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Homework Help: Calculate arcsin(sin√5)

  1. Aug 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Calculate arcsin(sin√5).

    2. Relevant equations


    3. The attempt at a solution

    f(1/2) = sin(1/2)
    = π / 6

    f'(x) = cosx

    f'(1/2) = cos(1/2)
    = π / 3

    y = π / 6 + π / 3(x - 1/2)
    = π / 3 x

    f(√5) = π / 3 (√5)
    = √5π / 3

    I tried using linear approximation, but didnt get the correct answer of π - √5
  2. jcsd
  3. Aug 2, 2010 #2


    Staff: Mentor

    Re: arcsin(sin√5)

    arcsin(sin x) = x, PROVIDED that x is in the interval [-pi/2, pi/2]. You need to find a number x such that sin(x) = sin(sqrt(5)), where x is in [-pi/2, pi/2].
  4. Aug 2, 2010 #3
    Re: arcsin(sin√5)

    TsAmE, I might add that Calculus *methods* are not appropriate for this problem (even though it might appear in a calc book/course).
    This is more about, as Mark44 has mentioned, coterminal angles and the domain of inverse trig functions.

    Some people have an easier time working with degrees at first. Could you calculate
  5. Aug 2, 2010 #4
    Re: arcsin(sin√5)

    you need to calculate arcsin(sin√5) ?

    arcsin and sin are inverse functions of each other, thus:
    arcsin(sin(x)) = x :)
  6. Aug 2, 2010 #5


    Staff: Mentor

    Re: arcsin(sin√5)

    This is not true in general, and definitely not true in this case arcsin(sin([itex]\sqrt{5}[/itex])) [itex]\neq[/itex] [itex]\sqrt{5}[/itex]. See post #2.
  7. Aug 2, 2010 #6
    Re: arcsin(sin√5)

    yup. my mistake.
  8. Aug 2, 2010 #7


    User Avatar
    Homework Helper

    Re: arcsin(sin√5)

    I don't know how much is assumed that you know in terms of numbers, but I remember that
    [tex]\sqrt{5} \approx 2.236[/tex]

    ... and I remember the radian equivalent of the angles for each quadrant (for example, the angles in Q1 go from 0 to ~1.571 rad), so I could figure out what quadrant [tex]\sqrt{5}\,rad[/tex] is.

    Also, it would be helpful to know one of the trig identities that deal with symmetry. (I hope that wasn't too many hints.)

  9. Aug 2, 2010 #8
    Re: arcsin(sin√5)

    Use the identity sin(x) = sin([itex]\pi[/tex] - x)
  10. Aug 3, 2010 #9
    Re: arcsin(sin√5)

    If that is the case then:

    arcsin(sin√5) = arc(sin(π - √5)), but π - √5 is still in the 2nd quadrant which doesnt satisfy arc(sinx) = x [-π/2, π/2]. :confused:
  11. Aug 3, 2010 #10


    User Avatar
    Homework Helper

    Re: arcsin(sin√5)

    Not quite. By intuition, [itex]\pi[/itex] is a bit more than 3 and [itex]\sqrt{5}[/itex] is a bit more than 2, so [itex]\pi-\sqrt{5}[/itex] is approximately 1. But [itex]\pi/2[/itex] is a number a bit more than 3 and halved, so a bit more than 1.5. Obviously [itex]\pi-\sqrt{5}<\pi/2[/itex] so it is in the first quadrant.
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