Sine and Cosine of i (imaginary unit)

The following is my working for the Sine and Cosine of the imaginary unit, $\displaystyle i$.

$\displaystyle e^{ix}=cos(x)+i(sin(x))$

Substituting $\displaystyle x=i$ gives

$\displaystyle e^{-1}=cos(i)+i(sin(i))$

Squaring both sides

$\displaystyle e^{-2}=(cos(i))^2-(sin(i))^2+2i(sin(i)cos(i))$

Substituing $\displaystyle 1-(sin(i))^2$ for $\displaystyle (cos(i))^2$ gives

$\displaystyle e^{-2}=(1-2((sin(i))^2)+2i(sin(i)cos(i))$

Rearranging the original equation in terms of $\displaystyle cos(i)$ gives

$\displaystyle cos(i)=e^{-1}-i(sin(i))$

Substituing this back into $\displaystyle e^{-2}=(1-2((sin(i))^2)+2i(sin(i)cos(i)) $gives

$\displaystyle e^{-2}=(1-2((sin(i))^2)+2i(sin(i))(e^{-1}-i(sin(i)))$

$\displaystyle e^{-2}=(1-2((sin(i))^2)+2i(sin(i))(e^{-1}) -2(i^2)((sin(i))^2)$

$\displaystyle e^{-2}=1+2i(sin(i))(e^{-1})$

Rearranging for sin(i) gives

$\displaystyle (e^{-2}-1)/2i(e^{-1})=sin(i)$

$\displaystyle sin(i)={{1 - e^2} \over 2ei}$

Then to find Cos(i): Substituting back into the original equation gives

$\displaystyle cos(i)=e^{-1}-(i(1-e^2))/(2ei)$

$\displaystyle cos(i)={{1 + e^2} \over 2e}$

The problem is that all the sources I can find for the sine and cosine of $\displaystyle i$ on the internet say

$\displaystyle sin(i) = {{e - 1/e} \over 2} \, i$ or $\displaystyle {{e^2 - 1} \over 2e} \, i$ and

$\displaystyle cos(i)={{e + 1/e} \over 2}$ or $\displaystyle {{e^2 + 1} \over 2e}$.

EDIT: I found my mistake in Cos(i), and now have the same as the internet for cos(i). BUT, I obtained my cos(i) from putting my sin(i) into the original equation and still got the right answer, which makes it even more puzzling.

As you can see, my answers are very close to the others, I just can't figure out where the difference in signs comes from... also, is there any difference if the $\displaystyle i$ is put outside the fraction like they have, or on the bottom like I work it to be?

If I've made any typos or not made anything clear enough, just point it out and I'll change the original post.

Thanks, Spud.