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The inequality is: $$ \frac{\sin{\theta}+1}{\cos{\theta}}\leq 1 \text{ with } \cos{\theta}\neq0 \land 0\leq \theta\lt 2\pi$$ I've tried splitting it up into cases of $\theta$ that make $\cos{\theta}$ positive so: $0\leq\theta<\frac{\pi}{2}$ and $\frac{3\pi}{2} \lt\theta\lt 2\pi$ thus allowing the inequality to simply to: $$ \sin{\theta}-\cos{\theta}\leq -1 $$ which after using the fact that $\sin{\theta}-\cos{\theta}=\sqrt{2}\sin{(\theta-\frac{\pi}{4})}$ and thus solving this inequality$$\sqrt{2}\sin{(\theta-\frac{\pi}{4})}\leq -1 \text{ for } 0\leq\theta<\frac{\pi}{2} \text{ or }\frac{3\pi}{2} \lt\theta\lt 2\pi $$ and this inequality $$ \sqrt{2}\sin{(\theta-\frac{\pi}{4})}\geq -1 \text{ for } \frac{\pi}{2}\lt\theta<\frac{3\pi}{2} $$ which give the solutions: $$ \theta=0 \text{ and } \frac{\pi }{2}<\theta<\frac{3 \pi }{2}\text{ and } \frac{3 \pi }{2}<\theta<2 \pi $$

I'm just curious if there's another way to do it that isn't quite so case based since it seems that I'm just solving it by plugging in values and testing. So I'm wondering if there's a better more insightful or efficient way to to this?

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  • $\begingroup$ $x$ should be $\theta$, I guess... $\endgroup$ – Hans Lundmark Feb 8 '14 at 21:16
  • $\begingroup$ yes that's right, I'm not quite sure why I decided to change $\theta$ to $x$ $\endgroup$ – Jay Feb 8 '14 at 21:18
  • $\begingroup$ Here is a variant crude calculation, which works with equations. (These are less treacherous than inequalities.) The quantity $\frac{1+\sin\theta}{\cos \theta}-1$ can only change sign at a singularity ($\pi/2$ and $3\pi/2$) and where $\frac{1+\sin \theta}{\cos \theta}=1$. This can be rewritten as $\cos\theta-\sin\theta=1$. If this holds, then squaring we get $-2\cos\theta\sin\theta=0$, which holds at our previously mentioned points, and $\pi$. Now check the signs in the intervals. There is no sign change at $\pi$. $\endgroup$ – André Nicolas Feb 8 '14 at 21:32
  • $\begingroup$ Note that the case $(\pi/2,3\pi/2)$ requires no testing, as in that range the numerator is non-negative while the denominator is negative. $\endgroup$ – Martin Argerami Feb 9 '14 at 0:28
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\begin{align} 1 + \sin \theta &= ( \cos (\theta /2) + \sin (\theta /2))^2 \\ \cos \theta &= ( \cos ^2 (\theta /2) - \sin ^2 (\theta /2) \\ \frac{1 + \sin \theta }{\cos \theta } &= \frac{ \cos (\theta /2) + \sin (\theta /2) }{ \cos (\theta /2) - \sin (\theta /2)} = \frac{ 1+ \tan (\theta /2)}{1 - \tan (\theta /2)} \end{align}

Therefore the given inequation reduces to $\frac{2 \tan (\theta /2)}{1 - \tan (\theta /2)} \leq 0$ whose solution is $\tan (\theta /2) \leq 0$ or $\tan (\theta /2) \geq 1$.

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I realize this question was tagged algebra-precalculus, trigonometry, so perhaps my answer is a little off the map; but I thought this approach might be worth sharing in any event, so here goes (In my defense I might add though I use calculus, I avoid the Riemann hypothesis!):

The inequality

$\dfrac{\sin{\theta}+1}{\cos{\theta}}\leq 1, \tag{1}$

is manifestly false in the range $0 < \theta < \pi / 2$, and this is very easy to see: for $0 < \theta < \pi / 2$, $0 < \cos \theta, \sin \theta < 1$; thus $1 + \sin \theta > 1$ so we have

$\dfrac{\sin{\theta}+1}{\cos{\theta}} > 1 \tag{2}$

for $\theta$ in this range. However, the inequality

$\dfrac{\sin{\theta}+1}{\cos{\theta}} < 1 \tag{3}$

does apply for $-\pi / 2 < \theta < 0$, though things get a little more tricky. With a little bit of calculus and differential equations we can obtain a quick demonstration, thus: let us define $f(\theta)$ thusly:

$f(\theta) = \dfrac{\sin{\theta}+1}{\cos{\theta}} = \tan \theta + \sec \theta, \tag{4}$

