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Cosine graphs.

  1. Dec 4, 2011 #1
    I know that the following two graphs are the same:

    x(t)=0.15cos(10.21t+pi)
    x(t)=-0.15cos(10.21t)

    But when I try to solve for t given x=0.94 for the first function, when I plug it into my calculator I get -0.22s and the second function I get 0.22s. For the question I am doing I have to use the first function to get 0.22s, does anyone know how?
     
  2. jcsd
  3. Dec 4, 2011 #2

    eumyang

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    (Emphasis mine.) Not sure what you are saying here. Were you solving for t if x(t) = 0.94? If so, you won't find any (real) solutions. Do you see why?
     
  4. Dec 5, 2011 #3
    I got 0.22s, I drew the 4 quadrants and it worked out. What do you mean by real solutions?
     
  5. Dec 5, 2011 #4

    eumyang

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    Looks like you made a typo. You wanted to set x(t) = 0.094, not 0.94.

    I solved these graphically on my TI-84 and both 0.22 and -0.22 are solutions of both equations. If you were trying to solve by using the "cos-1" button, then you need to realize the limitations of doing that -- the range of the inverse cosine is only 0 to π.
     
  6. Dec 5, 2011 #5
    There are things called complex numbers (real numbers are a part of this group) which are made up of two numbers, the first number is the real part and the second number is the "imaginary" part, real numbers are just complex numbers with the imaginary part equal to 0. Real numbers are just everything you've used in your life so far.

    A real solution is one which can be expressed with real numbers, in the case of cosine with [itex]\displaystyle{x \in \Re}[/itex] (just a notation of saying that [itex]\displaystyle{x}[/itex] is a real number) it has a range of [itex]\displaystyle{-1 \leq \cos x \leq 1}[/itex] and in your case it would have to be [itex]\displaystyle{\geq 1}[/itex] to satisfy the equation which means the answer couldn't have been a real number.
     
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