Answer
$cosx=\frac{e^{ix}+e^{-ix}}{2}$
and
$sinx=\frac{e^{ix}-e^{-ix}}{2i}$
Work Step by Step
Use Euler's formula
$e^{iy}=cosy+isiny$
For $cosx$:
$cosx=\frac{(cosx+isinx)+(cos(-x)+isin(-x))}{2}$
$cosx=\frac{2cosx}{2}$
$cosx=cosx$
For $sinx$:
$sinx=\frac{(cosx+isinx)-(cos(-x)+isin(-x))}{2i}$
$sinx=\frac{2isinx}{2i}$
$sinx=sinx$
Hence, $cosx=\frac{e^{ix}+e^{-ix}}{2}$
and
$sinx=\frac{e^{ix}-e^{-ix}}{2i}$
