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!
 A: Try graphing both of the functions.


Also, this answer may help:


$$
\begin{align}
&\cos (\sin x) - \sin (\cos x) > 0\\ 
\implies &\cos (\sin x) - \cos ( π/2 - \cos x ) > 0\\ 
\implies &2 \sin \left[ \frac{\pi}{4} + \frac{1}{2}(\sin x - \cos x) \right]\cdot \sin \left[ \frac{\pi}{4} - \frac{1}{2}(\sin x - \cos x) \right] > 0 \tag{1} 
\end{align}
$$
  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 

Reference: Yahoo Answers
A: 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 $x$.
A: 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.
A: 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.
A: $\cos(\sin\theta)=\sin(\cos\theta))=\cos(\pi/2-\cos\theta)$ 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.
A: 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 $\sqrt{2} < \pi/2$.
A: 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 $\theta \in \mathbb{R}.$
