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

Mathematics

1 answer:

Hitman42 [59]3 years ago

8 0

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

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

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