# For all real $\theta$ prove that $\cos(\sin\theta) \gt \sin(\cos\theta)$

How do I prove this? Im not able to even start it. Help please!

The desired estimate holds at $\theta=0$ and we have \begin{eqnarray} \cos(\sin\theta)-\sin(\cos\theta) &=& \cos(\sin\theta)-\cos\left(\frac\pi2-\cos\theta\right) \\&=& 2\sin\left(\frac12\left(\frac\pi2-\cos\theta-\sin\theta\right)\right)\sin\left(\frac12\left(\frac\pi2-\cos\theta+\sin\theta\right)\right) \end{eqnarray} so it suffices to show that $\frac\pi2-\cos\theta\pm\sin\theta\neq0$ for all $\theta$ for both signs.

Let $f_\pm(x)=\frac\pi2-\cos x\pm\sin x$. Now $f'_\pm(x)=\sin x\pm\cos x$, so at extremal points of $f_\pm$ we have $\sin x=\mp\cos x$. This equation can be solved, and the corresponding values give the minimum value $\frac\pi2-\sqrt2>0$ for both functions. Since $f_\pm(0)>0$, we have $f_\pm(x)>0$ for all $x$.

• shouldn't that be $\frac \pi2 \pm cos\theta - sin \theta$ – Shubham Sep 28 '14 at 13:42
• @Shubham, no, since two signs were wrong on the second line of the long equation. I used $\cos x-\cos y=2\sin(\frac12(y-x))\sin(\frac12(y+x))$. The signs should be correct now. – Joonas Ilmavirta Sep 28 '14 at 13:44
• Understood. Thanks for the explanation. – Shubham Sep 28 '14 at 14:08
• It seems to me that what you want to show is not ${\pi\over2}-\cos\theta\pm\sin\theta\not=0$, but rather $|\cos\theta\pm\sin\theta|\le\sqrt2\lt{\pi\over2}$. – Barry Cipra Oct 8 '14 at 12:24
• @BarryCipra, true. But the two claims are essentially the same (for my purposes here) and I thought that $\frac\pi2-\cos\theta\pm\sin\theta\neq0$ relates more obviously to my first calculation for $\cos(\sin\theta)-\sin(\cos\theta)$. – Joonas Ilmavirta Oct 8 '14 at 12:31

Try graphing both of the functions. \begin{align} &\cos (\sin x) - \sin (\cos x) > 0\\ \implies &\cos (\sin x) - \cos ( π/2 - \cos x ) > 0\\ \implies &2 \sin \left[ \frac{\pi}{4} + \frac{1}{2}(\sin x - \cos x) \right]\cdot \sin \left[ \frac{\pi}{4} - \frac{1}{2}(\sin x - \cos x) \right] > 0 \tag{1} \end{align} If we could prove that both the factors on the left hand side of $(1)$ are positive then the result obtained above $(1)$ is proved. Since: $$\left| \sin x - \cos x \right| = \left| √2 \sin (x- \frac{\pi}{4}) \right| ≤ √2 < \frac{\pi}{2}$$ We have, $$- \frac{\pi}{2} < ( \sin x - \cos x ) < \frac{\pi}{2}\\ \implies - \frac{\pi}{4} < ( \sin x - \cos x )/2 < \frac{\pi}{4}$$ So that, $$0 < \frac{\pi}{4}+ \frac{1}{2}( \sin x - \cos x ) < \frac{\pi}{2}$$ $\therefore \space \sin [ \frac{\pi}{4} + \frac{1}{2} (\sin x - \cos x) ] > 0 \quad\text{(ie, Positive)}$
Similarly we can prove that $\sin [ \frac{\pi}{4} - \frac{1}{2}(\sin x - \cos x) ] > 0$
Hence $(1)$ is true. QED

• Good solution. Here is one thing I didn't get : $|sinx - cosx| = |√2 sin (x-π/4)| ≤ √2 < π/2$ How $\sqrt 2 \lt \pi/2$ ? – Shubham Sep 28 '14 at 13:19
• Graphing does not prove an inequality. Linking with no explanation is usually frowned upon around here. – Teepeemm Sep 28 '14 at 13:22
• Don't be harsh guys. This is daleaf's first answer. – Kim Jong Un Sep 28 '14 at 13:41
• @KimJongUn, true, it's a first answer. But even first answers should meet some criteria. I do expect a new user to edit their answer when they learn that that the answer does not qualify as such. In this case the edit should explain what is behind that link. – Joonas Ilmavirta Sep 28 '14 at 13:56
• @daleaf.M: Buddy, welcome to MSE. Please read the FAQ and help section. Try to improve the quality of answers and questions you post as well as try to interact with our fellow community members in a respectable fashion. – Nick Sep 28 '14 at 14:21

Let $\theta$ be in the first or fourth quadrant and $\sin\theta=t$, so that $\cos\theta=\sqrt{1-t^2}$.

Now in the range $-1\le t\le1$ the inequality is rewritten as $$\cos t\gt\sin\sqrt{1-t^2}.$$ Taking the $arcsin$, $$\frac\pi2-t>\sqrt{1-t^2},$$or $$(\frac\pi2-t)^2>1-t^2.$$ The discriminant value proves that the two parabolas do not intersect.

