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Trig function of arc trig functions and the reverse

  1. Sep 1, 2011 #1
    I know the sin(arccos(x)) = (1-x^2)^0.5

    I was wondering what some of the others are:

    cos(arcsin(X))
    tan(arcsin(X))
    tan(arccos(x))
    sin(arctan(x))
    cos(arctan(x))

    also the reverse:

    arcsin(cos(x))
    arcsin(tan(X))
    arccos(Sin(X))
    arccos(tan(X))
    arctan(sin(X))
    arctan(cos(X))
     
  2. jcsd
  3. Sep 1, 2011 #2

    micromass

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    Let me do two:

    [tex]\cos(arcsin(x))=\sqrt{1-\sin(arcsin(x))}=\sqrt{1-x^2}[/tex]

    and

    [tex]\tan(arccos(x))=\sqrt{\frac{1}{\cos^2(arccos(x))}-1}=\sqrt{\frac{1}{x^2}-1}[/tex]

    I'll let you find out the other ones...
     
  4. Sep 1, 2011 #3
    are these always possible? I mean take

    arccos(tan(x))

    for example, the cosine of an angle is always between 0 and 1, and so, the argument to the arccos function should be a number between 0 and 1...but the tangent of an angle can get pretty large...so, I think these is no solution here...same for others.
     
  5. Sep 1, 2011 #4

    LCKurtz

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    Not if arccos(x) is in the second quadrant.
     
  6. Sep 2, 2011 #5

    HallsofIvy

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    As long as your "angles" are in the first quadrant (so you don't have multi-value problems), you can get all of those formulas by constructing an appropriate right triangle.

    For example, to get sin(arctan(x)), imagine a right triangle with "opposite side" x and "near side" 1 (so that the tangent of the angle opposite side "x" is x/1= x and the angle is arctan(x)). By the Pythagorean theorem, it will have "hypotenuse" [itex]\sqrt{x^2+ 1}[/itex]. Sine is "opposite side over hypotenuse" so [itex]sin(arctan(x))= \frac{x}{\sqrt{x^2+ 1}}[/itex].
     
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