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Math Help - Show that the equation sec θ + cosec θ = c has two roots...?

  1. #1
    Super Member fardeen_gen's Avatar
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    Show that the equation sec θ + cosec θ = c has two roots...?

    Show that the equation sec θ + cosec θ = c has two roots between 0 and 2π if 2 < c^2 <8 ?
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  2. #2
    Moo
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    Hi,
    Quote Originally Posted by fardeen_gen View Post
    Show that the equation sec θ + cosec θ = c has two roots between 0 and 2π if 2 < c^2 <8 ?
    This one looks easy at first sight I think I'll just give you help, not sure for the solution.

    \frac{1}{\sin \theta}+\frac{1}{\cos \theta}=c (note that c is a real number since the RHS is a real number)

    Square the equation :

    \overbrace{\frac{1}{\sin^2 \theta}+\frac{1}{\cos^2 \theta}}^{A}+\overbrace{\frac{2}{\sin \theta \cos \theta}}^{B}=c^2

    --------------------------------------
    Recall that \sin(2x)=2 \cos(x)\sin(x) \implies  \cos^2(x)\sin^2(x)=\tfrac{\sin^2(2x)}{4}

    A=\frac{\sin^2 \theta+\cos^2 \theta}{\sin^2 \theta \cos^2 \theta}=\frac{1}{\tfrac{\sin^2(2 \theta)}{4}}=\frac{4}{\sin^2(2 \theta)}


    B=\frac{2}{\sin \theta \cos \theta}=\frac{2}{\tfrac{\sin(2 \theta)}{2}}=\frac{4}{\sin(2 \theta)}
    --------------------------------------

    The equation is now :

    \frac{4}{\sin^2(2 \theta)}+\frac{4}{\sin(2 \theta)}=c^2

    Multiply both sides by \sin^2(2 \theta) (which is \neq 0, otherwise, the first equation would be nonsense)

    4+4 \sin(2 \theta)=c^2 \cdot \sin^2(2 \theta) \implies c^2 \cdot \sin^2(2 \theta)-4 \sin(2 \theta)-4=0

    X=\sin(2 \theta). The equation is now : c^2 X-4X-4=0

    The discriminant is : \Delta=16+4 \cdot 4 \cdot c^2=16(1+c^2), which is always positive (because c is a real number).

    Solutions are X=\frac{4 \pm 4 \sqrt{1+c^2}}{2c^2}=\frac{2 \pm 2 \sqrt{1+c^2}}{c^2}

    We know that X=\sin(2 \theta) \implies -1 \le X \le 1 \Leftrightarrow |X| \le 1.

    So find the range for c, such that both the solutions are in this interval :

    \left|2 \pm 2 \sqrt{1+c^2}\right| \le c^2
    Last edited by Moo; August 26th 2008 at 07:16 AM. Reason: added two missing =0
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  3. #3
    Moo
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    Ok, I have a *better* (and more correct) sequence for you

    You're asked : prove "******** has 2 roots if 2<c<8"

    So actually, you just have to prove that 2<c^2<8 \implies ********* \text{ has 2 roots }

    ---------------------------------------------
    Write the first inequality :

    We want c to satisfy this : 2+2 \sqrt{1+c^2} \le c^2 \implies 2+2 \sqrt{1+c^2}-c^2 \le 0

    2<c^2<8 \implies 1+c^2<9 \implies \sqrt{1+c^2}<3 \implies 2+2\sqrt{1+c^2}<8

    And 2<c^2<8 \implies -8<-c^2<-2 \implies -c^2<-2

    ------> 2+2 \sqrt{1+c^2}-c^2 \le 8-2=6

    Thus this inequality leads to nothing.

    ---------------------------------------------
    Write the second inequality :

    We want c to satisfy this : \left|2-2 \sqrt{1+c^2} \right| \le c^2

    Note that \sqrt{1+c^2} \ge 1 \implies 2-2\sqrt{1+c^2} \le 0

    Therefore \left|2-2 \sqrt{1+c^2} \right|=-(2-2\sqrt{1+c^2})=2 \sqrt{1+c^2}-2

    The inequality is now : 2 \sqrt{1+c^2}-2 \le c^2 \implies 2 \sqrt{1+c^2}-c^2 \le 2





    and this leads to nothing too... so I'm quite puzzled here


    ~~~~~~~~~~~~~~~~~~~
    we both don't see any mistake, and the text is correct.... does anyone know what's wrong ?
    Last edited by Moo; August 26th 2008 at 08:41 AM.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Hello,

    I didn't find the mistake your looking for but I've another proof :

    Using the graph (see attachment) it seems that for |c|<2\sqrt{2}\Longleftrightarrow c^2<8 the equation

    \frac{1}{\sin x}+\frac{1}{\cos x}=c\quad (1)

    has two solutions. If we manage to show this then we'll get the result for 2<c^2<8.


    Let f(x)= \frac{1}{\sin x}+\frac{1}{\cos x} for x\in \left(0,\frac{\pi}{2}\right)\cup\left( \frac{\pi}{2},\pi\right)\cup\left(\pi,\frac{3\pi}{  2}\right)\cup \left(\frac{3\pi}{2},2\pi\right). The derivative of f is


    f'(x)=-\frac{\cos x}{\sin^2x}+\frac{\sin x}{\cos^2x}=\frac{\sin^3x-\cos^3x}{\sin^2x\cos^2x}=\frac{(\sin x-\cos x)(1+\sin x\cos x)}{\sin ^2x\cos^2x}

    As \sin x \cos x \neq -1, f'(x) equals 0 iff \sin x=\cos x that is iff x=\frac{\pi}{4} or x=\frac{5\pi}{4}.

    As \lim_{x\to 0^+}f(x)=\lim_{x\to \frac{\pi}{2}^-}=\infty<br />
, f\left(\frac{\pi}{4}\right)=2\sqrt{2} is the minimum of f on \left(0,\frac{\pi}{2}\right). Similarly f\left(\frac{5\pi}{4}\right)=-2\sqrt{2} is the maximum of f on \left(\pi,\frac{3\pi}{2}\right). At this point we've shown that (1) has no solution for x\in\left(0,\frac{\pi}{2}\right)\cup\left(\pi,\fra  c{3\pi}{2}\right) if c<2\sqrt{2}.

    Let's show that (1) has exactly one solution in  \left(\frac{\pi}{2},\pi\right). Let x\in \left(\frac{\pi}{2},\pi\right). We have


    f'(x) =\frac{(\sin x-\cos x)(1+\sin x\cos x)}{\cos^2x\sin ^2x}

    with 1+\cos x \sin x>0 and \cos^2x\sin ^2x>0 so f' has same sign as \sin x-\cos x=-\sqrt{2}\cos \left(x+\frac{\pi}{4}\right)>0. Thus f is increasing on \left(\frac{\pi}{2},\pi\right). As f is continuous and


    \lim_{x\to \frac{\pi}{2}^+}f(x)=-\infty and \lim_{x\to \pi^-}f(x)=\infty

    f(x)=c has at least one solution on this interval (~intermediate value theorem). f being increasing, there is exactly one solution.

    Similarly one can show that f(x)=c has one solution in \left(\frac{3\pi}{2},2\pi\right) hence if |x|<2\sqrt{2}, (1) has two solutions.
    Attached Thumbnails Attached Thumbnails Show that the equation sec &#952; + cosec &#952; = c has two roots...?-sec_cosec.png  
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