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Find the minimum value of \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) for all real numbers \(\displaystyle x\).
Well for starters, we know it can never be any less than 0. Whether or not 0 is the minimum is another story though...Find the minimum value of \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) for all real numbers \(\displaystyle x\).
It is easy to verify that the sum of the derivatives of the functions $\sin x$ and $\cos x$, $\tan x$ and $\cot x$ , $\sec x$ and $\csc x$ vanishes for $x= \frac{\pi}{4}$...Find the minimum value of \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) for all real numbers \(\displaystyle x\).
Thanks for participating in this problemIt is easy to verify that the sum of the derivatives of the functions $\sin x$ and $\cos x$, $\tan x$ and $\cot x$ , $\sec x$ and $\csc x$ vanishes for $x= \frac {\pi}{4}$...
Kind regards
$\chi$ $\sigma$
To be fair, $\chi$ $\sigma$ ended his remark with dots, meaning that the real solution will come in a later post, which I'm sure he'll make.Thanks for participating in this problembut chisigma, I believe what you cited here, i.e. \(\displaystyle x={\pi}{4}\) only gives us the local minimum value but the question is asking for the absolute minimum value...
If we ignore the 'absolute value' the trigonometric function can be written as...It is easy to verify that the sum of the derivatives of the functions $\sin x$ and $\cos x$, $\tan x$ and $\cot x$ , $\sec x$ and $\csc x$ vanishes for $x= \frac{\pi}{4}$...
Hi I like Serena,To be fair, $\chi$ $\sigma$ ended his remark with dots, meaning that the real solution will come in a later post, which I'm sure he'll make.
If only the domain were limited to ($0, \frac \pi 2$)...
It's an interesting observation though.
Setting the derivative of (1) to zero does not become too bad.... and it is evident that the minimum of the absolute value of f(*) is in the range $\displaystyle \frac{\pi}{2} \le x \le 2\ \pi$. The minima can be obtained forcing to zero the derivative of (1) but it seems a complicated procedure, so that the task is delayed to next posts...
No harm done. Part of math is pointing out potential flaws in a reasoning.I think I should clear this up because when I made my previous remark, I did not intend to offend chisigma by saying his solution wasn't correct, but it was my mistake because I should have noticed chisigma is going to post for more to this thread that will eventually give us the final answer to this problem, I'm sorry if it came across that way.
The OP is offline now, but she did share with me a link to the solution you gave that she found online a few hours ago. It is the same, almost word for word:This is a Putnam problem that I accidentally assigned to my AP Calculus AB students. Only one of them solved it. I'm guessing this solution is out on the Internet somewhere, because it involved two brilliant insights that I don't think my student actually dreamed up himself. Anyhoo, here's the solution:
x | y' | This gives us a minimum point at (π/4, 6.242640687). |
2π/9 (40°) | negative | |
π/4 (45°) | 0 | |
5π/18 (50°) | positive |
x | y' | This gives us a minimum point at (5π/4, 6.242640687) |
11π/9 (220°) | negative | |
5π/4 (225°) | 0 | |
23π/18 (230°) | positive |
x | y' | This gives us a maximum point at (0.844611π, -1.828427125) |
5π/6 (150°) | positive | |
0.844611π (152.03°) | 0 | |
31π/36 (155°) | negative |
x | y' | This gives us a maximum point at (1.655389π, -1.828427125) |
29π/18 (290°) | positive | |
1.655389π (297.97°) | 0 | |
5π/3 (300°) | negative |
I just saw another really brilliant way to tackle this problem and since I have a habit to share all that I know with the members of the forum, (ahem, I am quite generous myselfFind the minimum value of \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) for all real numbers \(\displaystyle x\).
