
=1$-$ $(θ)^2$$/2!$ + $(θ)^4$$/4!$ $-$ $(θ)^6$$/6!$ +... =$cosθ$
now for $sinθ$
$\frac{e^{iθ}-e^{-iθ}}{2!}$ =$\frac{(1+ iθ+ (iθ)^2/2! + (iθ)^3/3!+ (iθ)^4/4! +...) - ( 1- iθ+ (iθ)^2/2! -(iθ)^3/3!+ (iθ)^4/4! -... )}{2!}$
=θ$-$ $(θ)^3$$/3!$ + $(θ)^5$$/5!$ $-$ $(θ)^7$$/7!$ +... =$sinθ$
