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Math Help - how do you represent cos in power form?

  1. #1
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    how do you represent cos in power form?

    for example, cos(m\pi) = (-1)^m because cos alternates between 1 and -1 every m\pi.

    so how can you represent cos(\frac{m\pi}{2}) in power form?
    this one is a bit confusing to me because when:

    m=1, cos(\frac{m\pi}{2})=cos(\frac{1\pi}{2})= 0
    m=2, cos(\frac{m\pi}{2})=cos(\frac{2\pi}{2})= -1
    m=3, cos(\frac{m\pi}{2})=cos(\frac{3\pi}{2})= 0
    m=4, cos(\frac{m\pi}{2})=cos(\frac{4\pi}{2})= 1
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mathfied View Post
    for example, cos(m\pi) = (-1)^m because cos alternates between 1 and -1 every m\pi.

    so how can you represent cos(\frac{m\pi}{2}) in power form?
    this one is a bit confusing to me because when:

    m=1, cos(\frac{m\pi}{2})=cos(\frac{1\pi}{2})= 0
    m=2, cos(\frac{m\pi}{2})=cos(\frac{2\pi}{2})= -1
    m=3, cos(\frac{m\pi}{2})=cos(\frac{3\pi}{2})= 0
    m=4, cos(\frac{m\pi}{2})=cos(\frac{4\pi}{2})= 1
    First off you have to state that m\in\mathbb{N}

    But how about...hmm...Let me think here...you need...ok

    Well after some thought I have come up with

    \cos\left(\frac{n\pi}{2}\right)=\frac{1}{2}(-1)^{\frac{n(n+1)}{2}}\bigg[(-1)^n+1\bigg]\quad{n\in\mathbb{N}}
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  3. #3
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    I will include my thought process as well..

    \cos m\pi = (-1)^m = 2\cos^2\left(\frac{m\pi}2\right) - 1

    \cos^2\left(\frac{m\pi}2\right) = \frac{(-1)^m + 1}2

    \cos\left(\frac{m\pi}2\right) = \pm\sqrt{\frac{(-1)^m + 1}2}

    Now we see that when an even m is a multiple of 4, its a + else it is a -

    We can capture this idea by (-1)^{\frac{m}2}

    When m is odd, anyway the expression is 0. The sign wont matter then. Thats why I choose (-1)^{\lfloor\frac{m}2\rfloor}

    Where I have used the floor function on \frac{m}2.

    Thus \cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\sqrt{\frac{(-1)^m + 1}2}

    Note that functionally, \sqrt{\frac{(-1)^m + 1}2} = \frac{(-1)^m + 1}2

    Thus a simplified form is:

    \cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\left(\frac{(-1)^m + 1}2\right)
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    I will include my thought process as well..

    \cos m\pi = (-1)^m = 2\cos^2\left(\frac{m\pi}2\right) - 1

    \cos^2\left(\frac{m\pi}2\right) = \frac{(-1)^m + 1}2

    \cos\left(\frac{m\pi}2\right) = \pm\sqrt{\frac{(-1)^m + 1}2}

    Now we see that when an even m is a multiple of 4, its a + else it is a -

    We can capture this idea by (-1)^{\frac{m}2}

    When m is odd, anyway the expression is 0. The sign wont matter then. Thats why I choose (-1)^{\lfloor\frac{m}2\rfloor}

    Where I have used the floor function on \frac{m}2.

    Thus \cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\sqrt{\frac{(-1)^m + 1}2}

    Note that functionally, \sqrt{\frac{(-1)^m + 1}2} = \frac{(-1)^m + 1}2

    Thus a simplified form is:

    \cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\left(\frac{(-1)^m + 1}2\right)
    Nice!

    I should probably include my thought process as well.

    I did not work off of the trig, rather I worked off of the sequence 0,-1,0,1\cdots


    I firstly saw that

    \frac{1}{2}\left((-1)^n+1\right) gave 0,1,0,1

    So I wanted then that for even value of n that ever other was positive.

    So I saw that this was described by (-1)^{\frac{n(n+1)}{2}}

    A little less eloquent than Iso
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  5. #5
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    Another nice formula is: \cos\left(\frac{n\,\pi}2\right)=\frac12\big(i^n+(-i)^n\big) where i=\sqrt{-1}, n \in \mathbb N.

    Also, the generating function is \frac1{1+x^2}.

    Edit: Also \cos\left(\frac{n\,\pi}2\right)=\Re\{i^n\}
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