I need help with the calculation of the following function $$\sin(\omega t)=\frac{(e^{iωt}-e^{-iωt})}{2j}\implies \sin(30^{\circ})= \sin\left(\frac \pi 6 \right) = \frac{e^{\frac \pi 6 i}-e^{-\frac \pi 6i}}{2i} $$...when I plug in the values for the function in my program code it replays $0.53\dots$ But it should do $0.5000000$
Is it because the approximation with two e's and the other numbers with commas divided by $2$ include an such a high error in it(the numbers are not ending) so that i would have to use a Taylor polynomial (it is the aim to use function that can be written as only + and - for schemas. Functions with a sine in it are too abstract and not allowed) for calculating the sine without precalculated values like in a cordic function.
Ps. Please do not write it as Taylor polynomial or cordic.
$\sin(30^\circ) = \sin \frac{\pi}{6}$. So in your formula, use $\omega t = \frac{\pi}{6}$.
\begin{align} e^{j\pi/6} &\approx 0.8660254040+j \;0.5000000000 \\ e^{-j\pi/6} &\approx 0.8660254040-j \;0.5000000000 \\ \frac{e^{j\pi/6}-e^{-j\pi/6}}{2j} &\approx 0.5000000000 \end{align}