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

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.

• 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 $\arctan(\sqrt2)$. – Michael Hoppe Jun 21 at 8:13
• @Michael Hoppe I am unable to understand what you want to say. Could you elaborate? Thanks. – AryanSonwatikar Jun 21 at 12:55
• @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 $\sqrt2$. – Michael Hoppe Jun 21 at 13:08
• @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: $$\frac{2}{3\sqrt 3}<\frac{2}{3\times 1.7}=\frac{2}{5.1}<\frac{2}{5}=0.4<\sqrt 2 -1$$ – AryanSonwatikar Jun 21 at 17:14
• Done, thank you. – Michael Hoppe Jun 21 at 17:30

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 $$\sin(2x)+\cos(x)-\cos^3(x)\leq\sqrt2.$$

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.

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.

• I appreciate your help, I'm unacquainted with Sturm sequences and they are very much out of the scope of our syllabus. – AryanSonwatikar Jun 21 at 5:57

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- $$T(2(\sqrt2-1))>0$$.

• 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. – Atticus Stonestrom Jun 21 at 16:57
• @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. – l1mbo Jun 22 at 2:29

$$f(x) = sin(2x) + \frac{sin(2x)sin(x)}{2} = sin(2x)(1 + \frac{sin(x)}{2})$$

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 $$10^{-2}$$.

• $\sqrt(2)$ = 1.414 and $\frac{9}{16}$ < 1...Am I missing something? – user3052655 Jun 21 at 6:39
• 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. – AryanSonwatikar Jun 21 at 7:02
• 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 $\frac{\pi}{2}$. – user3052655 Jun 21 at 7:05
• 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. – user3052655 Jun 21 at 7:13
• 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. – AryanSonwatikar Jun 21 at 7:39

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.

• It's very difficult to get that $0.259$ without a calculator, so do you have anything else? Appreciate it, anyway. – AryanSonwatikar Jun 21 at 7:46
• 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. – fleablood Jun 21 at 15:38
• Thank you nonetheless, appreciate the time and effort. – AryanSonwatikar Jun 21 at 17:00

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.

• Uhh, solving the cubic isn't very feasible by hand, but thanks anyway. – AryanSonwatikar Jun 21 at 5:24