# Thread: Sine and Cosine of i (imaginary unit)

1. ## Sine and Cosine of i (imaginary unit)

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

$e^{ix}=cos(x)+i(sin(x))$
Substituting $x=i$ gives
$e^{-1}=cos(i)+i(sin(i))$
Squaring both sides
$e^{-2}=(cos(i))^2-(sin(i))^2+2i(sin(i)cos(i))$
$e^{-2}=(1-2((sin(i))^2)+2i(sin(i)cos(i))$
Rearranging the original equation in terms of $cos(i)$ gives
$cos(i)=e^{-1}-i(sin(i))$
Substituing this back into $e^{-2}=(1-2((sin(i))^2)+2i(sin(i)cos(i))$gives
$e^{-2}=(1-2((sin(i))^2)+2i(sin(i))(e^{-1}-i(sin(i)))$
$e^{-2}=(1-2((sin(i))^2)+2i(sin(i))(e^{-1}) -2(i^2)((sin(i))^2)$
$e^{-2}=1+2i(sin(i))(e^{-1})$
Rearranging for sin(i) gives
$(e^{-2}-1)/2i(e^{-1})=sin(i)$
$sin(i)=(1-e^2)/2ei$
Then to find Cos(i): Substituting back into the original equation gives
$cos(i)=e^{-1}-(i(1-e^2))/(2ei)
cos(i)=(1-e^2)/2e$

The problem is that all the sources I can find for the sine and cosine of $i$ on the internet say
$sin(i) = ((e-e^{-1})/2)i$ or $((e^2-1)/2e)i$ and
$cos(i)=(e+e^{-1})/2$ or ( $e^2+1)/2e$.

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 $i$ 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.

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

$(1) + (2)$ gives

$e^{ix} + e^{-ix} = 2 \cos x$

$\cos x = \frac{1}{2} \left ( e^{ix} + e^{-ix} \right )$

$\cos i = \frac{1}{2} \left ( e^{-1} + e^{1} \right )$

I will leave you to do $\sin i$.

- Bobak