# Why $\frac{1}{2i}(e^{i\omega t} - e^{-i\omega t}) = \frac{i}{2} (e^{-i \omega t} - e^{i\omega t})$

Let $i := \sqrt{-1}$, $f$ be the frequency ($\frac1p$), and $\omega := 2 \pi f$.

From page 3 here, why does $\frac{1}{2i}(e^{i\omega t} - e^{-i\omega t}) = \frac{i}{2} (e^{-i \omega t} - e^{i\omega t})$?

I understand that the LHS's $\frac{1}{2i}$ is multiplied by $\frac ii$ to get the RHS's $\frac i2$, but I don't understand how the contents of the parentheses changes?

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Because $$\frac1i=-i$$ –  DonAntonio May 24 '13 at 13:57
I get it now thanks guys. –  Jase May 24 '13 at 13:58

Just that $\frac{1}{i} = - i$, so by distributing the minus sign, $$\frac{1}{2i}(e^{i\omega t} - e^{-i\omega t}) = -\frac{i}{2}(e^{i\omega t} - e^{-i\omega t}) = \frac{i}{2}(e^{-i\omega t} - e^{i\omega t}).$$