# Intervals of Increase/Decrease

1. May 16, 2015

### Cosmophile

1. The problem statement, all variables and given/known data
Identify the open intervals on which the function $f(x) = \cos \frac {x}{2}, 0 < x < 2 \pi$ is increasing or decreasing.

2. Relevant equations
$f(x) = \cos \frac {x}{2}$

$f'(x) = -\frac {1}{2} \sin \frac {x}{2}$

3. The attempt at a solution
I'm attempting to do these types of problems using inequalities, as Serge Lang does in his text, and some problems are easier than others (as one would expect). Here's my attempt so far:

$f$ is increasing/decreasing when $f' > 0$ and $f' < 0$ respectively.

$-\frac {1}{2} \sin \frac {x}{2} > 0$

$\sin \frac{x}{2} < 0$
Here, I am torn. If I take $\arcsin \frac {x}{2} < \arcsin 0$ I end up with $x < 0$. This essentailly means that $f$ is increasing when $x$ is negative. There are no $x < 0$ on the interval, so $f(x)$ is never increasing on the interval $0 < x < 2 \pi$. This happens to be the case, but I don't think this is the reason. For instance, if I look for the interval on which $f$ is decreasing, using the same method, I get:

$-\frac {1}{2} \sin \frac {x}{2} < 0 \to \sin \frac {x}{2} > 0 \to x > 0$

Which says that $f$ is decreasing when $x>0$, which is true on this interval, but fails beyond the interval.

I have tried doing this without taking the $\arcsin$ of both sides:

$-\frac {1}{2} \sin \frac {x}{2} < 0 \to \sin \frac {x}{2} > 0$

This is less upsetting for me, as I can see this and see which values of $x$ hold. Ultimately, I do not know which is the best method, or if there is another method entirely that I should be using.

2. May 16, 2015

### Ray Vickson

Drawing a graph always helps. What does the graph of $y = \cos(x/2), \; 0 \leq x \leq 2 \pi$ look like? Alternatively, you can also look at the unit circle in $\mathbb{R}^2$, and pinpoint the location of the points $(X,Y)$, where $X = \cos(x/2), Y = \sin(x/2), \; 0 \leq x \leq 2 \pi$ for the various values of $x$.

3. May 16, 2015

### Cosmophile

I graphed it, which is how I know that $f$ is never increasing on the interval $0 < x < 2\pi$. I just can't figure out how to come to these conclusions solely from the use of inequalities. That is, I'm looking for a purely analytical solution as opposed to a geometric one. (Is that the proper way to say it?) Thank you for the insights!

4. May 16, 2015

### HallsofIvy

You need to know what values sin(x) has! You should know that sin(x) is positive for $0< x< \pi$, then negative for $\pi< x< 2\pi$, then positive for $2\pi< x< 3\pi$, etc.

Last edited by a moderator: May 16, 2015
5. May 16, 2015

### Ray Vickson

So much depends on how you define the functions $\sin$ and $\cos$. For example, Rudin defines them through their MacLauren expansions, so that
$$\sin(x) = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}, \; \cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!},$$
by definition. He then goes on to prove things like $\sin^2(x) + \cos^2(x) = 1$, etc., and establishes the usual properties you get from the graphs. That is he proves there is a smallest $x > 0$ (call it $\pi$) such that $\sin(\pi) = 0$. Also, $\pi/2$ is the smallest positive root of $\cos(x) = 0$, etc. Furthermore, the functions are periodic, with period $2 \pi$. He gets these from the infinite series, plus the relations $\sin(x)' = \cos(x)$, $\cos(x)' = -\sin(x)$, together with mean value theorems and the like.

6. May 16, 2015

### JonnyG

Do you know the unit circle definition of sine? Ifso, use that to tell when sine is negative and positive. If not, look it up - it's very useful.

7. May 16, 2015

### Cosmophile

Indeed, and I do know this. I should probably make this more clear: My issue is not in solving the problem, but in solving the problem using inequalities, as Lang does (and as I've seen in the MIT 18.01 lecture series).

I've heard of Rudin's text before, and am looking forward to reading it. I think that I should probably have a better understanding of Calculus (at least series) before starting analysis.

I'm familiar with it, thank you!

8. May 17, 2015

### Cosmophile

I'm still struggling to come up with a sensible answer using inequalities.

Update: I'm beginning to think that this method is simply not the best for periodic functions. Feel free to show me why this isn't the case!

Last edited: May 17, 2015
9. May 17, 2015

### Ray Vickson

When you say "this" method, which method are you referring to?

10. May 17, 2015

### SammyS

Staff Emeritus

As you've noticed the inverse trig functions can introduce difficulties. Perhaps a larger problem is that the trig functions themselves are periodic.

For the function here, $\displaystyle \ f(x) = \cos \frac {x}{2},\ 0 < x < 2 \pi \,,\$ it's helpful to change that inequality to $\displaystyle \ 0 < \frac x2 < \pi \$

For $\ f(x)\$ increasing:
As you said that requires $\ f'(x)>0\$. → That eventually gives $\displaystyle \ \sin \frac {x}{2} < 0 \$.

However, sin(θ) > 0 for all θ such that 0 < θ < π . So $\ f'(x)<0\$ nowhere in the domain.
Using the inverse sine here can be problematical unless great care is used. In this case it does give an answer which is correct. (At least it seems that way.)​

The case of $\ f(x)\$ decreasing, shows some of the precautions involved with using the inverse sine in this way. Even better examples could be had using $\displaystyle \ g(x) = \sin \frac {x}{2}, 0 < x < 2 \pi \$ and/or by extending the domain of the function under consideration.

For $\ f(x)\$ decreasing:
For this $\displaystyle \ \sin \frac {x}{2} > 0 \$.

The direct result considering only the sine function: sin(θ) > 0 for all θ such that 0 < θ < π, so that ƒ(x) is increasing over the entire domain.

As you discovered, using the inverse sine only gives part of the answer. That's because the range of the inverse sine is [ -π/2 , π/2 ] , so sin-1(u) cannot return a value outside of this interval.​

Another problem may occur when using an inverse function to solve an inequality, if that function is decreasing.
If ƒ is a strictly decreasing continuous function, and its inverse function, ƒ-1 exists,. (Yes, of course it exists.) then the following is true, much as it is for multiplying or dividing by a negative.

$\displaystyle \ f(x) > b \$ implies that $\displaystyle \ x< f^{-1}(b) \$.
It is possible to develop some rules to allow using inequalities with trig functions and their inverses in a reasonably direct fashion. However, you might ask yourself if it's worth the trouble.