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Trignometric functions

  1. Aug 28, 2003 #1
    does anyone know how to find sin-1(1/3) + cos-1(1/2) in terms of pi?
     
  2. jcsd
  3. Aug 28, 2003 #2

    Integral

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    I am not sure what the argument of the sin is supposed to be. Is that -1*(1/3)= -1/3? That is the only way I can read it, it that what you mean?

    Even the meaning of the argument of the cos is not clear. Is that -1 degee? -1 radian?

    Generaly the results of trig functions are not expressed in terms of Pi but the arguments are. Radians, the natural unit of angle measure is usually given in terms of Pi, is that what you want?
     
  4. Aug 29, 2003 #3
    sorry the -1 was the inverse function.

    inverse of sine and cosine

    ie.

    inversesin (1/3) + inversecos (1/2) in terms of pi
     
  5. Aug 29, 2003 #4

    Hurkyl

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    For ascii text, the traditional names of these functions are:

    sin-1 = arcsin = asin
    cos-1 = arccos = acos


    Anyways, a quick calculation with my TI-89 hints that there is no way to express the answer as a fraction times π the fraction would have to equal .441506781303..., which is not one I recognize, and does not have a denominator less than 22.
     
  6. Aug 29, 2003 #5
    can you show me how to work it out by hand please?
     
  7. Aug 29, 2003 #6
    yeah I think I'd want it in radians
     
  8. Aug 29, 2003 #7

    Integral

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    OK,
    So the problem is asin(1/3) + acos(1/2)

    as Hurkyl pointed out, there is no standard fraction of pi for asin(1/3).

    acos(1/2)= π/3

    This is because I have memorized, as you should, the sin and cos of the primary angles.

    Double check your first value prehaps it should be ([squ](1/3))or something of that nature.
     
  9. Aug 29, 2003 #8
    the full question is:

    find the value of the following in terms of pi

    arcsin (1/3) + arccos (1/2)

    and the answer in the book is (1/2)pi

    how did the book get this answer?
     
  10. Aug 29, 2003 #9

    Hurkyl

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    The answer to what you wrote is certainly not π/2. Numerical calculation yields 1.387... while π/2 = 1.57...
     
  11. Aug 30, 2003 #10

    HallsofIvy

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    Is it at all possible that your problem said

    " asin([sqrt](3)/2)+ acos(1/2)" ?

    Since sin(pi/3)= [sqrt](3)/2 and cos(pi/3)= 1/2 that would is doable- although it does NOT give pi/2.

    asin([sqrt](3)/2)+ asin(1/2)= pi/2.
     
  12. Aug 30, 2003 #11
    no, it said:

    arcsin 1/3 + arccos 1/2

    so it must be an error because it can't be expressed in terms of pi
     
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