1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Modulous of a function

  1. Nov 26, 2012 #1
    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. jcsd
  3. Nov 26, 2012 #2
    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.
  4. Nov 26, 2012 #3
    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
  5. Nov 26, 2012 #4
    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
  6. Nov 26, 2012 #5
    Is it supposed to look something like this? 2mwtdfb.png

    and the solutions are at the troughs?

    It was supposed to reach 1 but I drew this in paint so...
  7. Nov 26, 2012 #6
    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.
  8. Nov 26, 2012 #7
    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 ?
  9. Nov 26, 2012 #8
    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°
  10. Nov 26, 2012 #9
    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.
  11. Nov 27, 2012 #10
    You're most welcome. I'm glad I could help and that you found the idea of making a sketch useful.
  12. Nov 27, 2012 #11
    I'm kind of confused though if I was to find the |cos| of like -61 it's not bigger than 1/2
  13. Nov 27, 2012 #12


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Of course, |cos(-61°)| < 1/2 .

    Look at the intervals given by Michael Redei .
  14. Nov 27, 2012 #13
    ... sorry.... the conditions hold true.
  15. Nov 27, 2012 #14
    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.
  16. Nov 27, 2012 #15


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Here's a graph from WolframAlpha.


    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook