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Math Help - [SOLVED] Complex Trig proof

  1. #1
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    [SOLVED] Complex Trig proof

    *This is exactly how the question reads from the book

    Using the fact that a complex equation is really two real equations, find the double angle formulas (for Sin2 \theta\, Cos2 \theta\ ) by using this equation: (e^i \theta\)^n = (Cos \theta\+ iSin \theta\)^n = Cos n \theta\ + iSin n \theta\


    I know if you put in 2 for n and just exapand it out I can get it but I figured that there might be some other way to prove it. Just seems a bit simple but any help or insight would be greatly appriciated.
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  2. #2
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    Hello,
    Quote Originally Posted by gidget View Post
    *This is exactly how the question reads from the book

    Using the fact that a complex equation is really two real equations, find the double angle formulas (for Sin2 \theta\, Cos2 \theta\ ) by using this equation: (e^i \theta\)^n = (Cos \theta\+ iSin \theta\)^n = Cos n \theta\ + iSin n \theta\


    I know if you put in 2 for n and just exapand it out I can get it but I figured that there might be some other way to prove it. Just seems a bit simple but any help or insight would be greatly appriciated.
    Nope, you'll just have to expan (\cos \theta + i \sin \theta)^2

    Note that \cos(2 \theta) is the real part of it, and \sin(2 \theta) the imaginary part.
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  3. #3
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    Hello, gidget!

    Using the fact that a complex equation is really two real equations,
    find the double-angle formulas for \sin2\theta,\;\cos2\theta
    by using this equation: (e^{i\theta})^n \:=\: (\cos\theta + i\sin\theta)^n \:=\: \cos(n]\theta) + i\sin(n\theta)

    I know if you put in n=2 and just expand it out. I can get it.
    But I figured that there might be some other way to prove it.
    . .
    I don't know of a simpler way.
    I'm truly grateful for this method of cranking out multiple-angle identities.


    If we want a formula for \cos3\theta, we'd normally proceed like this:

    \cos3\theta \;=\;\cos(\theta + 2\theta)

    . . . . = \;\cos\theta\cos2\theta - \sin\theta\sin2\theta

    . . . . = \;\cos\theta (2\cos^2\!\theta - 1) - \sin\theta(2\sin\theta\cos\theta)

    . . . . = \;2\cos^3\!\theta - \cos\theta - 2\sin^2\!\theta\cos\theta

    . . . . = \;2\cos^3\!\theta - \cos\theta - 2(1-\cos^2\!\theta)\cos\theta

    . . . . = \;2\cos^3\!\theta - \cos\theta - 2\cos\theta + 2\cos^3\!\theta

    Therefore: . \boxed{\cos3\theta \;=\;4\cos^3\!\theta - 3\cos\theta}

    Then we must repeat the process for \sin3\theta.



    Using Demoivre's Formula, we have:

    \cos(3\theta) + i\sin(3\theta) \;=\;(\cos\theta + i\sin\theta)^3

    . . . . . . . . . . . . =\;\cos^3\!\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\!\theta - i\sin^3\!\theta

    . . . . . . . . . . . . = \;\left(\cos^3\!\theta - 3\cos\theta\sin^2\!\theta\right) + i\left(3\cos^2\!\theta\sin\theta - \sin^3\!\theta\right)

    . . . . . . . . . . . . =\;\left(\cos^3\!\theta - 3\cos\theta[1-\cos^2\!\theta]\right) + i\left(3[1-\sin^2\!\theta]\sin\theta - \sin^3\!\theta\right)

    . . . . . . . . . . . . = \;\left(4\cos^3\!\theta - 3\cos\theta\right) + i\left(3\sin\theta - 4\sin^3\!\theta\right)


    Equating real and imaginary components, we have:

    . . . . \begin{array}{ccc} \boxed{\cos3\theta \;=\; 4\cos^3\!\theta - 3\cos\theta} \\ \\[-3mm] \boxed{\sin3\theta \;=\; 3\sin\theta - 4\sin^3\theta} \end{array}


    And we have both formulas . . . with much less work!

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