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Thread: [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$\displaystyle \theta\$, Cos2$\displaystyle \theta\$ ) by using this equation: (e^i$\displaystyle \theta\$)^n = (Cos$\displaystyle \theta\$+ iSin$\displaystyle \theta\$)^n = Cos n$\displaystyle \theta\$ + iSin n$\displaystyle \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$\displaystyle \theta\$, Cos2$\displaystyle \theta\$ ) by using this equation: (e^i$\displaystyle \theta\$)^n = (Cos$\displaystyle \theta\$+ iSin$\displaystyle \theta\$)^n = Cos n$\displaystyle \theta\$ + iSin n$\displaystyle \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 $\displaystyle (\cos \theta + i \sin \theta)^2$

    Note that $\displaystyle \cos(2 \theta)$ is the real part of it, and $\displaystyle \sin(2 \theta)$ 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 $\displaystyle \sin2\theta,\;\cos2\theta$
    by using this equation: $\displaystyle (e^{i\theta})^n \:=\: (\cos\theta + i\sin\theta)^n \:=\: \cos(n]\theta) + i\sin(n\theta)$

    I know if you put in $\displaystyle n=2$ 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 $\displaystyle \cos3\theta$, we'd normally proceed like this:

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

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

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

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

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

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

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

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



    Using Demoivre's Formula, we have:

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

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

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

    . . . . . . . . . . . .$\displaystyle =\;\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) $

    . . . . . . . . . . . .$\displaystyle = \;\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:

    . . . . $\displaystyle \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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