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Sketch arc trig

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Sketch y = cos-1(2x) and y = sin-1(3x)

    2. Relevant equations

    3. The attempt at a solution

    This is my attempt

    cos-1(2x) = y, which means cos y = 2x

    produce a table with y from (-pi/2 to pi/2)

    What I want to confirm is the 2x. I need do 1/2 x instead of 2x when we calculate the x?
    And 1/3x for the other problem?
  2. jcsd
  3. Apr 6, 2010 #2


    Staff: Mentor

    so (1/2) cos y = x
    No, y is in [0, pi]. That's the vertical axis. If you plot pairs of points (x, y), you will have the graph of x = (1/2)cos(y), which is also the graph of y = cos-1(2x).
  4. Apr 6, 2010 #3
    Hi, thank you for pointing it out the trend.

    So in general, I should always get x by itself?

    What about cases like this
    sin-1(1-1/x^2), which gives siny = 1-1/x^2
    do i solve it as 1-siny = 1/x^2, 1-siny = (1/x)^2, sqrt(1-siny) = 1/x ?
  5. Apr 6, 2010 #4


    Staff: Mentor

    Generally, yes.
    Take it all the way to solve for x.

    [tex]x = \pm \sqrt{\frac{1}{1 - sin(y)}}[/tex]

    There should be some restrictions on both x and y so that the +/- in the equation above aren't needed.
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