He tries to show thatMarsBarz said:For question 2 I lost you at the double angle formula. Can you clarify?
So does that mean that you would have two vertical asymptotes at -pi/2 and pi/2 ?gman03 said:1: since the image of arctan (inverse tan) is from -pi/2 to pi/2 exclusive, you have a straight interval of y=x between and exclude the two points (-pi/2, -pi/2) and (pi/2, pi/2)
MarsBarz said:I'm still not confident with those graphs and I am still unsure that the solution which you provided is correct.
I did it again and this is what I found:
y = tan-1(tanx)
Consider y = tan-1A
The range is always -PI/2 < y < PI/2.
So we have determined that the range of the original curve is -PI/2 < y < PI/2.
Now we must find the domain of this curve.
Consider y= tan-1A
The domain is all A.
So all tanx for the original curve.
Does that correspond to all x? What happens at asymptotes of tanx, and does that mean that the graph would not just stop at the x values of -PI/2 and PI/2 like in your diagram?
Thinking about this I think that we still get vertical asymptotes at x values of PI/2, 3PI/2 etc...
Somebody help lol!! I'm getting more confused by the minute.