Is Sin(x) Always Less Than x?

[Karol]

Homework Statement

Homework Equations

$$\sin\,x=a/r$$

The Attempt at a Solution

$$\sin\,x=a/r, ~~x=\mbox{arc}(a)/r$$
$$\frac{a}{r}<\frac{\mbox{arc}(a)}{r}~\rightarrow~\sin\,x<x$$
$$\frac{2x}{\pi}=\frac{2\,\mbox{arc}(a)}{\pi r}>\frac{2a}{\pi r}$$
$$\sin\,x=a/r>\frac{2}{\pi}\frac{a}{r}$$
But i am not sure this is what they wanted

 

[I like Serena]

Hi Karol,

What is arc(a)?
It seems to be the arc length that belongs to an angle of x at radius r, but then there should not be an 'a' should there?
Certainly the equation at the end with 2x/pi cannot be right.

Alternatively, how about using these function graphs?

 

[Karol]

In $~\mbox{arc}(a)~$ i just meant the arc enclosed by the edge a, not to multiply the arc length and a.
This question is from the chapter about derivatives. can i prove on the basis of derivatives?

 

[I like Serena]

Okay. Still, arc(a) is not a notation that is generally used.
And yes, you can use derivatives.
Any thoughts on how to do that?

 

[Karol]

I don't know the common notation for arc length.
The derivative of sin in $~0<x<\pi/2~$ is $~\sin'\,x<1~$ but smaller, in part of the domain, than $~\frac{2x}{\pi}$.
And it also gives only the rate of change, not the value of the function.
I don't know how to use derivatives to solve the inequality

 

[I like Serena]

Well, in this context, I'm not aware of a common notation for arc length either.
So if we want to use it, we have to write it out in words, or invent our own notation with an explanation what we did.
If we wanted to, we could for instance define $arc_r(x)$ for the arc length on a circle with radius r and angle x.
Anyway, I propose we focus on the derivatives instead of the arc length.

Let's start with $x \overset ?> \sin x$.
The derivative indeed identifies the rate of change of the function, also called slope.
Initially, at x=0, both sides of the inequality are 0 (which is outside of the interval, so the inequality can still hold).
And indeed, for the right side of the inequality we have $0 < \sin' x = \cos x < 1$ on our interval.
Which rate of change do we have on the left side of the inequality?
Does it tell us how it compares?

 

[Karol]

The slope of 2x/π is 2/π=0.64
cos x varies between 0 and 1

 

[I like Serena]

The slope of 2x/π is 2/π=0.64
cos x varies between 0 and 1

Yes. So zooming in on $x \overset ?> \sin x$, we know that the right hand side has a slope between 0 and 1.
What is the slope of the left hand side?

We'll get to $\sin x > \frac{2x}\pi$ later.

 

[Karol]

$$x'=1\geqslant \sin'\,x$$
So the slope of sin is smaller than the slope of x, so what does it say?
Maybe: they both start from 0 and since the slope of sin is smaller, it says sin(x) is under x?

 

[I like Serena]

$$x'=1\geqslant \sin'\,x$$
So the slope of sin is smaller than the slope of x, so what does it say?
Maybe: they both start from 0 and since the slope of sin is smaller, it says sin(x) is under x?

Yep. That's exactly it. Since they start from the same value, the one with the lower slope must be below the other.

For $\sin x > \frac{2x}\pi$ we will need to look a bit further to the second derivative though.

 

[Karol]

$$\sin''\,x=-\sin\,x=0 \leftrightarrow 1,~~\left( \frac{2x}{\pi} \right)''=0$$
The book didn't explain yet the graphical meaning of the second derivative.

 

[I like Serena]

$$\sin''\,x=-\sin\,x=0 \leftrightarrow 1,~~\left( \frac{2x}{\pi} \right)''=0$$
The book didn't explain yet the graphical meaning of the second derivative.

Hmm... that is a problem. Does your book suggest any other solution to your problem then?

 

[Karol]

No, but tell me in short what does the second derivative help here, please

 

[I like Serena]

No, but tell me in short what does the second derivative help here, please

We have a second derivative that is negative everywhere on our interval.
It means that the slope, which is positive everywhere, is decreasing.
That means in turn that the function value is everywhere above the line $y=\frac{2x}\pi$ as we can see in the function graph.

 

[Karol]

The slope of sin is bigger, so sin starts to shoot above 2x/π, but the slope is decreasing, so maybe the function itself goes down and under 2x/π.
I see from the graph that that isn't so, but can we deduce from the second derivative that it can't be?

 

[I like Serena]

y=2x/pi begins and ends on the graph of the sine. That is, for x=pi/2 its value is 1. And because the 2nd derivative is negative, sine is rounded upwards, putting it above the line.

 

[Karol]

So because there aren't any other points, apart of the beginning and ending, that sin and 2x/π equal, and because of the second derivative, you deduce that sin is above, right?

 

[I like Serena]

Yep

 

[Eclair_de_XII]

We'll get to $\sin x > \frac{2x}\pi$ later.

Is that when he learns about $\lim_{x \rightarrow 0} \frac{sinx}{x}=1$?

 

[Karol]

Thank you "I like Serena"