Let x be an element of R. Find which term has a larger value, sin(cos x) or cos(sin x)?
Any help or detailed working out would be appreciated!
Here is another view.
$\displaystyle sinx$ ranges from $\displaystyle -1\rightarrow\ 1$
$\displaystyle cos(sinx)$ therefore ranges from $\displaystyle cos(-1)=cos(1)=0.54\rightarrow\ cos(0)=1$
$\displaystyle cosx$ ranges from $\displaystyle -1\rightarrow\ 1$
$\displaystyle sin(cosx)$ therefore ranges from $\displaystyle sin(-1)=-0.84\rightarrow\ sin(1)=0.84$
A little more analysis will show they are never equal.
Here is a way we can develop the proof...
$\displaystyle cos(sinx)>0$
hence we need not be concerned with $\displaystyle sin(cosx)<0$
and the maximum value of $\displaystyle sin(cosx)$ is <1.
We only need consider both functions >0 which is in quadrant 1, where $\displaystyle 0<x<\frac{\pi}{2}$
In quadrant 1, $\displaystyle \displaystyle\ sin(cosx)=cos\left(\frac{\pi}{2}-cosx\right)$
Hence we are examining $\displaystyle \displaystyle\ cos(sinx)-cos\left(\frac{\pi}{2}-cosx\right)$
so we want to know if $\displaystyle sinx<\left[\frac{\pi}{2}-cosx\right]$
since in quadrant 1, $\displaystyle cos(angle)>cos(larger\;angle)$
$\displaystyle \frac{\pi}{2}>sinx+cosx$ ?
Hence we find the maximum value of $\displaystyle sinx+cosx$ by differentiating...
$\displaystyle cosx-sinx=0\Rightarrow\ cosx=sinx$ which happens at $\displaystyle x=\frac{\pi}{4}$ radians.
The maximum value of $\displaystyle sinx+cosx=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\f rac{2}{\sqrt{2}}=\sqrt{2}$
$\displaystyle \frac{\pi}{2}>\sqrt{2}\Rightarrow\ cos(sinx)>sin(cosx)$
The graph of the two functions $\displaystyle sin(cosx)$ and $\displaystyle cos(sinx)$ or $\displaystyle cos(sinx)-sin(cosx)$ are a nice guide.
If we are presented with just the 2 functions, $\displaystyle cos(sinx)$ and $\displaystyle sin(cosx)$
and we want to compare them, how do we answer "what stands out ?"
This tread is getting old, but Mr. Ryan still seemed confused.
This may be a Lemma to BobP's theorem.
I hope you can agree that: If two functions, f(x) and g(x) are continuous, and if they are not equal for any x, then one of them is greater than the other for all x.
In this case let's see if we can solve:
$\displaystyle \sin(\cos(x))=\cos(\sin(x))$.
$\displaystyle \cos(\theta)=\sin(\theta+{\pi\over2})$,
so $\displaystyle \sin(\cos(x))=\sin(\sin(x)+{\pi\over2})$.
Then $\displaystyle \cos(x)=\sin(x)+{\pi\over2}$. (O.K., you could add integer multiple of $\displaystyle 2\pi$ to this, but that won't help.)
$\displaystyle \cos(x)-\sin(x)={\pi\over2}$
Use the identity: $\displaystyle \cos(x)-\sin(x)=\sqrt{2}\cos(x+{\pi\over4})$.
This gives us $\displaystyle \sqrt{2}\cos(x+{\pi\over4})={\pi\over2}$
The maximum value of the cosine is 1, and $\displaystyle \sqrt{2}<{\pi\over2}$, so $\displaystyle \sin(\cos(x))=\cos(\sin(x))$ has no solution.
Since $\displaystyle \sin(\cos(\pi))=\sin(-1)\approx -0.8415$ (It's definitely negative!)
and $\displaystyle \cos(\sin(\pi))=\cos(0)=1$,
it must be that $\displaystyle \sin(\cos(x))<\cos(\sin(x))$, for all x.
Maybe simplest and without reference to graphs is:
$\displaystyle cos(A+B)=cosAcosB-sinAsinB\Rightarrow\ cos\left(\frac{\pi}{2}-A\right)=sinA$
$\displaystyle cos(sinx)-sin(cosx)=cos(sinx)-cos\left(\frac{\pi}{2}-cosx\right)$
This is the difference of cosines.
$\displaystyle cosA-cosB=-2sin\left[\frac{A+B}{2}\right]sin\left[\frac{A-B}{2}\right]$
$\displaystyle \Rightarrow\ cos(sinx)-sin(cosx)=-2sin\left[\frac{sinx-cosx+\frac{\pi}{2}}{2}\right]sin\left[\frac{sinx+cosx-\frac{\pi}{2}}{2}\right]$
One could go further with trigonometry here.
An alternative is to examine the contents of the square brackets on the previous line.
We can examine the maximum and minimum values of these expressions.
$\displaystyle \frac{d}{dx}(sinx-cosx)=cosx+sinx=0\Rightarrow\ cosx=-sinx$
Referring to the unit-circle, this occurs when $\displaystyle x=\frac{3{\pi}}{4},\;\;x=\frac{7{\pi}}{4}$
(1) maximum value within the first square brackets
$\displaystyle x=\frac{3{\pi}}{4}\Rightarrow\frac{\frac{1}{\sqrt{ 2}}+\frac{1}{\sqrt{2}}+\frac{\pi}{2}}{2}=\frac{1}{ \sqrt{2}}+\frac{\pi}{4}$
$\displaystyle 0<\frac{1}{\sqrt{2}}+\frac{\pi}{4}<\frac{\pi}{2}\R ightarrow\ sin\left[\frac{sinx-cosx+\frac{\pi}{2}}{2}\right]>0$
(2) minimum value within the first square brackets
$\displaystyle x=\frac{7{\pi}}{4}\Rightarrow\frac{-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+\frac{\pi}{2}}{2}=-\frac{1}{\sqrt{2}}+\frac{\pi}{4}$
$\displaystyle 0<-\frac{1}{\sqrt{2}}+\frac{\pi}{4}<\frac{\pi}{2}\Rig htarrow\ sin\left[\frac{sinx-cosx+\frac{\pi}{2}}{2}\right]>0$
$\displaystyle \frac{d}{dx}(sinx+cosx)=cosx-sinx=0\Rightarrow\ cosx=sinx\Rightarrow\ x=\frac{\pi}{4},\;\;x=\frac{5{\pi}}{4}$
(3) maximum value within the second square brackets
$\displaystyle x=\frac{\pi}{4}\Rightarrow\frac{\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}}-\frac{\pi}{2}}{2}=\frac{1}{\sqrt{2}}-\frac{\pi}{4}$
$\displaystyle -\frac{\pi}{2}<\frac{1}{\sqrt{2}}-\frac{\pi}{4}<0\Rightarrow\ sin\left[\frac{sinx+cosx-\frac{\pi}{2}}{2}\right]<0$
(4) minimum value within the second square brackets
$\displaystyle x=\frac{5\pi}{4}\Rightarrow\frac{-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{\pi}{2}}{2}=-\frac{1}{\sqrt{2}}-\frac{\pi}{4}$
$\displaystyle -\frac{\pi}{2}<-\frac{1}{\sqrt{2}}-\frac{\pi}{4}<0\Rightarrow\ sin\left[\frac{cosx+sinx-\frac{\pi}{2}}{2}\right]<0$
Therefore, it follows that
$\displaystyle cos(sinx)-sin(cosx)>0\Rightarrow\ cos(sinx)>sin(cosx)$