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Determining Arcsin(Sin[7Pi/5])

  1. May 25, 2009 #1
    The problem is... What to put as my answer :P arcsin(Sin(7Pi/5))



    http://mathworld.wolfram.com/TrigonometryAnglesPi5.html



    The solution is -2Pi/5

    The problem I'm having is this appears to me as a trick question. Would it be acceptable to say

    sin(7pi/5) = -(sqrt(5/8+sqrt(5)/8)) make a note of an identity. Then simply say arcsin(-(sqrt(5/8+sqrt(5)/8))) =-2Pi/5?

    Assuming this is correct, would not all the work I just cut out be acceptable because I'm not trying to prove the identity over again?

    Comments suggestions would be appreciated.

    Thanks.


    PS: Sorry for lack of tex.
     
  2. jcsd
  3. May 26, 2009 #2
    Re: Arcsin(Sin[7Pi/5])

    I think you need to look at where the arcsin is defined.
     
  4. May 26, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Arcsin(Sin[7Pi/5])

    This is much easier than you are making it. There is no reason to even think about the actual value of sin(7pi/5). As Hogger suggested, think about the definition of "arcsin". The only "tricky" point is getting the correct range for the answer. 7pi/5 itself is larger than pi. What's wrong with that?
     
  5. May 26, 2009 #4
    Re: Arcsin(Sin[7Pi/5])

    I don't understand, am I missing an identity somewhere?

    ArcSin[7Pi/5] is imaginary

    Sin[7Pi/5] is not

    And ArcSin[Sin[7Pi/5]] is not imaginary.
     
  6. May 26, 2009 #5

    Mark44

    Staff: Mentor

    Re: Arcsin(Sin[7Pi/5])

    You are not asked to find arcsin(7pi/5).

    If x is in the domain of the Sin function (which has a restricted domain, unlike the sin function), then ArcSin(Sin(x)) = x. 7pi/5 is not in the domain of the Sin function, so what you need to do is find a value of x in this restricted domain for which Sin(x) = sin(7pi/5).
     
  7. May 26, 2009 #6

    HallsofIvy

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    Re: Arcsin(Sin[7Pi/5])

    Mark44, you mean "You are not being asked to find sin(7pi/5) don't you?
     
  8. May 26, 2009 #7

    Mark44

    Staff: Mentor

    Re: Arcsin(Sin[7Pi/5])

    No, I was responding to something Beer w/Straw wrote in post 4:
     
  9. May 26, 2009 #8
    Re: Arcsin(Sin[7Pi/5])

    Well the definition ArcSin[Sinx]=x where -Pi/2=<x=<Pi/2 ...

    But it equals -x ...

    sin(7π/5), being sin (π + α) = - sinα and - sinα = sin(-α)

    So, just slap a minus sign infront of 2Pi/5?
     
  10. May 26, 2009 #9

    Mark44

    Staff: Mentor

    Re: Arcsin(Sin[7Pi/5])

    You're looking at the wrong identity. The sine function is symmetric about the line x = π/2. This means that sin(π/2 + a) = sin(π/2 - a). Find a value of x in [-π/2, π/2] such that sin(x) = sin(7π/5).
     
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