# Modulous of a function

1. Nov 26, 2012

### lionely

1. The problem statement, all variables and given/known data
Would the graph of |cos x| for values of x between 0 and 360 be a normal cosine graph?

2nd question

For what values of θ between -2∏ and 2∏ is |cosθ| > 1/2?

2. Relevant equations

3. The attempt at a solution

Well... I know Cos 0 is 1, Mod Cos ∏ would be 1, Mod Cos 2∏ would be 1. Am I on the right track?

2. Nov 26, 2012

### Michael Redei

Definitely on the right track. Now, you know that |cos π|=1, but what is cos π? If it's not the same as |cos π|, then the graphs aren't the same, are they?

For the second question you might first work out when |cos x| is exactly 1/2.

3. Nov 26, 2012

### lionely

I figured out the answer to the first question the graph looks like umm how do I describe like no part of it goes to the negative side of the graph. the Amplitude is 1.

|Cos 60| = 1/2

4. Nov 26, 2012

### Michael Redei

Right, and there are other values x with |cos x| = 1/2. Try sketching the graph (a rough sketch is all you need to get an idea of what we're looking for) of |cos x| between -2π and +2π, and draw a horizontal line 1/2 unit above the x axis. Then you should be able to guess the other solutions of |cos x| = 1/2.

Let's call those solutions a, b, c, d, e, etc. See how the curve forms "hills" and "valleys" between these points? And they alternate, so if there's a hill between a and b, you get

valley between b and c
hill between c and d
valley between d and e
hill between e and f
etc.

5. Nov 26, 2012

### lionely

Is it supposed to look something like this?

and the solutions are at the troughs?

It was supposed to reach 1 but I drew this in paint so...

6. Nov 26, 2012

### Michael Redei

That sketch is just what I meant. Easily done, and you can get a feeling for what the graph looks like. You need to check the values at -5π and +5π though. And you need to decide whether you're going to work in radians -- using π a lot -- or degrees -- then you'd need to change the scale to read

-360°, -270°, -180°, -90°, 0°, 90°, 180°, 270°, 360°.

Mark the first colution of |cos x| = 1/2 at x = π/3 (i.e. 60°) and you should be able to find the others.

Finally, since you want |cos x| > 1/2 (not < 1/2) you're looking for the hilltops, and not the valleys.

7. Nov 26, 2012

### lionely

lol it's not a -5 it's a 2 my penmanship is extra poor on paint.

So I marked |cosx| = 1/2 at -60,-120,-240,-300, 60,120,240,300. So is the answer like should I like split the parts of the graph into like it's own each inequality, I don't know how to explain what I mean... like.. one part be -360<x<-300?? and then the next one is between 240 and 120 or w/e ?

8. Nov 26, 2012

### Michael Redei

That's right; you get separate ranges which you could write like this:

We have |cos θ| > 1/2 for
• -360° < θ < -300°
• -240° < θ < -120°
• -60° < θ < 60°
• 120° < θ < 240°
• 300° < θ < 360°

9. Nov 26, 2012

### lionely

Wow sketching the graph really make these questions easier.. my teacher said he didn't think anyone would get this question out, but it's not like I did it by myself thank you for the help.

10. Nov 27, 2012

### Michael Redei

You're most welcome. I'm glad I could help and that you found the idea of making a sketch useful.

11. Nov 27, 2012

### lionely

I'm kind of confused though if I was to find the |cos| of like -61 it's not bigger than 1/2

12. Nov 27, 2012

### SammyS

Staff Emeritus
Of course, |cos(-61°)| < 1/2 .

Look at the intervals given by Michael Redei .

13. Nov 27, 2012

### lionely

... sorry.... the conditions hold true.

14. Nov 27, 2012

### Michael Redei

You can see that in your sketch. Start from 0° (height 1) and go left, then you'll be going "downhill". At -60° you've reached the height 1/2, and if you continue further, say to -61°, you'll be below the level 1/2.

15. Nov 27, 2012

### SammyS

Staff Emeritus
Here's a graph from WolframAlpha.

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