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Thread: Sine and Cosine of i (imaginary unit)

  1. #1
    Mar 2008

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

    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.
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  2. #2
    Super Member
    Oct 2007
    London / Cambridge
    $\displaystyle e^{ix} = \cos x + i \sin x \ \ \ \ (1)$
    $\displaystyle e^{-ix} = \cos x - i \sin x \ \ \ (2)$

    $\displaystyle (1) + (2)$ gives

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

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

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

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

    - Bobak
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