Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Functions - the fallacy in this argument

  1. Dec 25, 2011 #1
    I'm asked to find the fallacy in this argument

    sin ≣ sin(π - x)
    Hence, f(sin x) = f(sin(π-x))
    Let f(x) = x sin x
    Then x sin x = (π - x)sin(π-x)
    x = π - x
    Hence π = 2x and since x is any value we choose, so is π

    The fourth line, where the author says the x in front of the sin x is also equal to π - x is where the error is, but in the answers, he says, ''the fallacy is we cannot choose f(sin x) to be x sin x, this is not a function of sin x alone but of x and sin x'

    I suspect he's talking about the same thing, but I don't understand what it has to do with the fourth line.
  2. jcsd
  3. Dec 25, 2011 #2
    From what I understand,

    if you let

    f(x) = xsinx

    and u = sinx


    f(u) = u*sin(u)

    = sin(x)*sin(sin(x))

    and likewise for sin(pi - x)

    If you look at the proof, you see that the fourth line should be:

    sin(x)*sin(sin(x)) = sin(pi - x)*sin(sin(pi - x))

    which seems correct.
    Last edited: Dec 25, 2011
  4. Dec 25, 2011 #3
    So it's basically two functions in one?
  5. Dec 26, 2011 #4


    Staff: Mentor

    That's pretty meaningless in this problem.

    The problem is that the 4th line does not evaluate f correctly. f maps a number into that number time the sin of that number.

    For example,
    f(x) = xsin(x)
    f(y) = y sin(y)
    f(a + b) = (a + b) sin(a + b)
    f(sin(π-x)) = sin(π-x) sin(sin(π-x))
  6. Dec 26, 2011 #5
    Are you sure you typed it out correctly? Because the answers only make sense if either the function was defined as [itex] f(\sin x) = x \sin x [/itex] or if the author is deciding to be bizarrely pedantic about the scope of the variable x, otherwise the problem is definitely in the 4th line.
  7. Dec 26, 2011 #6
    JHamm is correct, the author either made a typo or the OP did.
  8. Jan 4, 2012 #7
    Yeah, if I understand you correctly, that's what I thought. It's just his answer confused me.

    @JHamm: Yeah, that's how the questioning and answer were given. I was wondering why he said we can't use f(sin x) to be x sin x because in the question he let f(x) = x sin x.

    When I see f(sin x) = x sin x, I interpret that as sining the x before inputing it's sin(x)sin(sin(x)).
  9. Jan 5, 2012 #8
    If you take [itex]f(x) = x\sin x [/tex] you can then go
    [tex]f(\sin x) = \sin x \sin \sin x [/tex]
    However if you try to define your function as
    [tex]f(\sin x) = x\sin x [/tex]
    Then you're in trouble because that isn't surjective.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Functions - the fallacy in this argument