$\begin{align*}\sin x&=\sin (a-135^{\circ})\\&=\dfrac{-\sin a-\cos a}{\sqrt{2}}\end{align*}$ | $\begin{align*}\cos x&=\cos (a-135^{\circ})\\&=\dfrac{-\cos a+\sin a}{\sqrt{2}}\end{align*}$ | $\begin{align*}\sin x+\cos x&=\dfrac{-\sin a-\cos a}{\sqrt{2}}+\dfrac{-\cos a+\sin a}{\sqrt{2}}\\&=-\sqrt{2}\cos a\end{align*}$ |
$\begin{align*}\tan x&=\tan(a-135^{\circ})\\&=\dfrac{\sin a+\cos a}{\cos a-\sin a}\end{align*}$ | $\begin{align*}\cot x&=\cot(a-135^{\circ})\\&=\dfrac{\cos a-\sin a}{\sin a+\cos a}\end{align*}$ | $\begin{align*}\tan x+\cot x&=\dfrac{\sin a+\cos a}{\cos a-\sin a}+\dfrac{\cos a-\sin a}{\sin a+\cos a}\\&=\dfrac{2}{\cos^2 a-\sin^2 a}\end{align*}$ |
$\begin{align*}\sec x&=\dfrac{1}{\cos x}\\&=-\dfrac{\sqrt{2}}{\sin a+\cos a}\end{align*}$ | $\begin{align*}\csc x&=\dfrac{1}{\sin x}\\&=\dfrac{\sqrt{2}}{\sin a-\cos a}\end{align*}$ | $\begin{align*}\sec x+\csc x&=-\dfrac{\sqrt{2}}{\sin a+\cos a}+\dfrac{\sqrt{2}}{\sin a-\cos a}\\&=\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}\end{align*}$ |
You can only conclude that if there is a particular value of $a$ for which your function of $a$ actually achieves $2 \sqrt{2}-1$. Showing that $A \ge B$ does not imply $A=B$, or even that the minimum value of $A$ is $B$. Example: $x^{2}\ge -1$, but this does not mean the minimum value of $x^{2}$ is $-1$. For $x^{2} \ge -1$, we say the inequality is not tight. But if we have the inequality $x^{2} \ge 0$, then the inequality is tight, because you get equality for at least one value of $x$.We can conclude at this point the minimum value for \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) is $2 \sqrt{2}-1$.
Thanks, Ackbach for reminding me that my post isn't complete and isn't very convincing.You can only conclude that if there is a particular value of $a$ for which your function of $a$ actually achieves $2 \sqrt{2}-1$. Showing that $A \ge B$ does not imply $A=B$, or even that the minimum value of $A$ is $B$. Example: $x^{2}\ge -1$, but this does not mean the minimum value of $x^{2}$ is $-1$. For $x^{2} \ge -1$, we say the inequality is not tight. But if we have the inequality $x^{2} \ge 0$, then the inequality is tight, because you get equality for at least one value of $x$.
So: is your last inequality there tight or not, according to this line of reasoning?
$\begin{align*}\sin x&=\sin (a-135^{\circ})\\&=\dfrac{-\sin a-\cos a}{\sqrt{2}}\end{align*}$ | $\begin{align*}\cos x&=\cos (a-135^{\circ})\\&=\dfrac{-\cos a+\sin a}{\sqrt{2}}\end{align*}$ | $\begin{align*}\sin x+\cos x&=\dfrac{-\sin a-\cos a}{\sqrt{2}}+\dfrac{-\cos a+\sin a}{\sqrt{2}}\\&=-\sqrt{2}\cos a\end{align*}$ |
$\begin{align*}\tan x&=\tan(a-135^{\circ})\\&=\dfrac{\sin a+\cos a}{\cos a-\sin a}\end{align*}$ | $\begin{align*}\cot x&=\cot(a-135^{\circ})\\&=\dfrac{\cos a-\sin a}{\sin a+\cos a}\end{align*}$ | $\begin{align*}\tan x+\cot x&=\dfrac{\sin a+\cos a}{\cos a-\sin a}+\dfrac{\cos a-\sin a}{\sin a+\cos a}\\&=\dfrac{2}{\cos^2 a-\sin^2 a}\end{align*}$ |
$\begin{align*}\sec x&=\dfrac{1}{\cos x}\\&=-\dfrac{\sqrt{2}}{\sin a+\cos a}\end{align*}$ | $\begin{align*}\csc x&=\dfrac{1}{\sin x}\\&=\dfrac{\sqrt{2}}{\sin a-\cos a}\end{align*}$ | $\begin{align*}\sec x+\csc x&=-\dfrac{\sqrt{2}}{\sin a+\cos a}+\dfrac{\sqrt{2}}{\sin a-\cos a}\\&=\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}\end{align*}$ |
If we have $(\sqrt{2}\cos a+1)$ and $\dfrac{2}{\sqrt{2}\cos a+1}$ where $\sqrt{2}\cos a+1>0$, by applying AM-GM inequality to these two terms yields $\dfrac{(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1}}{2}\ge \sqrt{(\sqrt{2}\cos a+1)\left(\dfrac{2}{\sqrt{2}\cos a+1} \right)}\ge \sqrt{2}$, or simply $(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1}\ge 2 \sqrt{2}$ $\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \ge 2 \sqrt{2}-1$ | If we have $(\sqrt{2}\cos a+1)$ and $\dfrac{2}{\sqrt{2}\cos a+1}$ where $\sqrt{2}\cos a+1<0$ which also implies $\sqrt{2}\cos a<0$, by applying AM-GM inequality to these two terms yields $\dfrac{-\left((\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \right)}{2}\ge \sqrt{(\sqrt{2}\cos a+1)\left(\dfrac{2}{\sqrt{2}\cos a+1} \right)}\ge \sqrt{2}$, or simply $-\left((\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \right)\ge 2 \sqrt{2}$ $(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \le -2 \sqrt{2}$ $\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \le -2 \sqrt{2}-1$ |
Very excellent! You've convinced me that the function actually achieves that minimum, hence the inequality is tight.Thanks, Ackbach for reminding me that my post isn't complete and isn't very convincing.
