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Show that $f(x)=\sqrt 2$ has no solutions when $$f(x)=\sin x\cos x(2+\sin x)$$ and $x\in [0,\frac{\pi}{2}]$.

My attempt:

Since $f(x)$ is continuous in its domain, it is enough to show that maximum value attained by $f$ in $[0,\frac{\pi}{2}]$ is less than $\sqrt 2$.

Also, since $f(0)=f(\pi/2)=0$, it must hold true that $f'(c)=0$ for some $c\in[0,\frac{\pi}{2}]$ (Rolle's Theorem). And since $f(\pi/6)>0$, there has to exist a maxima.

But differentiation doesn't help me here, because when I set $f'(x)$ to $0$, I get (on rearrangement and using basic trigonometry) a cubic in $\sin x$: $$3t^3+4t^2-2t-2=0$$ where $t=\sin x$.

The above cubic doesn't have any rational roots and I'm stuck.

Any help will be great. Thanks!

Edit:

I created this question by myself to solve in a pen-paper test. So methods that involve usage of calculators are useless. No offence.

P.S. Please keep in mind that I'm barely seventeen, so no highly advance math please!

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    $\begingroup$ As $$f(x)=2\sin(x)\cos(x)+\sin^2(x)\cos(x)=\sin(2x)+\cos(x)-\cos^2(x).$$ Now for $0\leq x\leq\pi/2$ we have $0\leq\sin(2x)\leq1$ and the function $x\mapsto\cos(x)-\cos^2(x)$ has its maximum at $\arctan(\sqrt2)$. $\endgroup$ Commented Jun 21, 2020 at 8:13
  • $\begingroup$ @Michael Hoppe I am unable to understand what you want to say. Could you elaborate? Thanks. $\endgroup$ Commented Jun 21, 2020 at 12:55
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    $\begingroup$ @AryanSonwatikar Sorry, typo. Here's the corrected attempt: We have $f(x)=\sin(2x)+\cos(x)-\cos^3(x)$. Now $\cos(x)-\cos^3(x)$ attains its maximum at $\arctan(\sqrt2)$, its value is $\frac{2}{3\sqrt3}\approx0.3849$, hence $$f(x)\leq1+\cos(\arctan(\sqrt2))-\cos^2(\arctan(\sqrt2)),$$ which is less than $\sqrt2$. $\endgroup$ Commented Jun 21, 2020 at 13:08
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    $\begingroup$ @Michael Hoppe Well, thank you! Could you convert that into an answer so that I can accept it? It's by far the simplest one. I also used the following to complete your attempt without any calculator usage: $$\frac{2}{3\sqrt 3}<\frac{2}{3\times 1.7}=\frac{2}{5.1}<\frac{2}{5}=0.4<\sqrt 2 -1$$ $\endgroup$ Commented Jun 21, 2020 at 17:14
  • $\begingroup$ Done, thank you. $\endgroup$ Commented Jun 21, 2020 at 17:30

8 Answers 8

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Here's a solution based on the remark of Angina Seng. It's quite annoying and as such might not be useful to you, but I found it interesting.

We see that it suffices to show that the polynomial $f(s)=2-s^2(1-s^2)(2+s)^2$ is a sum of squares. We seek a representation as the sum of two squares of integer polynomials.

Note that $f(-2)=f(-1)=f(0)=f(1)=2$, so if $f(s)=P(s)^2+Q(s)^2$ we require that $P(s),Q(s)\in\{-1,1\}$ for each $s\in\{-2,-1,0,1\}$. Since $f$ is a degree $6$ monic polynomial, we must have that one of $\{P,Q\}$ is degree $3$ (say that's $P$) and one is lower degree (say it's $Q$). Since we can find values of $s$ for which $f(s)-1<0$, we can't set $Q$ to be constant at $1$ or $-1$, so $Q$ must be linear or quadratic. However, since $Q(-2),Q(-1),Q(0),Q(1)\in\{-1,1\}$, some value must occur twice as a value of $Q$, so $Q$ can't be linear and must be quadratic. Since it can't switch direction too many times, its values must be $1,-1,-1,1$ at $-2,-1,0,1$ (or the negatives of these), and so we want $$Q(s)=s^2+s-1.$$ Now, we see that $$P(s)^2=s^6+4s^5+2s^4-6s^3-3s^2+2s+1.$$ By looking at the $s^6$, $s^5$, $s$, and $1$ terms, we should have that the polynomial should start with $s^3+2s^2$ and end with $\pm(s+1)$. Ending with $+s+1$ is bad because then $P$ would have all positive coefficients, so we should have $P(s)=s^3+2s^2-s-1$, which does actually square to the above. Thus, we have $$f(s)=(s^3+2s^2-s-1)^2+(s^2+s-1)^2.$$ Now all we need to do is show that $f(s)$ is never actually $0$. This is not too bad; assume that $s^2+s=1$ and $s^3+2s^2=s+1$. Then $$s+1=s^3+2s^2=(s+2)(s^2)=(s+2)(1-s)=-s^2-s+2=1\implies s=0,$$ a contradiction.

