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Homework Help: Inequalities with trigonometric functions

  1. Sep 27, 2007 #1
    1. The problem statement, all variables and given/known data

    Three functions are defined as follows:

    f:x> cos x for the domain 0< (or equal to) x < (or equal to) 180
    g:x> sin x for the domain 0< (or equal to) x < (or equal to) 90
    h:x>tan x for the domain p< (or equal to) x < (or equal to) q


    Find the range of f.

    -1<(or equal to) x < (or equal to) 1 (correct?)

    Given that the range of h is the same as the range of g, find a value of p and a value of q.

    this is the one i dont quite understand, i got p = 0, q= 90, is that right?

    if the domain is the same therefore the range is the same, yes?
     
  2. jcsd
  3. Sep 27, 2007 #2

    EnumaElish

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    No, not necessarily. And, isn't the question "given identical ranges, find the domain of h"?
     
  4. Sep 27, 2007 #3
    Yes, so I am right. ?
     
  5. Sep 27, 2007 #4

    EnumaElish

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    Can you graph sin and tan between 0 and 360? Or between 0 and 90? What is Sin(0)? What is Sin(90)?

    What is Tan(0)? Tan(90)?
     
    Last edited: Sep 27, 2007
  6. Sep 27, 2007 #5

    HallsofIvy

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    A bit peculiar, actually! I assume they mean 180 degrees, and 90 degrees.
    Normally, sine and cosine, as functions are interpreted as in radians. The way sine and cosine are defined, as functions, x is "dimensionless" but radians give the correct values. Anyway, I'll go with degrees.


    When p= 0, tan(p)= 0. That's not what you want is it?
    What is tan(90)? That's also not what you want is it?

    For what p is tan(p)= -1?
    For what q is tan(q)= 1?
     
  7. Sep 29, 2007 #6
    I think I got it

    Wait.. I think I got this..

    so sin(0) - tan(0

    sin(90) = 1
    tan(45) = 1

    therefore the P and Q are 0 and 45?
     
  8. Sep 29, 2007 #7

    HallsofIvy

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    I'm sorry, how did we get to talking about sin(x)- tan(x)? I thought the question was about the range of tan(x).
    "h:x>tan x for the domain p< (or equal to) x < (or equal to) q" and you were to find the domain given that the range was the same as the range of sin(x) (-1 to 1).
    You said you thought the domain would be the same as long as the range was the same. That is certainly not true! Different functions can take different domains (x-value) to the same range (y-value).
    Yes, it is true that tan(45)= 1 so the upper limit is 45. But since the lower limit on the range of sin(x) is -1, you need to determine where tan(x)= -1, not 0!
     
    Last edited by a moderator: Sep 29, 2007
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