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Homework Help: Arcsin x sin

  1. Jul 7, 2009 #1


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    1. The problem statement, all variables and given/known data

    arcsin(sin) = 1 right?

    2. Relevant equations

    3. The attempt at a solution

    Basically, I see arcsin as 1/sin

    is this correct?
  2. jcsd
  3. Jul 7, 2009 #2
    1/sin(x) = csc(x)

    arcsin is the function such that sin(arcsin(x)) = x
  4. Jul 7, 2009 #3


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    Oh I think I get it.

    So, I can take the arcsine of both sides in a problem such as:

    sin(2x) = (Root3 )/2

    and I would get arcsin(sin(2x)) = arcsin ((root3)/2)

    Which would get me to 2x = arcsin ((root3)/2)?

  5. Jul 7, 2009 #4


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    Staff Emeritus
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    Yes or no, depending on what you literally mean.

    The big overwhelming obstacle that you need to make sure you understand is that the equation
    has infinitely many solutions. (or zero solutions, if |x| > 1)

    If I'm to define a function Arcsin(x) that gives a solution to sin(y)=x, I can only pick one of them. (The solution lying in [itex]-\pi/2 \leq y \leq \pi/2[/itex] is traditional)

    So if I want all solutions to sin(y)=x, I have more work to do because Arcsin(x) gives me one of them. Fortunately, knowing one solution, it's easy to find all of the others. (If it's not obvious, study the graph of sin(y)=x for a while....)

    In otherwords, Arcsin(sin(y)) is not y. It is "the number in [itex][-\pi/2 , \pi/2][/itex] that is related to y".
  6. Jul 8, 2009 #5


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    so in this case would it be arcsin(sin(60)) = Arcsin (([tex]\sqrt{3}[/tex]/2

    Can you give me a problem that displays what you have just told me? I'd really like to see one ( as I have not been told that in my Precalculus class).

  7. Jul 8, 2009 #6
    [tex]\sin{0}=\sin{\pi}=0[/tex] but [tex]\pi \not= 0[/tex]. A function can only map one output to a given input, so we have to specify which solution we want when we say Arcsin(0). The solutions which are typically used are the ones between [tex]-\pi/2[/tex] and [tex]\pi/2[/tex]
    Last edited: Jul 8, 2009
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