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We have $$f(x)=2\sin(x)\cos(x)+\sin^2(x)\cos(x)=\sin(2x)+\cos(x)-\cos^3(x).$$ Now for $0\leq x\leq\pi/2$ we have $0\leq\sin(2x)\leq1$ and the function $x\mapsto\cos(x)-\cos^3(x)$ has its maximum at $\arctan(\sqrt2)$, its value is $$\frac{2}{3\sqrt3}=\frac{2\sqrt3}9<\frac{2\cdot1.8}9=0.4.$$ From here $$\sin(2x)+\cos(x)-\cos^3(x)\leq\sqrt2.$$

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If you want to avoid numerical methods, note that the problem is equivalent to proving $\sin^2x\cos^2x(2+\sin x)^2<2$ on $[0,\pi/2]$. This is the same as proving $s^2(1-s^2)(2+s)^2<2$ on $[0,1]$ where we set $s=\sin s$. You can prove this using rational arithmetic by considering the Sturm sequence of the polynomial $f(s)=2-s^2(1-s^2)(2+s)^2$, but I wouldn't like to do this by hand.

Plotting the graph of $f$ persuades me that $f(s)$ is positive for all real $s$. You may be able to prove this by writing $f(s)$ as a positive linear combination of squares of polynomials.

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    $\begingroup$ I appreciate your help, I'm unacquainted with Sturm sequences and they are very much out of the scope of our syllabus. $\endgroup$ Commented Jun 21, 2020 at 5:57
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Consider the function $g(x)=sin(x)cos(x)$ independently. It can easily seen with calculus that the maximum value it attains is $=1/2$ (at $x=\pi/4+2n\pi$).

So, it immediately follows that $f(x)\leq \frac{1}{2}(2+sinx)=p(x)$, with the equality holding when $x=\pi/4+2n\pi$. Thus it is sufficient to show that $p(x)$ itself is less than $\sqrt2$ in the given interval at the $x$ for which $f(x)$ was maximum (that is, at the roots of the cubic you formed). So, we have - $$\frac{1}{2}(2+t)<\sqrt2$$ $$t<2(\sqrt2-1)$$

Now, we have transformed the question into checking if the root of the cubic in the interval $[0,1]$ is $<2(\sqrt2-1)$.

Let the cubic be $T(t)$. By differentiating once and calculating the roots of the quadratic by the quadratic formula we find that one of its minima/maxima is necessarily $-ve$. Also, we notice that $T(0)$ is $-ve$ and $T(1)$ is $+ve$ . Thus, there are either $1$ or $3$ roots between $0$ and $1$. But, as we said right before one of the minima/maxima of the cubic is $-ve$, and for all $3$ roots to be in $[0,1]$ both the minima/maxima need to be in it as well. Thus, the cubic only has one root for $x\in[0,\frac{\pi}{2}]$.

Now, if the root was $<2(\sqrt2-1)$, $T(2(\sqrt2-1))$ would be $+ve$. This is fairly easy to compute by hand (especially since you only need to know the sign), and what you'll see is that it indeed is true- $T(2(\sqrt2-1))>0$.

