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Math Help - is this true?

  1. #1
    Member billym's Avatar
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    Question is this true?

    cos(n\Pi)=(-1)^n
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    Yes. Simply look at the unit circle and you should see this immediately.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Of course assuming that n\in\mathbb{N}. Otherwise, not so much. If you don't make that restriction you need to say that \cos(n\pi)=\mathfrak{R}\left((-1)^n\right) which is really just a highfalutin way to say \cos(n\pi). This is of course because (-1)^n=e^{\pi\cdot i\cdot n}=\cos(n\pi)+ i\sin(n\pi).
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  4. #4
    Member billym's Avatar
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    I take it sin(n\pi)=0 then?
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    Quote Originally Posted by billym View Post
    I take it sin(n\pi)=0 then?
    You don't know that?
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  6. #6
    Member billym's Avatar
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    But when you touch me like this
    And you hold me like that
    I just have to admit
    That it's all coming back to me
    When I touch you like this
    And I hold you like that
    It's so hard to believe but
    It's all coming back to me
    (It's all coming back, it's all coming back to me now)
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by billym View Post
    I take it sin(n\pi)=0 then?
    Whether or not that is true, what I said is still valid. To see this merely note that \cos(x),\sin(x) are both mappings from \mathbb{R}\mapsto\mathbb{R}. So \cos(n\pi),\sin(n\pi) is real for any n\in\mathbb{R}. Also, it should be apparent that if both a,b\in\mathbb{R} that \mathfrak{R}\left[a+bi\right]=a. Consequently, it follows that \cos(n\pi)=\mathfrak{R}\left[\left(-1\right)^n\right]=\mathfrak{R}\left[e^{n\pi i}\right]=\mathfrak{R}\left[\cos(n\pi)+i\sin(n\pi)\right]=\cos(n\pi)
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