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

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August 26th 2008, 06:33 AM
fardeen_gen

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 ?
August 26th 2008, 06:51 AM
Moo

Hi,

Quote:

Originally Posted by fardeen_gen

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 (Worried) 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 :

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

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

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

$B=\frac{2}{\sin \theta \cos \theta}=\frac{2}{\tfrac{\sin(2 \theta)}{2}}=\frac{4}{\sin(2 \theta)}$
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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 (Wondering) :

$\left|2 \pm 2 \sqrt{1+c^2}\right| \le c^2$
August 26th 2008, 08:27 AM
Moo

Ok, I have a *better* (and more correct) sequence for you (Giggle)

You're asked : prove "******** has 2 roots if 2<c²<8"

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

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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 (Crying)

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we both don't see any mistake, and the text is correct.... does anyone know what's wrong ? http://smileys.sur-la-toile.com/repo...sol%E9-987.gif
August 26th 2008, 10:56 AM
flyingsquirrel

1 Attachment(s)

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.