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Thread: Trigonometric "transformation"?

  1. #1
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    Trigonometric "transformation"?

    Rewrite $\displaystyle 6cos(3t) - 5sin(3t)$ in the form
    a. $\displaystyle A_{1}sin(3t + \phi_{1})$
    b. $\displaystyle A_{2}cos(3t - \phi_{2})$

    I'm a bit puzzled by this. Can anyone offer insight?

    --marc
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  2. #2
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    Hello, Marc!

    This takes some Olympic-level gymnastics . . .


    Rewrite $\displaystyle 6\cos3t - 5\sin3t$ in the form:

    $\displaystyle (a)\;A\sin(3t + \phi)$

    $\displaystyle (b)\;A\cos(3t - \phi)$
    I'll do part (a) . . .


    We have: .$\displaystyle 6\cos3t - 5\sin3t$ .[1]


    Use the two coefficients as legs of a right triangle.

    . . Find the hypotenuse: .$\displaystyle h \:=\:\sqrt{6^2 + 5^2} \:=\:\sqrt{61}$


    Multiply by $\displaystyle \frac{h}{h}\!:\quad\frac{\sqrt{61}}{\sqrt{61}}\big g[6\cos3t - 5\sin3t\bigg] \;=\;\sqrt{61}\bigg[\frac{6}{\sqrt{61}}\cos3t - \frac{5}{\sqrt{61}}\sin3t\bigg] $ .[2]


    Let $\displaystyle \phi$ be in a right triangle with: .$\displaystyle opp = 6,\; adj = 5,\; hyp = \sqrt{61}$
    Code:
                      *
                    * *
            __    *   *
           √61  *     * 6
              *       *
            * φ       *
          *  *  *  *  *
                  5
    Then: .$\displaystyle \sin\phi \,=\,\frac{6}{\sqrt{61}},\;\;\cos\phi \,=\,\frac{5}{\sqrt{61}} $


    Substitute into [2]: . $\displaystyle \sqrt{61}\bigg[\sin\phi\cos3t - \cos\phi\sin3t\bigg] $


    Therefore, we have: . $\displaystyle \sqrt{61}\,\sin(3t - \phi)$

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