Show that the equation sec θ + cosec θ = c has two roots between 0 and 2π if 2 < c^2 <8 ?
Hi,
This one looks easy at first sight I think I'll just give you help, not sure for the solution.
$\displaystyle \frac{1}{\sin \theta}+\frac{1}{\cos \theta}=c$ (note that c is a real number since the RHS is a real number)
Square the equation :
$\displaystyle \overbrace{\frac{1}{\sin^2 \theta}+\frac{1}{\cos^2 \theta}}^{A}+\overbrace{\frac{2}{\sin \theta \cos \theta}}^{B}=c^2$
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Recall that $\displaystyle \sin(2x)=2 \cos(x)\sin(x) \implies \cos^2(x)\sin^2(x)=\tfrac{\sin^2(2x)}{4}$
$\displaystyle A=\frac{\sin^2 \theta+\cos^2 \theta}{\sin^2 \theta \cos^2 \theta}=\frac{1}{\tfrac{\sin^2(2 \theta)}{4}}=\frac{4}{\sin^2(2 \theta)}$
$\displaystyle B=\frac{2}{\sin \theta \cos \theta}=\frac{2}{\tfrac{\sin(2 \theta)}{2}}=\frac{4}{\sin(2 \theta)}$
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The equation is now :
$\displaystyle \frac{4}{\sin^2(2 \theta)}+\frac{4}{\sin(2 \theta)}=c^2$
Multiply both sides by $\displaystyle \sin^2(2 \theta)$ (which is $\displaystyle \neq 0$, otherwise, the first equation would be nonsense)
$\displaystyle 4+4 \sin(2 \theta)=c^2 \cdot \sin^2(2 \theta) \implies c^2 \cdot \sin^2(2 \theta)-4 \sin(2 \theta)-4=0$
$\displaystyle X=\sin(2 \theta)$. The equation is now : $\displaystyle c^2 X-4X-4=0$
The discriminant is : $\displaystyle \Delta=16+4 \cdot 4 \cdot c^2=16(1+c^2)$, which is always positive (because c is a real number).
Solutions are $\displaystyle X=\frac{4 \pm 4 \sqrt{1+c^2}}{2c^2}=\frac{2 \pm 2 \sqrt{1+c^2}}{c^2}$
We know that $\displaystyle X=\sin(2 \theta) \implies -1 \le X \le 1 \Leftrightarrow |X| \le 1$.
So find the range for c, such that both the solutions are in this interval :
$\displaystyle \left|2 \pm 2 \sqrt{1+c^2}\right| \le c^2$
Ok, I have a *better* (and more correct) sequence for you
You're asked : prove "******** has 2 roots if 2<c²<8"
So actually, you just have to prove that $\displaystyle 2<c^2<8 \implies ********* \text{ has 2 roots }$
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Write the first inequality :
We want c to satisfy this : $\displaystyle 2+2 \sqrt{1+c^2} \le c^2 \implies 2+2 \sqrt{1+c^2}-c^2 \le 0$
$\displaystyle 2<c^2<8 \implies 1+c^2<9 \implies \sqrt{1+c^2}<3 \implies 2+2\sqrt{1+c^2}<8$
And $\displaystyle 2<c^2<8 \implies -8<-c^2<-2 \implies -c^2<-2$
------> $\displaystyle 2+2 \sqrt{1+c^2}-c^2 \le 8-2=6$
Thus this inequality leads to nothing.
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Write the second inequality :
We want c to satisfy this : $\displaystyle \left|2-2 \sqrt{1+c^2} \right| \le c^2$
Note that $\displaystyle \sqrt{1+c^2} \ge 1 \implies 2-2\sqrt{1+c^2} \le 0$
Therefore $\displaystyle \left|2-2 \sqrt{1+c^2} \right|=-(2-2\sqrt{1+c^2})=2 \sqrt{1+c^2}-2$
The inequality is now : $\displaystyle 2 \sqrt{1+c^2}-2 \le c^2 \implies 2 \sqrt{1+c^2}-c^2 \le 2$
and this leads to nothing too... so I'm quite puzzled here
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we both don't see any mistake, and the text is correct.... does anyone know what's wrong ?
Hello,
I didn't find the mistake your looking for but I've another proof :
Using the graph (see attachment) it seems that for $\displaystyle |c|<2\sqrt{2}\Longleftrightarrow c^2<8$ the equation
$\displaystyle \frac{1}{\sin x}+\frac{1}{\cos x}=c\quad (1)$
has two solutions. If we manage to show this then we'll get the result for $\displaystyle 2<c^2<8$.
Let $\displaystyle f(x)= \frac{1}{\sin x}+\frac{1}{\cos x}$ for $\displaystyle x\in \left(0,\frac{\pi}{2}\right)\cup\left( \frac{\pi}{2},\pi\right)\cup\left(\pi,\frac{3\pi}{ 2}\right)\cup \left(\frac{3\pi}{2},2\pi\right)$. The derivative of $\displaystyle f$ is
$\displaystyle f'(x)=-\frac{\cos x}{\sin^2x}+\frac{\sin x}{\cos^2x}=\frac{\sin^3x-\cos^3x}{\sin^2x\cos^2x}=\frac{(\sin x-\cos x)(1+\sin x\cos x)}{\sin ^2x\cos^2x}$
As $\displaystyle \sin x \cos x \neq -1$, $\displaystyle f'(x)$ equals 0 iff $\displaystyle \sin x=\cos x$ that is iff $\displaystyle x=\frac{\pi}{4}$ or $\displaystyle x=\frac{5\pi}{4}$.
As $\displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to \frac{\pi}{2}^-}=\infty
$, $\displaystyle f\left(\frac{\pi}{4}\right)=2\sqrt{2}$ is the minimum of $\displaystyle f$ on $\displaystyle \left(0,\frac{\pi}{2}\right)$. Similarly $\displaystyle f\left(\frac{5\pi}{4}\right)=-2\sqrt{2}$ is the maximum of $\displaystyle f$ on $\displaystyle \left(\pi,\frac{3\pi}{2}\right)$. At this point we've shown that (1) has no solution for $\displaystyle x\in\left(0,\frac{\pi}{2}\right)\cup\left(\pi,\fra c{3\pi}{2}\right)$ if $\displaystyle c<2\sqrt{2}$.
Let's show that (1) has exactly one solution in $\displaystyle \left(\frac{\pi}{2},\pi\right)$. Let $\displaystyle x\in \left(\frac{\pi}{2},\pi\right)$. We have
$\displaystyle f'(x) =\frac{(\sin x-\cos x)(1+\sin x\cos x)}{\cos^2x\sin ^2x}$
with $\displaystyle 1+\cos x \sin x>0$ and $\displaystyle \cos^2x\sin ^2x>0$ so $\displaystyle f'$ has same sign as $\displaystyle \sin x-\cos x=-\sqrt{2}\cos \left(x+\frac{\pi}{4}\right)>0$. Thus $\displaystyle f$ is increasing on $\displaystyle \left(\frac{\pi}{2},\pi\right)$. As $\displaystyle f$ is continuous and
$\displaystyle \lim_{x\to \frac{\pi}{2}^+}f(x)=-\infty$ and $\displaystyle \lim_{x\to \pi^-}f(x)=\infty$
$\displaystyle f(x)=c$ has at least one solution on this interval (~intermediate value theorem). $\displaystyle f$ being increasing, there is exactly one solution.
Similarly one can show that $\displaystyle f(x)=c$ has one solution in $\displaystyle \left(\frac{3\pi}{2},2\pi\right)$ hence if $\displaystyle |x|<2\sqrt{2}$, (1) has two solutions.