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  • $\begingroup$ Nice answer, +1! I made a few small edits if you don't mind; first, as you correctly express later, you're looking for roots of the cubic in the interval $[0, 1]$ and not $[0, \pi/2]$, since $t=\sin(x)$ ranges from $0$ to $1$ as $x$ ranges from $0$ to $\pi/2$. Second, there was a small logic error in the last paragraph; instead of assuming the desired result (that the root is less that $2(\sqrt 2-1)$, you want to assume its negation and show a contradiction. Finally I also changed out the $-ve$ and $+ve$ symbols to the full words, since I think it's much clearer to read. $\endgroup$ Commented Jun 21, 2020 at 16:57
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    $\begingroup$ @Atticus Stonestrom I've changed the first error, that is, my inconsistent use of intervals. I don't think the second one is necessary so I've left it as it is for now. For the last one, I actually had it the way you proposed earlier but changed it because it made the text too wordy. $\endgroup$
    – Amadeus
    Commented Jun 22, 2020 at 2:29
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$f(x) = sin(2x) + \frac{sin(2x)sin(x)}{2} = sin(2x)(1 + \frac{sin(x)}{2})$

Since $x \in [0,\frac{\pi}{2}]$, $sin(x)$ and $sin(2x)$ are both positive.

So, we can apply, AM-GM inequality to say that : $\sqrt(f(x)) <= \frac{(sin(2x) + \frac{sin(x)}{2} + 1)}{2}$

So, now we try to find the max value of $g(x) = sin(2x) + \frac{sin(x)}{2} + 1$

$g'(x) = 2cos(2x) + \frac{cos(x)}{2} = 2(2cos^2(x) - 1) + \frac{cos(x)}{2}$

$g'(t) = 4t^2 - 2 + \frac{t}{2}, t \in [0,1]$

Now, you have a quadratic on the right, whose positive root lies between 0 and 1, let this root be $t_{0}$. You can evaluate it but it involves $\sqrt(129)$

So, $g(t)$ is decreasing from 0 to $t_{0}$ and increasing from $t_{0}$ to 1.

So, $g(x)$ is increasing from $0$ to $cos^{-1}(t_{0})$ and decreasing from $cos^{-1}(t_{0})$ to $0$, so max value occurs at $x = cos^{-1}(t_{0})$.

Unfortunately, at this point, you do need to evaluate $t_{0}$ as $\frac{-1 + \sqrt(129)}{16}$, which can be approximated as 0.62, if you take $\sqrt(129)$ as 11. So, $cos(x) = 0.62$ and $sin(x) = 0.78$.

This makes the RHS of the AM-GM inequality as $\frac{(2*0.62*0.78 + 0.39 + 1)}{2}$, which becomes 1.1786. Square of which is less than 1.414.

The approximations are sufficiently small to not have an impact on the final result, I think since they are in the order of $10^{-2}$.

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    $\begingroup$ $\sqrt(2)$ = 1.414 and $\frac{9}{16}$ < 1...Am I missing something? $\endgroup$ Commented Jun 21, 2020 at 6:39
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    $\begingroup$ I'm sorry, I commented wrong. Actually, the minimum value of $f(x)$ is nearly $1.368$. Which invalidates your solution. But I don't know why it's wrong. $\endgroup$ Commented Jun 21, 2020 at 7:02
  • $\begingroup$ Yeah, I have changed it. It is wrong, because cos(x) takes 0 at $\frac{\pi}{2}$ and 1 at $0$. So, the rules for $t$ change for $x$. Basically, cos is a decreasing function from $0$ to $\frac{\pi}{2}$. $\endgroup$ Commented Jun 21, 2020 at 7:05
  • $\begingroup$ So, if you're going left to right w.r.t to $t$, w.r.t $x$, you'll be going from right to left, since as $x$ decreases, $t = cos(x)$ increases. $\endgroup$ Commented Jun 21, 2020 at 7:13
  • $\begingroup$ Hmmm. The maximum value of $g$ is nearly $1.402$ which is dangerously close to $\sqrt 2$. Also, squaring $1.1786$ by hand to show that it is less than $\sqrt 2$ is too painful without a calculator. $\endgroup$ Commented Jun 21, 2020 at 7:39
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Crazy idea.

We know the $\sin x\cos x$ reaches a maximum $\frac 12$ when $x=\frac {\pi}4$. But $2+\sin x$ is increasing so the maximum value must be for some $x>\frac \pi 4$. And as $\sin x\cos x < \frac 12$ at that later point we can't have a maximum unless $2+\sin x>2\sqrt 2$.

