Which is larger?

• November 28th 2010, 07:19 PM
Mr Rayon
Which is larger?
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!
• November 30th 2010, 05:29 AM
Sudharaka
Quote:

Originally Posted by Mr Rayon
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!

Dear Mr Rayon,

Please refer the attached file. The blue line represents y=sin(cosx) while the green curve represents y=cos(sinx). From this graph it is clear that sin(cosx)<cos(sinx). But for the moment I dont have any idea of how prove this result.
• November 30th 2010, 07:04 AM
Quote:

Originally Posted by Mr Rayon
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.

$sinx$ ranges from $-1\rightarrow\ 1$

$cos(sinx)$ therefore ranges from $cos(-1)=cos(1)=0.54\rightarrow\ cos(0)=1$

$cosx$ ranges from $-1\rightarrow\ 1$

$sin(cosx)$ therefore ranges from $sin(-1)=-0.84\rightarrow\ sin(1)=0.84$

A little more analysis will show they are never equal.
• November 30th 2010, 03:41 PM
Sudharaka
Quote:

Here is another view.

$sinx$ ranges from $-1\rightarrow\ 1$

$cos(sinx)$ therefore ranges from $cos(-1)=cos(1)=0.54\rightarrow\ cos(0)=1$

$cosx$ ranges from $-1\rightarrow\ 1$

$sin(cosx)$ therefore ranges from $sin(-1)=-0.84\rightarrow\ sin(1)=0.84$

A little more analysis will show they are never equal.

Just out of curiosity, with what software did you create your images???
• November 30th 2010, 03:49 PM
Quote:

Originally Posted by Sudharaka

Just out of curiosity, with what software did you create your images???

Hi Sudharaka,

I only use the "pages" application in Apple iWorks!
in conjunction with a Mathtype program (very basic).

AM
• December 1st 2010, 12:42 AM
BobP
Quote:

Originally Posted by Mr Rayon
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!

Both functions are continuous and you can't find a value of x for which they are equal.
So, if one is bigger than the other for a particular value of x, it will be bigger for all values of x.
• December 1st 2010, 04:39 AM
Here is a way we can develop the proof...

$cos(sinx)>0$

hence we need not be concerned with $sin(cosx)<0$

and the maximum value of $sin(cosx)$ is <1.

We only need consider both functions >0 which is in quadrant 1, where $0

In quadrant 1, $\displaystyle\ sin(cosx)=cos\left(\frac{\pi}{2}-cosx\right)$

Hence we are examining $\displaystyle\ cos(sinx)-cos\left(\frac{\pi}{2}-cosx\right)$

so we want to know if $sinx<\left[\frac{\pi}{2}-cosx\right]$

since in quadrant 1, $cos(angle)>cos(larger\;angle)$

$\frac{\pi}{2}>sinx+cosx$ ?

Hence we find the maximum value of $sinx+cosx$ by differentiating...

$cosx-sinx=0\Rightarrow\ cosx=sinx$ which happens at $x=\frac{\pi}{4}$ radians.

The maximum value of $sinx+cosx=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\f rac{2}{\sqrt{2}}=\sqrt{2}$

$\frac{\pi}{2}>\sqrt{2}\Rightarrow\ cos(sinx)>sin(cosx)$
• December 2nd 2010, 04:49 AM
Sudharaka
Quote:

Originally Posted by BobP
Both functions are continuous and you can't find a value of x for which they are equal.
So, if one is bigger than the other for a particular value of x, it will be bigger for all values of x.

Hi BobP,

Is this a theorem? Can you please tell me where you have encountered it?
• December 4th 2010, 12:52 AM
BobP
Quote:

Originally Posted by Sudharaka
Hi BobP,

Is this a theorem? Can you please tell me where you have encountered it?

If it isn't a theorem I'm happy to claim it !

Actually, I thought it was just common sense (Happy).
• December 4th 2010, 06:38 AM
Sudharaka
Hi everyone,

Below is a proof that I have formulated for the theorem BobP had given in post #6. If you find any mistake in it please do not hesitate to tell me. Using this result the above problem could be easily solved.
• December 4th 2010, 07:14 AM
The graph of the two functions $sin(cosx)$ and $cos(sinx)$ or $cos(sinx)-sin(cosx)$ are a nice guide.

If we are presented with just the 2 functions, $cos(sinx)$ and $sin(cosx)$

and we want to compare them, how do we answer "what stands out ?"
• December 4th 2010, 04:40 PM
Sudharaka
Quote:

The graph of the two functions $sin(cosx)$ and $cos(sinx)$ or $cos(sinx)-sin(cosx)$ are a nice guide.

If we are presented with just the 2 functions, $cos(sinx)$ and $sin(cosx)$

and we want to compare them, how do we answer "what stands out ?"

