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Math Help - sin^4 θ as linear combo of e^ikθ

  1. #1
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    sin^4 θ as linear combo of e^ikθ

    Use the Binomial Theorem to write sin^4\theta as a linear combo of functions of the form e^{ik\theta}, where k E Z.

    Attempt:

    sin^4 \theta
    = (\frac{e^{i\theta} - e^{-i \theta}}{2i})^4
    = \frac{1}{16}(e^{i\theta} - e^{-i\theta})^4

    (e^{i\theta} - e^{-i\theta})^4 <br />
= \sum_{k=0}^{4}\binom{4}{k} (e^{i\theta})^{4 - k}(-e^{-i\theta})^k
    =\binom{4}{0}(e^{i\theta})^4 + \binom{4}{1}(e^{i\theta})^3(-e^{-i\theta}) + \binom{4}{2}(e^{i\theta})^2(-e^{-i\theta})^2 + \binom{4}{3}(e^{i\theta})(-e^{-i\theta})^3 + (-e^{-i\theta})^4
    =e^{i4\theta} - 4e^{i3\theta - i\theta} + 6e^{i2\theta + i2\theta} - 4e^{i\theta - i3\theta} + e^{i4\theta}
    =e^{i4\theta} - 4e^{i2\theta} + 6e^{i4\theta} - 4e^{-i2\theta} + e^{i4\theta}
    =8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta}

    sin^4 \theta
    = \frac{1}{16}(8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta})
    =\frac{1}{2}e^{4\theta} - \frac{1}{4}e^{i2\theta} - \frac{1}{4}e^{-{i2\theta}}

    but I think I made a mistake somewhere
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  2. #2
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    It should be just 6 not 6e^{4\theta i}.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SyNtHeSiS View Post
    Use the Binomial Theorem to write sin^4\theta as a linear combo of functions of the form e^{ik\theta}, where k E Z.

    Attempt:

    sin^4 \theta
    = (\frac{e^{i\theta} - e^{-i \theta}}{2i})^4
    = \frac{1}{16}(e^{i\theta} - e^{-i\theta})^4

    (e^{i\theta} - e^{-i\theta})^4 <br />
= \sum_{k=0}^{4}\binom{4}{k} (e^{i\theta})^{4 - k}(-e^{-i\theta})^k
    =\binom{4}{0}(e^{i\theta})^4 + \binom{4}{1}(e^{i\theta})^3(-e^{-i\theta}) + \binom{4}{2}(e^{i\theta})^2(-e^{-i\theta})^2 + \binom{4}{3}(e^{i\theta})(-e^{-i\theta})^3 + (-e^{-i\theta})^4
    =e^{i4\theta} - 4e^{i3\theta - i\theta} + 6e^{i2\theta + i2\theta} - 4e^{i\theta - i3\theta} + e^{i4\theta}
    =e^{i4\theta} - 4e^{i2\theta} + 6e^{i4\theta} - 4e^{-i2\theta} + e^{i4\theta}
    =8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta}

    sin^4 \theta
    = \frac{1}{16}(8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta})
    =\frac{1}{2}e^{4\theta} - \frac{1}{4}e^{i2\theta} - \frac{1}{4}e^{-{i2\theta}}

    but I think I made a mistake somewhere
    you did. note that (-e^{-x})^2 = e^{-2x} and similarly, (-e^{-x})^4 = e^{-4x}

    EDIT: which explains Plato's comment
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  4. #4
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    Thanks. So would the correct answer be:

    sin^4\theta = \frac{1}{8}cos4\theta = \frac{1}{2}cos2\theta + \frac{3}{8}?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SyNtHeSiS View Post
    Thanks. So would the correct answer be:

    sin^4\theta = \frac{1}{8}cos4\theta = \frac{1}{2}cos2\theta + \frac{3}{8}?
    ??

    Did you read the instructions?
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    ??Did you read the instructions?
    Yeah I did. Here is my working out:

    sin^4 \theta
    = (\frac{e^{i\theta} - e^{-i \theta}}{2i})^4
    = \frac{1}{16}(e^{i\theta} - e^{-i\theta})^4

    (e^{i\theta} - e^{-i\theta})^4 <br />
= \sum_{k=0}^{4}\binom{4}{k} (e^{i\theta})^{4 - k}(-e^{-i\theta})^k
    =\binom{4}{0}(e^{i\theta})^4 + \binom{4}{1}(e^{i\theta})^3(-e^{-i\theta}) + \binom{4}{2}(e^{i\theta})^2(-e^{-i\theta})^2 + \binom{4}{3}(e^{i\theta})(-e^{-i\theta})^3 + (-e^{-i\theta})^4
    =e^{i4\theta} - 4e^{i3\theta - i\theta} + 6e^{i2\theta - i2\theta} - 4e^{i\theta - i3\theta} + e^{-i4\theta}
    =e^{i4\theta} - 4e^{i2\theta} + 6 - 4e^{-i2\theta} + e^{-i4\theta}

    sin^4 \theta
    =\frac{1}{16}(e^{i4\theta} - 4e^{i2\theta} + 6 - 4e^{-i2\theta} + e^{-i4\theta})
    =\frac{1}{8}(\frac{e^{i4\theta + e^{-i4\theta}}}{2}) - \frac{1}{2}(\frac{e^{i2\theta} + e^{-i2\theta}}{2}) + \frac{3}{8}
    =\frac{1}{8}cos4\theta = \frac{1}{2}cos2\theta + \frac{3}{8}
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