In the second and third quadrants, the inequality trivially holds because of signs.

• Wow! I was wondering how can I use inverse trigo to solve this. You did a beautiful job. So the inequality is true only for 2nd and 3rd quadrants right? – Shubham Sep 28 '14 at 15:54
• No, it's true everywhere. Needs careful discussion in Q1/Q4 and obvious in Q2/Q3. Look at the plot by @daleaf.M. – Yves Daoust Sep 28 '14 at 15:55
• Yes you right. Thanks! :) – Shubham Sep 28 '14 at 15:59

The inequality $\cos(\sin\theta)\gt\sin(\cos\theta)$ for all $\theta$ really boils down to the inequality $x\gt\sin x$ for $x\gt0$ and the fact that $\cos x$ is decreasing on the interval $0\le x\le\pi$. I'll take those as known. (They are easy enough to prove.)

Let's start by observing that we need only verify the inequality $\cos(\sin\theta)\gt\sin(\cos\theta)$ for $0\lt\theta\lt\pi/2$:

By the periodicity of sine and cosine, it suffices to prove the inequality for $-\pi\le\theta\le\pi$. Since cosine is an even function (and sine is odd), it suffices to prove it for $0\le\theta\le\pi$. The inequality is clearly satisfied on since $\pi/2\lt\theta\le\pi$, $\sin(\cos\theta)$ is negative there while $\cos(\sin\theta)$ is positive. Finally, it's easily checked for $\theta=0$ and $\pi/2$. So that leaves $0\lt\theta\lt\pi/2$

Since we're now on an inteval where $\cos\theta\gt0$, we can let $x=\cos\theta$ and get $\cos\theta\gt\sin(\cos\theta)$. But we also have $\pi\gt\theta\gt\sin\theta\gt0$ on this interval, which, by the fact that the cosine function in decreasing there, gives us $\cos(\sin\theta)\gt\cos\theta$. Putting it all together gives the desired inequality.

$\cos(\sin\theta)=\sin(\cos\theta))=\cos(\pi/2-\cos\theta)$ implies $0=\cos(\sin\theta)-\cos(\pi/2-\cos\theta)=-2\sin(\frac{\sin\theta+\pi/2-\cos\theta}{2})\sin(\frac{\sin\theta-\pi/2+\cos\theta}{2})$.

So this can only happen if $\sin(\frac{\sin\theta\pm(\pi/2-\cos\theta)}{2})=0$, and the latter can only happen if $\sin\theta\pm(\pi/2-\cos\theta)=0$. To rule the latter out, investigate the minima and the maxima of $f(t):=\sin t+\cos t$ (resp. $f(t)=\sin t-\cos t$); they are reached for $f'(t)=0=\cos t\pm\sin t$, and so $|f(t)|\leq\frac{\sqrt{2}}{2}<0.7072$. Thus you never have $\sin\theta\pm(\pi/2-\cos\theta)=0$. Therefore the 1st equality never happens. It suffices now to check the claimed inequality for $\theta=0$, which is trivial. QED.

The functions appearing on both sides of the inequality are $2\pi$-periodic. Therefore, it suffices to prove the inequality on the interval $[-\pi,\pi]$. Moreover, since both functions are even, we can in fact reduce this to $[0,\pi]$.

Assume $\theta \in [0,\pi]$. Then the inequality to be proved is equivalent to $$\sin(\pi/2 - \sin \theta) > \sin(\cos \theta).$$ But the sine function is strictly increasing on the interval $[-\pi/2,\pi/2]$, and both numbers $\cos \theta$ and $\pi/2 - \sin \theta$ belong to this interval. Thus the inequality to be proved is in turn equivalent to $$\pi/2 - \sin \theta > \cos \theta,$$ which follows immediately from the equality $\sin \theta + \cos \theta = \sqrt{2} \sin (\theta + \pi/4)$ and the fact that $\sqrt{2} < \pi/2$.

Let $$f(\theta)=\cos(\sin \theta)-\sin(\cos\theta).$$ Then clearly $f$ is continuous on $\mathbb{R}.$ Note that $f$ is even and $$f(0)=1-\sin(1)>0$$ Now according to the intermediate value theorem it is enough to show that $f(\theta)\not=0$ for all $\theta>0.$
Suppose $f(\theta)=0$ for some $\theta>0.$
Then $$\cos(\sin\theta)=\sin(\cos\theta)$$ $$\cos(\sin\theta)=\cos(\pi/2-\cos\theta)$$ $$\pi/2-\sin\theta=2n\pi±\cos\theta$$ for some $n\in \mathbb{Z}.$ $$\sin\theta±\cos\theta=2n\pi+\pi/2$$ $$\sin(\theta±\pi/4)=2\sqrt2n\pi+\pi/\sqrt2>1\,\,\,\,\,\text{or}\,\,\,\,\,\lt-1.$$ for all $n\in \mathbb{Z}.$ This is s contradiction.
Therefore $f(\theta)>0$ for all $\theta \in \mathbb{R}.$

• Where do you get your first equation? Shouldn't you have $\cos(\sin\theta)=\sin(\pi/2-\sin\theta)$? – Joonas Ilmavirta Sep 28 '14 at 13:18