I dont quite understand what you meant by saying ".......how do we answer what stands out?" Can you please explain this.
• December 4th 2010, 07:57 PM
SammyS
Which is larger?
Quote:

Originally Posted by BobP
Both functions are continuous and you can't find a value of x for which they are equal.
So, if one is bigger than the other for a particular value of x, it will be bigger for all values of x.

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:

$\sin(\cos(x))=\cos(\sin(x))$.

$\cos(\theta)=\sin(\theta+{\pi\over2})$,

so $\sin(\cos(x))=\sin(\sin(x)+{\pi\over2})$.

Then $\cos(x)=\sin(x)+{\pi\over2}$. (O.K., you could add integer multiple of $2\pi$ to this, but that won't help.)

$\cos(x)-\sin(x)={\pi\over2}$

Use the identity: $\cos(x)-\sin(x)=\sqrt{2}\cos(x+{\pi\over4})$.

This gives us $\sqrt{2}\cos(x+{\pi\over4})={\pi\over2}$

The maximum value of the cosine is 1, and $\sqrt{2}<{\pi\over2}$, so $\sin(\cos(x))=\cos(\sin(x))$ has no solution.

Since $\sin(\cos(\pi))=\sin(-1)\approx -0.8415$ (It's definitely negative!)

and $\cos(\sin(\pi))=\cos(0)=1$,

it must be that $\sin(\cos(x))<\cos(\sin(x))$, for all x.

• December 31st 2010, 05:52 PM
Maybe simplest and without reference to graphs is:

$cos(A+B)=cosAcosB-sinAsinB\Rightarrow\ cos\left(\frac{\pi}{2}-A\right)=sinA$

$cos(sinx)-sin(cosx)=cos(sinx)-cos\left(\frac{\pi}{2}-cosx\right)$

This is the difference of cosines.

$cosA-cosB=-2sin\left[\frac{A+B}{2}\right]sin\left[\frac{A-B}{2}\right]$

$\Rightarrow\ cos(sinx)-sin(cosx)=-2sin\left[\frac{sinx-cosx+\frac{\pi}{2}}{2}\right]sin\left[\frac{sinx+cosx-\frac{\pi}{2}}{2}\right]$

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.

$\frac{d}{dx}(sinx-cosx)=cosx+sinx=0\Rightarrow\ cosx=-sinx$

Referring to the unit-circle, this occurs when $x=\frac{3{\pi}}{4},\;\;x=\frac{7{\pi}}{4}$

(1) maximum value within the first square brackets

$x=\frac{3{\pi}}{4}\Rightarrow\frac{\frac{1}{\sqrt{ 2}}+\frac{1}{\sqrt{2}}+\frac{\pi}{2}}{2}=\frac{1}{ \sqrt{2}}+\frac{\pi}{4}$

$0<\frac{1}{\sqrt{2}}+\frac{\pi}{4}<\frac{\pi}{2}\R ightarrow\ sin\left[\frac{sinx-cosx+\frac{\pi}{2}}{2}\right]>0$

(2) minimum value within the first square brackets

$x=\frac{7{\pi}}{4}\Rightarrow\frac{-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+\frac{\pi}{2}}{2}=-\frac{1}{\sqrt{2}}+\frac{\pi}{4}$

$0<-\frac{1}{\sqrt{2}}+\frac{\pi}{4}<\frac{\pi}{2}\Rig htarrow\ sin\left[\frac{sinx-cosx+\frac{\pi}{2}}{2}\right]>0$

$\frac{d}{dx}(sinx+cosx)=cosx-sinx=0\Rightarrow\ cosx=sinx\Rightarrow\ x=\frac{\pi}{4},\;\;x=\frac{5{\pi}}{4}$

(3) maximum value within the second square brackets

$x=\frac{\pi}{4}\Rightarrow\frac{\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}}-\frac{\pi}{2}}{2}=\frac{1}{\sqrt{2}}-\frac{\pi}{4}$

$-\frac{\pi}{2}<\frac{1}{\sqrt{2}}-\frac{\pi}{4}<0\Rightarrow\ sin\left[\frac{sinx+cosx-\frac{\pi}{2}}{2}\right]<0$

(4) minimum value within the second square brackets

$x=\frac{5\pi}{4}\Rightarrow\frac{-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{\pi}{2}}{2}=-\frac{1}{\sqrt{2}}-\frac{\pi}{4}$

$-\frac{\pi}{2}<-\frac{1}{\sqrt{2}}-\frac{\pi}{4}<0\Rightarrow\ sin\left[\frac{cosx+sinx-\frac{\pi}{2}}{2}\right]<0$

Therefore, it follows that

$cos(sinx)-sin(cosx)>0\Rightarrow\ cos(sinx)>sin(cosx)$