Re: Graphing Inverse Circular Functions on Calculator

Posted by Joel on October 19, 2002 at 12:49:00:

In Reply to: Re: Graphing Inverse Circular Functions on Calculator posted by Liz on October 18, 2002 at 19:21:54:

: : : I have a quick question about how one would go about graphing arccot(x) in radians using a TI-83 Plus graphing calculator.
: : : Now, what exactly is it that I have to input into my Y₁ that would make the graph have a range of 0 [ < / = ] y [ < / = ] [pi], cross the y-axis at [pi]/2, and have a domain of infinity. I've tried 1/arctan(x), 1/(cos(x)/sin(x)), 1/(arccos(x)/arcsin(x)), and several other combinations that did not produce the desired graph. WHAT AM I DOING WRONG!?

: : : Thanks a bunch :)
: : : ~Liz

: : :First, set mode to RADIANS
: : :Next, press ZOOM and select 7 ~ ZTrig
: : :For Y1 enter 1/tan-1(X)
: : :Press GRAPH

: i'm still getting a different graph than what's in the textbook, their's is a curved line with an asymptote at [pi] and curving down to cross the y-axis at [pi] / 2 with an asymptote at the x-axis. it's like the arctan(x) graph flipped so that what was in quadrant one is in quadrant two and quadrant three is moved to quardrant four and then the graph is moved up to that the part that crosses at the origin now crosses at its asymptote of [pi]/2 ... if that helps with anything

: the graph i'm making on my calculator has a curve in quadrant 3 and another in quadrant 1, forming something like a hyperbola

: my mode is set to radians, my window is Xmin=-3, Xmax=3, 1, -[pi], [pi], 1, 1 (I changed it from that whole zoom=ZTrig thing)

Sorry, folks. Cotan(x) = 1/tan(x) is true but
arccotan(x) = 1/arctan(x) is definitely NOT TRUE.
Look:
tan(pi/6) radians = 1/sqrt(3)
1/(tan(pi/6)) radians = cotan(pi/6) radians = sqrt(3)

arctan(1/sqrt(3)) = pi/6 radians = "the ANGLE whose tangent is 1/sqrt(3)"
arccotan(1/sqrt(3)) = pi/3 radians = "the ANGLE whose cotangent is 1/sqrt(3)"

Actually, the SHAPE of the graph of y=arccotan(x) is the same as the SHAPE of the graph of y = -arctan(x). But the RANGE of the arctan function is generally defined to be (-pi/2, pi/2), while the RANGE of arccotan is generally defined as (0, pi).

So to get a graph of arccotangent, make a graph of:
y = pi/2 - arctan(x)

Follow Ups:

• Figured out a different way Liz 17:39:21 10/19/02 (2)
  • Who reads manuals? :) nt Joel 18:56:56 10/19/02 (1)
    • People who know how to use their calculators :) Liz 16:07:41 10/20/02 (0)
• Thanks Joel! Liz 14:32:43 10/19/02 (0)

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