Talk:Cosine Exponential Formulation

The current formulation is a bit strange and uncommon. I immediately thought a typo was made, but it had been applied consistently. I suggest that:


 * $\displaystyle \frac 1 2 i \left({e^{-i x} - e^{i x}}\right) \quad\rightsquigarrow\quad \frac 1 {2i} \left({e^{i x} - e^{-i x}}\right)$

effectively using that $\dfrac 1 i = -i$. --Lord_Farin (talk) 11:30, 3 December 2012 (UTC)


 * I don’t mind that, but you’ll have to fix Sine and Cosine of Sum/Proof using Exponential Formulation accordingly then. — Timwi (talk) 11:39, 3 December 2012 (UTC)