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Thread: 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;

    $\displaystyle \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;

    $\displaystyle \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

    $\displaystyle U_3 = U_2 + d = a + 2d$

    $\displaystyle U_2 = a + d$

    $\displaystyle a = \sqrt{3}cos\theta$

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

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

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


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


    sub [1] into [2]

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

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

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

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

    $\displaystyle \theta = arctan \left(\frac{1-\sqrt3}{1+\sqrt3}\right) = -15^{\circ}$
    Last edited by e^(i*pi); Dec 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: .$\displaystyle \sqrt{3}\cos\theta,\;\sin(\theta-30^o),\;\sin\theta$
    where $\displaystyle \theta$ is acute. . Find the value of $\displaystyle \theta.$
    The first three terms are: .$\displaystyle a,\;a+d,\;a+2d$

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


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

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


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

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


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

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

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


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

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