I think I should have mentioned first that $a$ is defined for all real numbers.
Next, I wanted a re-post to fix things right...
If we let $x=a-135^{\circ}$, where $a\in R$, we see that we have:
$\begin{align*}\sin x&=\sin (a-135^{\circ})\\&=\dfrac{-\sin a-\cos a}{\sqrt{2}}\end{align*}$ $\begin{align*}\cos x&=\cos (a-135^{\circ})\\&=\dfrac{-\cos a+\sin a}{\sqrt{2}}\end{align*}$ $\begin{align*}\sin x+\cos x&=\dfrac{-\sin a-\cos a}{\sqrt{2}}+\dfrac{-\cos a+\sin a}{\sqrt{2}}\\&=-\sqrt{2}\cos a\end{align*}$ $\begin{align*}\tan x&=\tan(a-135^{\circ})\\&=\dfrac{\sin a+\cos a}{\cos a-\sin a}\end{align*}$ $\begin{align*}\cot x&=\cot(a-135^{\circ})\\&=\dfrac{\cos a-\sin a}{\sin a+\cos a}\end{align*}$ $\begin{align*}\tan x+\cot x&=\dfrac{\sin a+\cos a}{\cos a-\sin a}+\dfrac{\cos a-\sin a}{\sin a+\cos a}\\&=\dfrac{2}{\cos^2 a-\sin^2 a}\end{align*}$ $\begin{align*}\sec x&=\dfrac{1}{\cos x}\\&=-\dfrac{\sqrt{2}}{\sin a+\cos a}\end{align*}$ $\begin{align*}\csc x&=\dfrac{1}{\sin x}\\&=\dfrac{\sqrt{2}}{\sin a-\cos a}\end{align*}$ $\begin{align*}\sec x+\csc x&=-\dfrac{\sqrt{2}}{\sin a+\cos a}+\dfrac{\sqrt{2}}{\sin a-\cos a}\\&=\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}\end{align*}$
$\begin{align*}\therefore|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|&=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}+\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}-\dfrac{2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{2\cos^2 a-1}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|\\&=|-\sqrt{2}\cos a+\dfrac{2(1-\sqrt{2}\cos a)}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|\\&=|-\sqrt{2}\cos a-\dfrac{2}{\sqrt{2}\cos a+1}|\\&=|-\left(\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \right)|\\&=|\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1}|\end{align*}$
If we have $(\sqrt{2}\cos a+1)$ and $\dfrac{2}{\sqrt{2}\cos a+1}$ where $\sqrt{2}\cos a+1>0$, by applying AM-GM inequality to these two terms yields
$\dfrac{(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1}}{2}\ge \sqrt{(\sqrt{2}\cos a+1)\left(\dfrac{2}{\sqrt{2}\cos a+1} \right)}\ge \sqrt{2}$, or simply
$(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1}\ge 2 \sqrt{2}$
$\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \ge 2 \sqrt{2}-1$If we have $(\sqrt{2}\cos a+1)$ and $\dfrac{2}{\sqrt{2}\cos a+1}$ where $\sqrt{2}\cos a+1<0$ which also implies $\sqrt{2}\cos a<0$, by applying AM-GM inequality to these two terms yields
$\dfrac{-\left((\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \right)}{2}\ge \sqrt{(\sqrt{2}\cos a+1)\left(\dfrac{2}{\sqrt{2}\cos a+1} \right)}\ge \sqrt{2}$, or simply
$-\left((\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \right)\ge 2 \sqrt{2}$
$(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \le -2 \sqrt{2}$
$\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \le -2 \sqrt{2}-1$
Now if we set $\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} = 2 \sqrt{2}-1$, this gives us $\sqrt{2}\cos a=\dfrac{\sqrt{2}-1}{\sqrt{2}}<1$, which means what we set to be true has solution(s), then we know we're dealing with tight inequality here, and hence, we can conclude at this point the minimum value for \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) is $2 \sqrt{2}-1$.
Does this look okay now, Ackbach?![]()