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Homework Help: Pre-Test on Trigonometric Equations and Applications

  1. May 14, 2005 #1
    I have a math test on the chapter on Tuesday, and my teacher handed out the pre-test on Thursday. There are a few problems I am totally stumped on, and figured the math geniuses here could give me some help. Some I can get somewhere with, some I don't know where to begin.

    Here is one problem I am stuck on:

    At CI State Park, the average height of the tide in feet t hours after hight tide is given by the function


    a. What is the smallest positive value of t corresponding to the low tide?

    The answer to a is 8:30. How did they come up with this answer?
  2. jcsd
  3. May 14, 2005 #2


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    Hmm,i dunno,calculus,maybe...?Or even simpler,what's the minimum value of the function (HINT:it has to do with the inferior boundedness of the cosine).....?

  4. May 15, 2005 #3
    It's better to express t as 8.5 rather than 8:30 I think.
  5. May 15, 2005 #4


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    Incidentally,he'll get the result as 8.5...:wink:If he wants to make the conversion,or not,that's another issue.

  6. May 15, 2005 #5
    Here's a problem I got somewhere with, but I still get the wrong answer.


    I have to simplify this, and the answer ends up being \cot\theta

    I did the following:


    [tex]\tan\theta = \frac{\sin\theta}{\cos\theta}[/tex] , [tex]\cot\theta = \frac{cos\theta}{\sin\theta}[/tex]

    So [tex]\frac{\sin\theta}{\cos\theta} + \frac{cos\theta}{\sin\theta} = \frac{sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}[/tex] • [tex]\frac{\cos^2\theta}{1} (\sec^2})[/tex] which was our denominator to begin with

    I get stuck here.
  7. May 15, 2005 #6


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    [tex] \frac{\sin^{2}\theta+\cos^{2}\theta}{\sin\theta\cos\theta}\frac{\cos^{2}\theta}{1}=\frac{1}{\sin\theta\cos\theta}\cdot\cos^{2}\theta=\frac{\cos\theta}{\sin\theta}=\cot\theta [/tex]


  8. May 15, 2005 #7
    Oh yeah

  9. May 16, 2005 #8
    All right, the test is tomorrow and one or two of these problems are still giving me trouble.


    So I did the following: took the square of everything to get rid of the square root, and I got:


    Then I changed [tex]\sin^2(x)[/tex] to be [tex]1-\cos^2(x)[/tex] and got:


    After I rearranged it, I got:


    I have tried breaking this down into something like:


    but it does not seem to work. I am supposed to get the values for [itex]\cos[/itex] that allow me to get as answers [itex]2\pi/3[/itex] and [itex]5\pi/3[/itex]

    Where did I go wrong?
  10. May 16, 2005 #9
    What's the lowest value the cosine function could possibly have?
  11. May 16, 2005 #10
    your first step is wrong, you can't square the terms individually to get rid of the root(3).

    try just looking at it like this:
    [tex] sinx= -\sqrt{3}cosx[/tex]
    then think of the unit circle values that would let this be true.
  12. May 16, 2005 #11
    Ah, gale thanks, that helped a lot. If I can't solve another problem I'll head it your way, but I don't think I will, so just wish me luck on my test tomorrow everyone!
  13. May 16, 2005 #12
    How about cos has a lowest value of -1 and that is when [tex]x=\pi[/tex]. Therfore, [tex]\frac{2\pi}{13}\left(t-2\right)= \pi[/tex]
  14. May 16, 2005 #13


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    Even easier :

    Let [itex]t = \tan\theta[/itex]

    [tex]\frac{t + \frac{1}{t}}{1 + t^2}[/tex]
    [tex]=\frac{1}{t} = \cot\theta[/tex]
  15. May 17, 2005 #14


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    It would help if here were to divide through [itex] \cos x [/itex] and then solve this

    [tex] \tan x=-\sqrt{3} [/tex]

    in the reals...

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