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Homework Help: 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

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    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

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    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

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    "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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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

    PeterO

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    sketch graph!!! the answer is not (a)
     
  22. Jul 29, 2011 #21
    Oops, i ticked the wrong answer in my exam!! :redface:

    I won't be able to sketch the graph correctly and i am not that comfortable in adding graphs.
     
  23. Jul 29, 2011 #22

    PeterO

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    The sketch graph for the sine partial and the cosine partial is just a series of off-set bumps [it would be great if you could draw real sine curves, but even if each bump looks like an inverted parabola will show you. heck, even a bunch of half circles like tracing round a dime].
     
  24. Jul 29, 2011 #23

    eumyang

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    Maybe not. I have looked over this problem again and I had misguided you all this time.

    Previously, I had asked,
    if |sin x| = 0 and |cos x| = 0, then what is |sin x| + |cos x|?

    Well, it's not possible for both to equal 0 at the same time, isn't it?

    If |sin x| = 0, then |cos x| must equal 1. And if |cos x| = 0 then |sin x| must equal 1. Then the minimum value for |sin x| + |cos x| is not 0, is it?
     
  25. Jul 29, 2011 #24
    Oops sorry, that was my mistake too that i didn't noticed it. :redface:
    So If |sin x| = 0, then |cos x|= 1 and |sin x|+|cos x|=1.
    But what is the answer then? :confused:
     
  26. Jul 29, 2011 #25

    PeterO

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    Taking you through the sketch graphs.

    A normal sine function is a bump beteen 0 and pi [180degrees if you like], then a dip under the axis between pi and 2xpi.
    The absolute value of the sine function is just two bumps beside each other.

    Please acknowlege you understand that.
     
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