MHB Is $\sin^2 x$ Less Than $\sin x^2$ for $0 \leq x \leq \sqrt{\frac{\pi}{2}}$?

[anemone]

Hi MHB,

When I first saw the problem (Prove that $\sin^2 x<\sin x^2$ for $0\le x\le \sqrt{\dfrac{\pi}{2}}$), I could tell that is one very good problem, but, a good problem usually indicates it is also a very difficult problem and after a few trials using calculus + trigonometry method, I failed and the only conclusion that I could draw by now is one cannot use calculus route in this problem. (But I could be wrong.)

So, I decided to post it here and I would appreciate it if you could tell me what method you would use in your attempt, thanks so much in advance.

Edit:

Oops..now that I redo the problem and I noticed I could use some trick in the calculus method that I overlooked before...so, I finally proved it! Yeah! I will post my solution later.

But, I still hope to see how others would attempt at the problem and I would deeply appreciate it if you could share your solution with me too!

 

[MarkFL]

You originally gave me the weak inequality:

$$\sin^2(x)\le\sin\left(x^2\right)$$ where $$0\le x\le\sqrt{\frac{\pi}{2}}$$

Indeed, we see that the inequality must be weak because of equality at $x=0$. Now, at the other end-point, we find:

$$\sin^2\left(\sqrt{\frac{\pi}{2}}\right)<\sin\left(\frac{\pi}{2}\right)=1$$

Using a numeric technique, we find the smallest positive root of:

$$f(x)=\sin^2(x)-\sin\left(x^2\right)=0$$

is:

$$x\approx1.36441296312529>\sqrt{\frac{\pi}{2}}$$

Given that both functions in the original inequality are monotonic inside the given interval, we may then conclude the inequality is true.

 

[Nono713]

You originally gave me the weak inequality:

$$\sin^2(x)\le\sin\left(x^2\right)$$ where $$0\le x\le\sqrt{\frac{\pi}{2}}$$

Indeed, we see that the inequality must be weak because of equality at $x=0$. Now, at the other end-point, we find:

$$\sin^2\left(\sqrt{\frac{\pi}{2}}\right)<\sin\left(\frac{\pi}{2}\right)=1$$

Using a numeric technique, we find the smallest positive root of:

$$f(x)=\sin^2(x)-\sin\left(x^2\right)=0$$

is:

$$x\approx1.36441296312529>\sqrt{\frac{\pi}{2}}$$

Given that both functions in the original inequality are monotonic inside the given interval, we may then conclude the inequality is true.

I am not sure this follows: two functions being monotonic on the same interval doesn't mean they intersect at most once, or twice. For instance, $f(x) = x$ and $g(x) = x + \frac{1}{2} \sin x$ are both monotonic yet intersect infinitely many times. I think concavity is required, or am I missing something? Thanks!

 

[MarkFL]

I agree that monotonicity is not relevant here...given that the two functions are equal at $x=0$, and have no other points of intersection in the interval, and given that:

$$\sin^2\left(\sqrt{\frac{\pi}{2}}\right)<\sin\left(\frac{\pi}{2}\right)=1$$

the result then follows.

 

[anemone]

... I noticed I could use some trick in the calculus method that I overlooked before...so, I finally proved it! Yeah! I will post my solution later.

Hello again, MHB!

When I thought I've solved the problem and hence I edited the original post to mention so, I didn't, it only dawned on me hours ago that my solution is flawed.:(

You originally gave me the weak inequality:

$$\sin^2(x)\le\sin\left(x^2\right)$$ where $$0\le x\le\sqrt{\frac{\pi}{2}}$$

Indeed, we see that the inequality must be weak because of equality at $x=0$. Now, at the other end-point, we find:

$$\sin^2\left(\sqrt{\frac{\pi}{2}}\right)<\sin\left(\frac{\pi}{2}\right)=1$$

Using a numeric technique, we find the smallest positive root of:

$$f(x)=\sin^2(x)-\sin\left(x^2\right)=0$$

is:

$$x\approx1.36441296312529>\sqrt{\frac{\pi}{2}}$$

Given that both functions in the original inequality are monotonic inside the given interval, we may then conclude the inequality is true.

Thanks MarkFL for your reply, I suppose you're right, if we have proved the smallest positive root of $$f(x)=\sin^2(x)-\sin\left(x^2\right)=0$$ is $$x\approx1.36441296312529>\sqrt{\frac{\pi}{2}}$$, then the inequality holds true in that given interval.(Yes)

And yes, the problem should be a weak inequality, I will edit my original post. Thanks for catching this!

 

[anemone]

Hi MHB,

I was told by someone his way of proving this particular inequality to be right for that given interval and I thought it would be nice if I share it here. After all, my philosophy is I want to share all that I can give!

[TABLE="class: grid, width: 800"]
[TR]
[TD]For $x=0$,
we see that $\sin^2 0=\sin 0^2=0$.

[/TD]
[TD]For $0<x< 1$, if we can show that the derivative of $\sin x^2$ is greater than the derivative of $\sin^2 x$ for that interval, then, of course $\sin x^2>\sin^2 x$.

$\dfrac{d}{dx}(\sin x^2)=\color{yellow}\bbox[5px,green]{2x}\color{yellow}\bbox[5px,red]{\cos x^2}$

$\dfrac{d}{dx}(\sin^2 x)= \color{yellow}\bbox[5px,green]{2\sin x}\color{yellow}\bbox[5px,red]{\cos x}$

It's well known that $x>\sin x$ in $\left(0,\,\dfrac{\pi}{2}\right)$, therefore, we have that $\color{yellow}\bbox[5px,green]{2x}\color{black}{>}\color{yellow}\bbox[5px,green]{2\sin x}$.

It's also obvious that $\color{yellow}\bbox[5px,red]{\cos x^2}\color{black}{>}\color{yellow}\bbox[5px,red]{\cos x}$ for $0<x< 1$.

Hence, we can say that $\sin x^2>\sin^2 x$ in $0<x< 1$.[/TD]
[TD]For $1<x<\sqrt{\dfrac{\pi}{2}}$, it's easy to see that $\sin x^2>\sin^2 x$.[/TD]
[/TR]
[/TABLE]

Therefore, we can conclude that $\sin x^2\ge\sin^2 x$ in $1<x<\sqrt{\dfrac{\pi}{2}}$.