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How to sketch: $\csc^{(-1)}(x) \in (0,\frac{\pi}{2}]$

I can sketch the graph of $f(x) = \csc(x)$ fine, and I can see that an inverse does exist on the given interval as a bijection occurs over it; however, I can't seem to see a method to graphing the inverse. The only step I know to take is to make the domain of $\csc(x)$ the range of the inverse and the range of $\csc(x)$ the domain of the inverse. Is it just the half parabola on its side pointing to the right? or how to I graph the inverse correctly?

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You only have to restrict the domain of $\csc$ so that it's one-to-one on that restricted domain, yes? Then you just reflect things over the line $y=x$... –  J. M. Apr 28 '12 at 13:14
    
@J. M.: How do I do this without graphing $\csc(x)$ and $y = x$ and then just sketching in the reflection? How do I find points etc.? –  stariz77 Apr 28 '12 at 13:24
    
...you can rotate your paper by ninety degrees and pretend you're drawing the cosecant backwards? –  J. M. Apr 28 '12 at 13:27
    
@J.M.: Hmm ok, I guess I confused myself. I put a plot command into wolfram alpha and got a weird looking graph, figured it was more in-depth than what I was doing :S –  stariz77 Apr 28 '12 at 13:32
    
You don't even need to draw $y=x$. Draw the cosecant as usual, fold the paper along the appropriate line, and trace... –  J. M. Apr 28 '12 at 13:35

1 Answer 1

As J.M. said, one can visualize the graph of inverse function by plotting the direct function (on the interval $(0,\pi/2]$) and reflecting in the axis $y=x$. The sine should be familiar:

sine

Taking reciprocal turns the graph into something like a parabola, but horizontal at the end over $\pi/2$:

1/sine

The range is $[1,\infty)$. After reflecting in the bisector $y=x$, the range and domain trade places:

inverse

The graph has horizontal asymptote $0$, because the previous one had a vertical asymptote at $0$.

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