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Math Help - 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; September 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

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

    where ...

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

    and

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

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

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

    I'd let 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:

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

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



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

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

    We have:

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


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

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


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

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


    Equate "corresponding parts":

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


    [2] and [1] becomes: . \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]: . \dfrac{A\sin b}{A\cos b} \;=\;\dfrac{2\sin1+3\sin3}{2\cos1+3\cos3} \;=\;\tan b

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


    \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 <br />
\\ \text{Square [4]:} & A^2\cos^2b &\!\!=\!\!& (2\cos1+3\cos3)^2 &\!\!=\!\!& 4\cos^21 + 12\cos1\cos3 + 9\cos^23  \end{array}


    Add:

    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: . A^2 \;=\;4 + 12\cos 2 + 9

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