# Inverse trigonometric functions

MartynaJ
Homework Statement:
Create one equation of a reciprocal trigonometric function that has the following:
Domain: ##x\neq \frac{5\pi}{6}+\frac{\pi}{3}n##
Range: ##y\le1## or ##y\ge9##
Relevant Equations:
Create one equation of a reciprocal trigonometric function that has the following:
Domain: ##x\neq \frac{5\pi}{6}+\frac{\pi}{3}n##
Range: ##y\le1## or ##y\ge9##

I think the solution has to be in the form of ##y=4sec( )+5## OR ##y=4csc( )+5##, but I am not sure on what to include in the brackets.
I got the ##4## and the ##5## from the range, since ##5-4=1## and ##5+4=9##. Here I put ##5## as the equation of axis and ##4## as the amplitude.

Gold Member
I assume at the prohibited x points with ##\pi/3## distance the function diverge to ##\pm \infty##.
The function is an even function. ##f(2n\pi/3),f((2n+1)\pi/3)##=1 or 9.

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MartynaJ
I assume at the prohibited x points with ##\pi/3## distance the function diverge to ##\pm \infty##.
The function is an even function. f(0)=1 or 9.
why ##\frac{\pi}{3}## and not ##\frac{5\pi}{6}+\frac{\pi}{3}##?

Gold Member
I took n is any integer and the prohibited points are
$$x=... ,-5\pi/6,-\pi/2,-\pi/6,\pi/6,\pi/2,5\pi/6,...$$

MartynaJ
I took n is any integer and the prohibited points are
x=...,−5π/6,−π/2,−π/6,π/6,π/2,5π/6,...
Ya sorry I now understand how you got these values... But how did you know it is an even function?

Gold Member
Values of ##5\pi/6+n\pi/3## for n=..., -4,-3,-2,-1,0,...

##cosec\ 0=\pm \infty## but f(0) is a definite value.

$$\frac{y-5}{4}=\pm \sec mx$$
Find appropriate m to adjust period.

Last edited:
Staff Emeritus
Homework Helper
Homework Statement:: Create one equation of a reciprocal trigonometric function that has the following:
Domain: ##x\neq \frac{5\pi}{6}+\frac{\pi}{3}n##
Range: ##y\le1## or ##y\ge9##

Create one equation of a reciprocal trigonometric function that has the following:
Domain: ##x\neq \frac{5\pi}{6}+\frac{\pi}{3}n##
Range: ##y\le1## or ##y\ge9##

I think the solution has to be in the form of ##y=4sec( )+5## OR ##y=4csc( )+5##, but I am not sure on what to include in the brackets.
I got the ##4## and the ##5## from the range, since ##5-4=1## and ##5+4=9##. Here I put ##5## as the equation of axis and ##4## as the amplitude.
Consider ##y = 4 \sec \theta + 5##. It's undefined at the points ##\theta = m\pi + \frac \pi 2##. The function ##f(x)## you want is undefined at points ##x = n \frac \pi 3 + \frac{5 \pi}6##. What you want to do is find how to map the values of ##x## to the values of ##\theta##. Let's take ##m=n## for simplicity. Then you want to find a function ##\theta(x)## that takes ##x = 5\pi/6## to ##\theta = \pi/2##, ##x = 5\pi/6 + \pi/3## to ##\theta=\pi/2 + \pi##, and so on. Every time ##x## increases by ##\pi/3##, you want ##\theta## to increase by ##\pi##.

If you can't see where this is headed, try plotting a few pairs of ##(x,\theta)##.