Trigonometry: Negative cosine to polar form

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I want to convert a cosine to it's polar form in the format $Xe^{i\alpha t}$. The way I tried to do it resulted in a negative exponent in the $X$/phase of the imaginary part, where the answer provided had a positive phase. See the calculations below.

How i did it:

$-2\cos(2\pi t) = 2\cos(2\pi t + \pi) = e^{i\pi}e^{i2\pi t} + e^{-i\pi}e^{-i2\pi t}$

Answer:

$-2\cos(2\pi t) = -2\cos(2\pi t) = -(e^{i2\pi t} + e^{i2\pi t}) = e^{i\pi}e^{i2\pi t} + e^{i\pi}e^{i2\pi t}$

If someone could point out my mistake, it would be much appreciated.

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Your calculation seems correct, indeed

$$e^{i\pi}e^{i2\pi t} + e^{-i\pi}e^{-i2\pi t}=e^{i(2\pi t+\pi)} + e^{-i(2\pi t+\pi)}=2\cos(2\pi t + \pi) $$