Posted by Liz on October 18, 2002 at 19:21:54:
In Reply to: Re: Graphing Inverse Circular Functions on Calculator posted by Jack on October 18, 2002 at 18:39:09:
: : 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)
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