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Solving Sinusoidal Equations involving Cosine Inverse

  1. Jan 7, 2012 #1
    Suppose that a spaceship is fired into orbit from Cae Canerveral. Ten minutes after it leaves Cape, it reaches its farthest distance north of the equator, 4000 kilometers. Half a cycle later it reaches its farthest distance south of the equator (on the other side of the Earth, of course!), also 4000 kilometers. The spaceship completes an orbit once every 90 minutes.
    Let y be the number of kilometers the spaceship is north of the equator (you may consider distances south of the equator to be negative). Let t be the number of minutes that have elapsed since lift off.

    b. write an equation expressing y in terms of t.
    c. use your equation to predict the distance of the spaceship from the equator when
    (i.) t= 25, (ii.) t= 41, (iii.) t= 163

    Relevant equations
    y=4000cos∏/45(x-10)
    y=-4000sin∏/45(x-32.5)

    25=4000cos∏/45(x-10) divide by (x-10) then what?
     
  2. jcsd
  3. Jan 7, 2012 #2

    eumyang

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    Homework Helper

    I assume you mean
    [itex]y = 4000\cos \left( \frac{\pi}{45}(t - 10)\right)[/itex],
    and that it should be t, not x.

    Do you mean
    [tex]25=\frac{4000\cos \left( \frac{\pi}{45}(x - 10)\right)}{x - 10}[/tex]?
    If so, that's wrong. All you need to do is to plug in for t. For t = 25,
    [itex]y = 4000\cos \left( \frac{\pi}{45}(25 - 10)\right) = ?[/itex]
     
  4. Jan 7, 2012 #3
    y=4000cos(∏/45(25-10))
    y=4000cos(.0698131701(15))
    y=4000cos(1.04719751)
    y=4000(.5)
    y=2000 is this math correct ?
     
  5. Jan 7, 2012 #4

    eumyang

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    Looks right to me.
     
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