How Does the Conjugate of e^(it) Extend to e^(-it)?

[Dissonance in E]

Hi I have a parametrisation of the unit circle
z = cos(t) + isin(t) = e^(it)
now i guess that the conjugate of that is the same but with a negative coefficient on the imaginary part, namely:
conj z = cos(t) - isin(t)
How does that extend to the e^(it) form? is the following true:
cos(t) - isin(t) = e^(-it)

Thanks

 

[Office_Shredder]

$$e^{-it}=\cos(-t)+i\sin(-t)$$. cosine is an even function, an sine is an odd function, so

$$e^{-it}=\cos(t)-i\sin(t)$$

 

[Dissonance in E]

nice, thanks.