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!

Trig question

  1. Jul 29, 2011 #1
    1. The problem statement, all variables and given/known data
    This is a question from my test paper.

    If sec [itex]\theta[/itex]=[itex]\frac{4xy}{(x+y)^2}[/itex], where x,y [itex]\in[/itex] R, then

    (a)x,y [itex]\in[/itex] R+
    (b)x,y [itex]\in[/itex] R-
    (c)x=y[itex]\neq[/itex]0
    (d)x[itex]\in[/itex]R+, y[itex]\in[/itex]R-

    2. Relevant equations



    3. The attempt at a solution

    I don't understand what i have to do and where should i start with? :confused:
     
  2. jcsd
  3. Jul 29, 2011 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Remembering that sec a is by definition 1/(cos a),
    then what values for x and y can you throw into that quotient term
    and still yield a value that is a valid value for sec ? Do you need to
    restrict the permissible values for x and/or y to a narrower domain than all Reals ?
     
  4. Jul 29, 2011 #3
    What i did is [itex]cos \theta=\frac{(x+y)^2}{4xy}[/itex].
    I substituted different values according to the given conditions in my equation.
    I found my answer to be (c) option. Is that correct?
    But what should be the correct method to find out the answer? :confused:
     
  5. Jul 29, 2011 #4

    eumyang

    User Avatar
    Homework Helper

    What's wrong with guess-and-check as a method? All you need is one example that would make an answer choice wrong.

    In any event, you could try using the definition sec θ = R/X in the Cartesian plane (I'm using capital letters to distinguish from the lower case x and y that you already used), and that R2 = X2 + Y2. Find Y, and then look at the answer choices again to see what must hold.
     
  6. Jul 29, 2011 #5
    What's this "R"? :confused:
     
  7. Jul 29, 2011 #6

    eumyang

    User Avatar
    Homework Helper

    "R" stands for radius. See the attached diagram, and the website below (look under the heading "TRIGONOMETRIC FUNCTIONS").
    http://www.mathacademy.com/pr/prime/articles/trig_ident/index.asp" [Broken]
     

    Attached Files:

    Last edited by a moderator: May 5, 2017
  8. Jul 29, 2011 #7
    Ok, i got it, thanks!! :smile:

    I have one more question.
    If f(x)=|sin x|+|cos x|, x[itex]\in[/itex]R, then
    (a)f(x)[itex]\in[/itex][0,2]
    (b)f(x)[itex]\in[/itex][0,[itex]\sqrt{2}[/itex]]
    (c)f(x)[itex]\in[/itex][0,1]
    (d)f(x)[itex]\in[/itex][1,[itex]\sqrt{2}[/itex]]

    How should i start with in this question? :confused:
     
    Last edited by a moderator: May 5, 2017
  9. Jul 29, 2011 #8

    PeterO

    User Avatar
    Homework Helper

    I like that you have identified that it is unsuitable for both x and y to be 0, but if x and y are alternately positive or negative real numbers, then the possibility of them both being zero is still excluded.

    Think about the maximum and minimum value the cosine function can have, and the significance that would have to the relative sizes of the numerator and denominator of the fraction you have.
     
  10. Jul 29, 2011 #9

    eumyang

    User Avatar
    Homework Helper

    Start by finding the range of sin x and cos x. In other words, what are all the possible values of y in y = sin x, and what are all the possible values of y in y = cos x?
     
  11. Jul 29, 2011 #10

    PeterO

    User Avatar
    Homework Helper

    Perhaps a sketch graph may help? [use adding ordinates if necessary]
     
  12. Jul 29, 2011 #11
    Range of sin x and cos x is [-1,1].
    What next?
     
  13. Jul 29, 2011 #12

    eumyang

    User Avatar
    Homework Helper

    Well, what happens when you take the absolute value of any number between -1 and 1 inclusive? What does the range become?
     
  14. Jul 29, 2011 #13
    Taking the absolute value the range becomes [0,1].
     
  15. Jul 29, 2011 #14

    eumyang

    User Avatar
    Homework Helper

    Okay, so the range of |sin x| is [0, 1], and the range of |cos x| is [0, 1]. What would be the range of f(x)=|sin x| + |cos x|, then?
     
  16. Jul 29, 2011 #15

    PeterO

    User Avatar
    Homework Helper

    That is the separate range of each function, which you have to add together.

    Can they both be zero at the same time? can they both be 1 at the same time?

    You might need that sketch graph now.
     
  17. Jul 29, 2011 #16
    That only i don't know. I don't understand how to do that? :confused:
     
  18. Jul 29, 2011 #17

    eumyang

    User Avatar
    Homework Helper

    EDIT: Big mistake...
     
    Last edited: Jul 29, 2011
  19. Jul 29, 2011 #18
    I add the max and min values of the range of sin x and cos x.
    If |sin x| = 0 and |cos x| = 0, then
    |sin x| + |cos x| = 0.
    If |sin x| = 1 and |cos x| = 1, then
    |sin x| + |cos x| = 2

    Is the answer (a) option?
     
  20. Jul 29, 2011 #19

    eumyang

    User Avatar
    Homework Helper

    EDIT: Another big mistake...
     
    Last edited: Jul 29, 2011
  21. Jul 29, 2011 #20

    PeterO

    User Avatar
    Homework Helper

    sketch graph!!! the answer is not (a)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trig question
  1. Trig question (Replies: 10)

  2. Trig question (Replies: 4)

  3. Trig Question (Replies: 1)

  4. Trig cosx question (Replies: 5)

  5. Trig question (Replies: 2)

Loading...