Euler's Formula states that for any real $x$,
\[e^{i x}=\cos{x}+i \sin{x},\]
where $i$ is the imaginary unit, $i=\sqrt{-1}$.
Euler's formula can be used to derive the following identities for the trigonometric functions $\sin{x}$ and $\cos{x}$ in terms of exponential functions:
\begin{align} \cos{x} &= \frac{ e^{i x} + e^{-i x} }{2} \\ \sin{x} &= \frac{ e^{i x} - e^{-i x} }{2i} \end{align}
First note that:
\[e^{-i x} = \cos(-x)+i \sin(-x).\]
Now recall that $\cos{x}$ is an even function, so $\cos(-x)=\cos{x}$. Furthermore, $\sin{x}$ is an odd function, so $\sin(-x)=-\sin{x}$.
Hence,
\begin{align} e^{-i x} &= \cos(-x)+i \sin(-x), \\ &= \cos{x} - i\sin{x}. \end{align}
An expression for $\cos{x}$ is found by taking the sum of $ e^{i x}$ and $ e^{-i x}$:
\[e^{i x} + e^{-i x} = (\cos{x}+i \sin{x}) + (\cos{x}-i \sin{x}).\]
Notice that the $i \sin{x}$ terms cancel, giving
\[e^{i x} + e^{-i x} = 2\cos{x}.\]
Dividing both sides of this expression by $2$ gives the identity for $\cos{x}$
\[\cos{x}=\frac{ e^{i x} + e^{-i x} }{2}.\]
An expression for $\sin{x}$ is now found by taking the difference of $ e^{i x}$ and $ e^{-i x}$
\[e^{i x} - e^{-i x} = (\cos{x}+i\sin{x}) - (\cos{x}-i\sin{x}).\]
Notice that now the $\cos{x}$ terms cancel, giving
\begin{align} e^{i x} - e^{-i x} &= i \sin{x} - (-i \sin{x}) \\ &= 2i\sin{x}. \end{align}
Dividing both sides of this expression by $2i$ gives the identity for $\sin{x}$
\[\sin{x} = \frac{ e^{i x} - e^{-i x} }{2i}.\]
Euler's Identity is a special case of Euler's Formula, obtained from setting $x=\pi$:
\begin{align} e^{i\pi} &=\cos{\pi}+i\sin{\pi} \\ &= -1, \end{align}
since $\cos{\pi}=-1$ and $\sin{\pi}=0$.
Euler's Identity is conventionally written in the form
\[e^{i\pi}+1=0.\]
It is not necessary to memorise Euler's Identity. It is included here, however, as it is regarded by many to be an object of mathematical beauty.