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=(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=(e^(iat)-e^(-iat))/(2i)`