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

See above please...

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.

 

[anuttarasammyak]

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.

 

[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 $\frac{5\pi}{6}+\frac{\pi}{3}$?

 

[anuttarasammyak]

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?

 

[anuttarasammyak]

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.

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

 

[vela]

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:: See above please...

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 $(x,\theta)$.