$-\pi / 2 < \theta < \pi / 2$; then

$f'(\theta) = \sec^2 \theta + (\sec \theta)(\tan \theta) = \sec \theta(\tan \theta + \sec \theta) = (\sec \theta) f, \tag{5}$

and the unique solution to (5) with $f(0) = 1$ is

$f(\theta) = \exp(\int_0^\theta \sec \tau \; \mathrm d \tau). \tag{6}$

For $-\pi / 2 < \theta < 0$, $\int_0^\theta \sec \tau \; \text{d} \tau < 0$, so we have $0 < f(\theta) < 1$ for this interval, hence for $3\pi / 2 < \theta < 2\pi$. So we've covered the cases $3\pi / 2 < \theta < 2\pi$ and $0 \le \theta < \pi /2$. What happens for $\pi / 2 < \theta < 3\pi /2$? Well, it is pretty clear that $f(\pi) = -1$, and the unique solution with this initial condition is

$f(\theta) = -\exp(\int_\pi^\theta \sec \tau \; \mathrm d \tau), \tag{7}$

holding for $\pi /2 < \theta < 3\pi / 2$. Since $\exp(\int_\pi^\theta \sec \tau) \; \text{d} \tau > 0$ for all $\theta \in (\pi / 2, 3\pi / 2)$, we have $f(\theta) < 0 < 1$ on this interval as well. Summarizing, we see that $f(\theta) = \tan \theta + \sec \theta = (\sin \theta + 1) / \cos \theta$ satisfies:

$f(0) = 1, \tag{8}$

$f(\theta) > 1 \; \text{for} \; 0 < \theta < \dfrac{\pi}{2}, \tag{9}$

$f(\theta) < 0 < 1 \; \text{for} \; \dfrac{\pi}{2} < \theta < \dfrac{3\pi}{2}, \tag{10}$

$0 < f(\theta) < 1 \; \text{for} \; \dfrac{3\pi}{2} < \theta < 2\pi. \tag{11}$

Apparently $f(\theta) \le 1$ for $\theta = 0$, $\theta \in (\pi / 2, 3\pi / 2)$, $\theta \in (3\pi / 2, 2\pi)$.

Note that if the question had been to find $\theta$ with

$\mid \dfrac{\sin \theta + 1}{\cos \theta} \mid \le 1 \tag{12}$

then a little more work would be needed, but I think the ODE-based analysis I have given here would help for that case as well. I leave the details to my readers as a teaser.

I must say in closing that I find equations such as

$\exp(\int_0^\theta \sec \tau \; \mathrm d \tau) = \tan \theta + \sec \theta \tag{13}$

most amusing; that's why it's tagged with a lucky $13$!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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Here's yet another idea. Let $\theta=2x+\pi/2$. Then the inequality becomes $$ \frac{2 \cos^2 x}{-2\sin x \cos x} \le 1 , $$ so $$ -\cot x \le 1 \quad (\text{and } \sin x \cos x \neq 0) . $$ This gives $0<x \le 3\pi/4$ and $x \neq \pi/2$ (and translates of that interval by integer multiples of $\pi$). Multiply by 2 and add $\pi/2$ to get the result in terms of $\theta$.

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Answer by Nghi H Nguyen, author of the new Transforming Method to solve quadratic equations (Google Search).

  1. Transform the given inequality $F(x)$ into the form $F(x) = f(x)/g(x) < 0$. The transformation was completed by S. Thilakan on Feb 8 (see Answer 1 above). $F(x) = f(x)/g(x) < 0$, with $f(x) = 2 \tan (x/2)$ and $g(x) = 1 - \tan (x/2)$.
  2. The common period of the inequality is: $2\pi$. Find the solution set within this period $2\pi$.
  3. Solve the inequality algebraically. Determine the variation of $f(x)$ and $g(x)$ within the period $2\pi$. Make a sign chart (table) with $x$ varies from $0$ to $2\pi$ ($x/2$ varies from $0$ to $\pi$). The values of $x$ create many intervals between them. Put the sign (+) or (-) accordingly inside these intervals by considering the various positions of the arc $x$ that varies on the trig unit circle. The signs of $F(x)$ will be the resultant signs of the quotient $f(x)/g(x)$. Solution set for $F(x) < 0$ are the intervals: $(\pi/2 , \pi)$ and $(3\pi/2 , 2\pi)$. Sorry that I can't figure the sign chart due to the answer format. Note: Since there is an equal sign in the inequality (F(x) less or equal to zero), the solution set for F(x) - less or equal to zero - will be the half closed intervals: [Pi/2 , Pi) and [3Pi/2 , 2Pi).
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