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Homework Help: Is my reasoning correct?

  1. Aug 21, 2005 #1

    Tau

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    This is the problem:

    Sketch the graph of:

    [tex]\sin\Theta[/tex] = [tex]\frac{\pi}{2}[/tex] - [tex]\Theta[/tex]


    Now the graph (between 0 and pi) is symmetrical about the line

    [tex]\Theta[/tex] = [tex]\frac{\pi}{2}[/tex]

    So I can write the equation as

    [tex]\sin\Theta[/tex] = [tex]\frac{\pi}{2}[/tex] + [tex]\Theta[/tex]

    This will be a graph of [tex]\sin\Theta[/tex] pulled [tex]\frac{\pi}{2}[/tex] units to the left.

    Is this reasoning correct?

    If it is, would you do it a different way?


    Thanks
    Jeremy

    (Noone wants a gmail invite?lol!)
     
  2. jcsd
  3. Aug 22, 2005 #2
    What I am not understanding is how you sketch it on a coordinate plane if there is only one variable (theta).
     
  4. Aug 22, 2005 #3

    Tau

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    I am SUCH AN IDIOT!

    I mistyped the entire question. You all now see what a dufus I am.

    Anyway, here is the question (typed properly this time):

    Sketch the graph of:

    y = [tex]\sin(\frac{\pi}{2}[/tex] - [tex]\Theta[/tex])

    and using the same reasoning above I end up with

    y = [tex]\sin(\Theta[/tex] + [tex]\frac{\pi}{2}[/tex])

    Which is a curve of [tex]sin\Theta[/tex] pulled half pi units to the left
    Now- is this reasoning correct.

    If it is, would any of you done it a different way?
     
  5. Aug 22, 2005 #4
    Both of the expressions are equal to cos theta.
     
  6. Aug 22, 2005 #5

    Tau

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    I know that both curves are the same.

    But I wanna know if I reasoned it out correctly.

    Were the actions I took (see first post) the correct ones. Was I thinking along the correct lines?
     
  7. Aug 22, 2005 #6

    Tau

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    I can see that noone is interested in giving me an answer.

    Oh well.
     
  8. Aug 22, 2005 #7

    AKG

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    Science Advisor
    Homework Helper

    Yes, your reasoning is fine.
     
  9. Aug 22, 2005 #8
    Well, both equations do simplify to cos theta, so yeah they are the same function. If you meant, "how do I see for sure that they are the same function?" you can always try substitution.

    sin theta when theta equals zero is zero. Substituting (using degrees because I don't have those pretty graphics), your first equation becomes:

    y = sin (90 - 0) = sin (90) - sin (0) = 1 - 0 = 1

    Then your next equation becomes:

    y = sin (0 + 90) = sin (0) + sin (90) = 0 + 1 = 1

    To check to see that both equations can be substituted for simply y = cos (theta).

    y = cos (0) = 1

    The math police would probably arrest me if they found me dumping in numbers like this when you are probably asking for a formal proof, but it's good enough for me so I hope it helped.
     
    Last edited: Aug 22, 2005
  10. Aug 22, 2005 #9

    Tau

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    Thanks guys

    I thought for a moment that I was alone in the dark.

    More questions to come soon!
     
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