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Math Help - trig arithmetic series..

  1. #1
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    trig arithmetic series..

    The first three terms of an arithmetic series are √3cosθ, sin(θ−30) and sin θ, where θ is acute. Find the value of θ.
    So far I have expanded sin(θ−30)
    to get;

     \sqrt{3}cos\theta + \frac{\sqrt{3}}{2}sin\theta -\frac{1}{2}cos\theta + sin\theta

    how do I find theta? Is there a formula?
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Tweety View Post
    So far I have expanded sin(θ−30)
    to get;

     \sqrt{3}cos\theta + \frac{\sqrt{3}}{2}sin\theta -\frac{1}{2}cos\theta + sin\theta

    how do I find theta? Is there a formula?
    You can use the usual arithmetic sequence formulae. It will require simultaneous equations though

    U_3 = U_2 + d = a + 2d

    U_2 = a + d

    a = \sqrt{3}cos\theta

    ================================================== ======

    \sqrt{3}cos\theta + \frac{\sqrt{3}}{2}sin\theta -\frac{1}{2}cos\theta + sin\theta = \sqrt{3}cos\theta + 2d

    \therefore \: \: \frac{2+\sqrt{3}}{2}\,sin\theta -\frac{1}{2}cos\theta = d..........[1]


    sin \theta = \sqrt{3}cos \theta + 2d \:\: \rightarrow \:\: d = \frac{1}{2}(sin \theta - \sqrt{3}cos\theta)............[2]


    sub [1] into [2]

    \frac{2+\sqrt{3}}{2}\,sin\theta -\frac{1}{2}cos\theta = \frac{1}{2}(sin \theta - \sqrt{3}cos\theta) = \frac{1}{2}sin\theta - \frac{\sqrt3}{2}cos\theta

    \frac{1+\sqrt3}{2}sin\theta + \frac{\sqrt3 -1}{2}cos\theta=0

    (1+\sqrt3)sin\theta = -(\sqrt3 -1)cos\theta = (1-\sqrt3)cos\theta

    \frac{sin\theta}{cos\theta} = tan\theta =  \frac{1-\sqrt3}{1+\sqrt3}

    \theta = arctan \left(\frac{1-\sqrt3}{1+\sqrt3}\right) = -15^{\circ}
    Last edited by e^(i*pi); December 26th 2009 at 06:33 AM. Reason: changing from radians to degrees as per the question
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  3. #3
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    HGello, Tweety!

    The first three terms of an arithmetic sequence are: . \sqrt{3}\cos\theta,\;\sin(\theta-30^o),\;\sin\theta
    where \theta is acute. . Find the value of \theta.
    The first three terms are: . a,\;a+d,\;a+2d

    We have: . \begin{Bmatrix}a &=& \sqrt{3}\cos\theta  & [1] \\<br />
a + d &=& \sin(\theta-30^o) & [2] \\<br />
a+2d &=& \sin\theta & [3] \end{Bmatrix}


    Subtract [2] from [1]: . d \:=\:\sin(\theta-30^o) - \sqrt{3}\cos\theta \;\;[4]

    Subtract [3] from [2]: . d \;=\;\sin\theta - \sin(\theta-30^o)\;\;[5]


    Equate [4] and [5]: . \sin(\theta-30^o) - \sqrt{3}\cos\theta \;=\;\sin\theta - \sin(\theta-30^o) . \Rightarrow\quad 2\sin(\theta-30^o) \;=\;\sin\theta + \sqrt{3}\cos\theta

    Divide by 2: . \sin(\theta-30^o) \;=\;\tfrac{1}{2}\sin\theta + \tfrac{\sqrt{3}}{2}\cos\theta


    Since \sin30^o = \tfrac{1}{2}\,\text{ and }\,\cos30^o = \tfrac{\sqrt{3}}{2}

    . . we have: . \sin(\theta-30^o) \;=\;\sin30^o\sin\theta + \cos30^o\cos\theta

    . . . That is: . \sin(\theta-30^o) \;=\;\cos(\theta -30^o) \quad\Rightarrow\quad \frac{\sin(\theta-30^o)}{\cos(\theta-30^o)} \;=\;1


    Therefore: . \tan(\theta-30^o) \:=\:1 \quad\Rightarrow\quad \theta-30^o \:=\:45 \quad\Rightarrow\quad\boxed{ \theta \:=\:75^o}

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