\[e^{ix}=cos x+isinx\]
to obtain the basic trigonometric functions \[sinx, \; cosx, \: tanx\]
in exponential form.\[e^{ix}=cos x+isinx \rightarrow e^{-ix}=cos(-x)+isin(-x)=cos x-isinx\]
Then
\[\begin{equation} \begin{aligned} e^{ix}+e^{-ix} &=(cos x+isinx)+(cos x-isinx) \\ &= 2cos x \rightarrow cosx \\ &= \frac{e^{ix}+e^{-ix}}{2} \end{aligned} \end{equation}\]
\[\begin{equation} \begin{aligned} e^{ix}-e^{-ix} &= (cos x+isinx)-(cos x-isinx) \\ &= 2isin x \rightarrow sinx \\ &= \frac{e^{ix}-e^{-ix}}{2i} \end{aligned} \end{equation}\]
\[\begin{equation} \begin{aligned} tanx &= \frac{sinx}{cosx} \\ &= \frac{(e^{ix}-e^{-ix})/(2i)}{(e^{ix}+e^{-ix})/(2)}=\frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})} \\ &= \frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})} \times \frac{e^{ix}}{e^{ix}} \\ &= \frac{e^{2ix}-1}{i(e^{2ix}+1)} \end{aligned} \end{equation}\]