Question

How did he get this 1/2i ? I dont remember studying this definition in calc 1 neither calc 2!

Answer 1

The $i$ in the denominator is missing in the second expression. It should be

$\sin at=\dfrac{e^{iat}-e^{-iat}}{2i}$

This follows from

$e^{iat}=\cos at+i\sin at$
$e^{-iat}=\cos at-i\sin at$
$\implies e^{iat}-e^{-iat}=2i\sin at$
$\implies \sin at=\dfrac{e^{iat}-e^{-iat}}{2i}$