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Math Help - Complex Number Trigonometry

  1. #1
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    Complex Number Trigonometry

    I understand the fact:

    \frac{e^{x}+e^{-x}}{2} = \mathrm{cosh}x

    But why does:

    \frac{e^{jx}+e^{-jx}}{2} = \cos x?
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  2. #2
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    Oops, just remembered:

    e^{jx} = \cos x + j \sin x
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Air View Post
    I understand the fact:

    \frac{e^{x}+e^{-x}}{2} = \mathrm{cosh}x

    But why does:

    \frac{e^{jx}+e^{-jx}}{2} = \cos x?
    The reason is due to Euler's Formula:

    e^{jx}=\cos(x)+j\sin(x)

    e^{-jx}=\cos(-x)+j\sin(-x)=\cos(x)-j\sin(x) [Due to the even and odd properties of sine and cosine]

    Thus, \frac{e^{jx}+e^{-jx}}{2}=\frac{\cos(x)+j\sin(x)+\cos(x)-j\sin(x)}{2}=\frac{2\cos(x)}{2}=\color{red}\boxed{  \cos(x)}

    Does this make sense?

    --Chris
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