If we solve for $2+\sin x= 2\sqrt 2$ we get that $\sin x =2(\sqrt 2 -1)$. And at this point we have $\sin x\cos x = 2(\sqrt 2-1)(\sqrt{ 1 - 4(\sqrt 2-1)^2})\approx 0.46399943604436523012148335988572$.

But if $x > \arcsin(2(\sqrt 2-1)$ then $\sin x\cos x < 0.46399943604436523012148335988572$. Now the most that $2+\sin x$ can be is $3$ And $0.46399943604436523012148335988572*3 = 1.3919983081330956903644500796571 < 1.414 \approx \sqrt 2$.

We can never make it.

But annoyingly I wouldn't have been able to estimate witthout a caluculator.

If $\sqrt 2\approx 1.4$ then $2(\sqrt 2- 1)\approx 0.8$ so $\cos x \approx \sqrt {1-0.8^2}=\sqrt{1-0.64}=\sqrt{0.36} = 0.6$ and $0.6\cdot 0.8 =0.48$ and that margin of error is enough to through my calculations off as $3*0.48= 1.44 > \sqrt 2$.

Maybe two digit approximation will work but I imagine that too much work for inaccuracy.

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    $\begingroup$ It's very difficult to get that $0.259$ without a calculator, so do you have anything else? Appreciate it, anyway. $\endgroup$ Commented Jun 21, 2020 at 7:46
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    $\begingroup$ I was wrong in the calculation but you can estimate it as $\sqrt 2\approx 1.4$ so $2(\sqrt 2-1)\approx .8$ so $\cos x approx \sqrt{1-0.64}=\sqrt{0.36}=0.6$ as $\sin x \cos x \approx 0.48$ and ... that margin of error is another to push $0.48\times 3=1.5-0.06=1.44>\sqrt 2$. In actuality $\cos(\arcsin (2(\sqrt 2-1)))\sin(\arcsin(2(\sqrt 2-1))) = 0.464$ and times $3$ is $1.319$ so my reasoning if valid but not really possible without a calculator. $\endgroup$
    – fleablood
    Commented Jun 21, 2020 at 15:38
  • $\begingroup$ Thank you nonetheless, appreciate the time and effort. $\endgroup$ Commented Jun 21, 2020 at 17:00
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You can check that the only positive, hence feasible, root of the cubic is at $\sin x=t\approx0.748$ (I got this using Mathematica, but this is achievable rigorously using e.g. the cubic formula). At that value of $x$, we then have $\cos x\approx0.663$, so we bound $$f(x)=\sin x\cdot \cos x\cdot(2+\sin x)<0.75\cdot0.665\cdot2.75<1.4<\sqrt{2}$$ and since this is the maximum the desired result is true.

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    $\begingroup$ Uhh, solving the cubic isn't very feasible by hand, but thanks anyway. $\endgroup$ Commented Jun 21, 2020 at 5:24
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$$\sin x\cos x(2+\sin x)=\frac12\sin 2x \cdot (2+\sin x)=\sin 2x \cdot \left(1+\frac12\sin x\right)$$ $0\le\sin2x\le 1$, $\ \ 0\le\sin x\le 1$ for $x\in[0, \pi/2]$

$$0\le\sin (2x)\left(1+\frac12\sin x\right)<\sqrt2$$

$\therefore \sin x\cos x(2+\sin x)=\sqrt2$ has no solutions.

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    $\begingroup$ Why do we have $0\le\sin2x\le \frac1{\sqrt2}$? $\sin\left(2\frac{\pi}{4}\right)=1>\frac{\sqrt{2}}{2}$ $\endgroup$ Commented Jun 21, 2020 at 5:25
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    $\begingroup$ Uhh, I'm stuck on proving that it is less than $\sqrt 2$ as outlined in my method. $\endgroup$ Commented Jun 21, 2020 at 5:49
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    $\begingroup$ This answer is inadequate. The bounds $|\sin(2x)|\leq 1$ and $|\sin(x)|\leq 1$ are enough to show only that $\sin(2x)(1+\frac{1}{2}\sin(x))\leq\frac{3}{2}$, which does not prove the desired result. $\endgroup$ Commented Jun 21, 2020 at 5:52

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