# Who reads manuals? :) nt

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Posted by Joel on October 19, 2002 at 18:56:56:

In Reply to: Figured out a different way posted by Liz on October 19, 2002 at 17:39:21:

: : : : : 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)

: i took a gander on over into my calculator's manual (more like textbook) and i foudn that if you input the function and graph it then go to the draw menu and select 8:DrawInv and Y1 it'll graph the entire graph .. maybe i should sit down with my manual and calculator for a while and see what i should have read 3 years ago

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