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Homework Help: Need Urgent Math Help

  1. Jun 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi guys,
    I was working on some few Trigonometric questions and since my book doesn't have an answer sheet, I thought it would be a good idea to scan it and then upload so I someone could review my answers and tell me where I went wrong. There are 14 questions but I left some blank because I didn't get what they were asking... Please correct my mistakes and reply so I can understand better.

    2. Relevant equations



    3. The attempt at a solution
    http://www.facebook.com/?sk=media#!/photo.php?pid=1254245&id=1371620807
     
  2. jcsd
  3. Jun 6, 2010 #2

    vela

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    Can you put it somewhere that doesn't require one to register with Facebook to see it?
     
  4. Jun 6, 2010 #3
    Urghh.. I didn't know it would happen that way.. Do you have any suggestions as where or how I can upload the page??
     
  5. Jun 6, 2010 #4
    You didn't put an answer for 1. What is sin(3pi/2)? What is cos(3pi/2)? What is cos(0) and sin(0)? 2. Is wrong? The sine function max is 1 and min is -1. So 5+2*(-1)=3. 3. Not the y-axis. Sin(pi/2)=1;whereas sin(-pi/2)=-1. So symmetric with the origin. Being symmetric with the origin means f(-x)=-f(x). 4. Correct. 5. The question was what x value, not what is the minimum value. -3 is incorrect. 6. Correct
     
  6. Jun 6, 2010 #5
    8. Correct
    9. Correct
    10. Correct
    11. Max of cosine is 1 and min is -1. Use that to answer this question.
    12. When x=pi, it reaches it's max. Minimum of sine is -1. When is sine -1?
    13. If the function was 2/3sin(theta), then it would be 1 cycle. If it was 2/3sin(2*theta), then it would be 2 cycles. But instead it 2/3sin(4*theta). So how many cycles?
    14. Correct
     
  7. Jun 6, 2010 #6

    vela

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    You should be able to attach it to a post here. Click the "manage attachment" button. You may have to go to the advanced options to see it.
     
  8. Jun 6, 2010 #7
    Thanks...needed that I will try posting it again. ( someone already posted answers to questions 8-14)
     
  9. Jun 6, 2010 #8
    Question 11- i figured my amplitude was '4' and so the range should be -4[tex]\subseteq[/tex]y[tex]\subseteq[/tex]4,... why is it not?
     
  10. Jun 6, 2010 #9
    Page re-posted

    1. The problem statement, all variables and given/known data

    I re-posted the questions from the previous post... The only questions I ma still confused about is 7, 11 and please someone check 1- 6..

    2. Relevant equations



    3. The attempt at a solution
     

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  11. Jun 6, 2010 #10
    The amplitude is 3. Max of cos2x is 1 and min of cos2x is -1. So max of y = (3*1+1) & min of y = (3*-1+1). The other answer are posted above 8-14 posting.
     
  12. Jun 7, 2010 #11
    Re: Page re-posted

    For 11., it may help to think that [tex]z = 2x[/tex]. Then we can re-write the function as [tex]y = 3cos(z) + 1[/tex]. What would the maximum and minimum values for the function [tex]cos(z)[/tex] be? Considering that, what would the maximum and minimum values of [tex]3cos(z)[/tex] be? Then you just add 1 to both of those to get the answer.

    (Do you understand why for a problem like this, you could have replaced 2x with anything and gotten the same answer? No matter if you're dealing with cos(4a) or cos(7d) or cos(6b), the function has a maximum of 1 and minimum of -1.)
     
  13. Jun 7, 2010 #12
    The answer to number 1 is (3)
    Just refer to the Unit Circle- Check the IV Quadrant and notice as the cos or "x value" changes
    as the angle approaches 360 degree line.
    And how did you get 3 for number 2?? I still think its "4"
     
    Last edited: Jun 7, 2010
  14. Jun 7, 2010 #13
    majormuss: look at Question 2 again. It asks "What is the minimum element in the range of the equation [tex]y = 5 + 2sin\theta[/tex]?" In other words, what is the lowest y-value for this function? Well, you know that the sine function goes from -1 to 1, and therefore has a minimum of -1. Knowing this, what would the minimum of the function [tex]2sin\theta[/tex] be?

    Then since the function is [tex]y = 5 + 2sin\theta[/tex], you would add 5 to the minimum of [tex]2sin\theta[/tex] (since [tex]y = 5 + 2sin\theta[/tex] is the same as [tex]y = 2sin\theta + 5[/tex]) to get the minimum of [tex]y = 5 + 2sin\theta[/tex].
     
    Last edited: Jun 7, 2010
  15. Jun 7, 2010 #14
    oh yes... I get your point...the answer is 3 ... thanks alot!!
     
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