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Thread: Sin and Cos

  1. #1
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    Sin and Cos

    How do I solve this? I have been working on this same problem for about an hour and can't come up with anything. I have to find the value of A, w, and b.

    2sin(12x + 1) + 3sin(12x + 3) = A sin(wx + b)

    Any help (with steps included) would be GREATLY appreciated! Thanks!
    Last edited by mr fantastic; Sep 6th 2010 at 07:45 PM. Reason: Moved, re-formatted.
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  2. #2
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    ok ... found an identity at this link that may help.

    List of trigonometric identities - Wikipedia, the free encyclopedia

    $\displaystyle a\sin{x} + b\sin(x+\alpha) = c\sin(x+\beta)$

    where ...

    $\displaystyle c = \sqrt{a^2+b^2+2ab\cos{\alpha}}$

    and

    $\displaystyle \beta = \arctan\left(\frac{b\sin{\alpha}}{a+b\cos{\alpha}} \right) + ...$

    $\displaystyle 0$ if $\displaystyle a+b\cos{\alpha} \ge 0$

    $\displaystyle \pi$ if $\displaystyle a+b\cos{\alpha} < 0$

    I'd let $\displaystyle x = 12\theta + 1$ and see where that goes.
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  3. #3
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    Hello, aeagan2!

    Wow . . . This took a lot of work!


    I assume you know the Compound Angle Identities:

    . . $\displaystyle \sin(A\pm B) \;=\;\sin A\cos B \pm \cos A\sin B$

    . . $\displaystyle \cos(A \pm B) \;=\;\cos A\cos B \mp \sin A\sin B$



    $\displaystyle 2\sin(12x +1) + 3\sin(12x+3) = A\sin(wx+b)$

    $\displaystyle \text{Find }A,\,w,\text{ and }b.$

    We have:

    . . $\displaystyle 2\bigg[\sin12x\cos1 + \cos12x\sin1\bigg] + 3\bigg[\sin12x\cos3 + \cos12x\sin3\bigg]$
    . . . . . . . . . . $\displaystyle =\;A\bigg[\sin wx\cos b + \cos wx\sin b\bigg] $


    . . $\displaystyle 2\cos1\sin12x + 2\sin1\cos12x + 3\cos3\sin12x + 3\sin3\cos12x$

    . . . . . . . . . . $\displaystyle =\;A\cos b\sin wx + A\sin b\cos wx$


    . . $\displaystyle (2\cos1 + 3\cos3)\sin12x + (2\sin1 + 3\sin3)\cos12x$

    . . . . . . . . . . $\displaystyle =\;A\cos b\sin wx + A\sin b\cos 2x$


    Equate "corresponding parts":

    . . $\displaystyle \begin{array}{cccc}(2\cos1 + 3\cos3)\sin12x &=& A\cos b\sin wx & [1] \\ (2\sin 1 + 3\sin 3)\cos12x &=& A\sin b\cos wx & [2]\end{array}$


    Let: .$\displaystyle \begin{Bmatrix}\sin12x &=& \sin wx \\ \cos12x &=& \cos wx\end{Bmatrix} \quad\Rightarrow\quad \boxed{w \:=\:12}$


    [2] and [1] becomes: .$\displaystyle \begin{array}{cccc}A\sin b &=&2\sin1 + 3\sin3 & [3] \\A\cos b &=& 2\cos1 + 3\cos 3 & [4] \end{array}$

    Divide [3] by [4]: .$\displaystyle \dfrac{A\sin b}{A\cos b} \;=\;\dfrac{2\sin1+3\sin3}{2\cos1+3\cos3} \;=\;\tan b$

    . . $\displaystyle \text{Hence: }\;\boxed{b \;=\;\tan^{-1}\left(\frac{2\sin1+3\sin3}{2\cos1+3\cos3}\right) } $


    $\displaystyle \begin{array}{cccccccc}\text{Square [3]:} & A^2\sin^2b &\!\!=\!\!& (2\sin 1 + 3\sin3)^2 &\!\!=\!\!& 4\sin^21 + 12\sin1\sin3 + 9\sin^23
    \\ \text{Square [4]:} & A^2\cos^2b &\!\!=\!\!& (2\cos1+3\cos3)^2 &\!\!=\!\!& 4\cos^21 + 12\cos1\cos3 + 9\cos^23 \end{array}$


    Add:

    $\displaystyle A\underbrace{(\sin^2\!b+\cos^2\!b)}_{\text{This is 1}} \;=\;4\underbrace{(\sin^2\!1+\cos^2\!1)}_{\text{Th is is 1}} + 12\underbrace{(\cos3\cos1 + \sin3\sin1)}_{\text{This is }\cos(3-1)} + 9\underbrace{(\sin^2\!3+\cos^2\!3)}_{\text{This is 1}} $

    And we have: .$\displaystyle A^2 \;=\;4 + 12\cos 2 + 9$

    . . . . Hence: .$\displaystyle \boxed{A \;=\;\sqrt{13 + 12\cos2}}$
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