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

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

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

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

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

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

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

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

cos(i)=(1-e^2)/2e$

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

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

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

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.