Euler's formula in a Fourier series

Question

How is this achieved

$$\frac{e^{2i\pi t}}{4(2i\pi + 1)} + \frac{e^{-2i\pi t}}{4(1 - 2i\pi )} = \frac{1}{2}[\cos(2\pi t) + 2 \sin(2\pi t)] \, ?$$

Answer 1

The result is: \begin{align} \frac{e^{2i\pi t}}{4(1 + 2i\pi)} + \frac{e^{-2i\pi t}}{4(1 - 2i\pi )} &= \frac{(1 - 2 \pi i) \, e^{2 \pi i t} + (1 + 2 \pi i) \, e^{-2 \pi i t}}{4 (1-2\pi i)(1 + 2 \pi i)} \\ &= \frac{2 \cos(2 \pi t) + 4 \pi \sin(2 \pi t)}{4 (1 + 4 \pi^{2})} \\ &= \frac{\cos(2 \pi t) + 2 \pi \sin(2 \pi t)}{2 (1 + 4 \pi^{2})} \end{align}