Question

How do I find the domain of arcsin(sec x)

Answer 1

by combining the domain of arcsin and secant

Answer 2

Dof arcsin is [-1,1]
Dof secx is all reals except pi/2+2kpi

Answer 3

How did you get the domain of sec x?

Answer 4

Recall that \(\sec x=\dfrac{1}{\cos x}\), which means \(\sec x\) is not defined for values of \(x\) that make \(\cos x=0\). This occurs for \(x=\dfrac{\pi}{2}+n\pi\), where \(n=0,\pm1,\pm2,...\)

This means the domain of \(\sec x\) is
\[\left\{x~|~x\not=\frac{\pi}{2}+n\pi,~n\in\mathbb{Z}\right\}~~~~\text{(set builder notation)}\]
or in interval notation:\[\cdots\cup\left(\frac{\pi}{2}-2\pi,\frac{\pi}{2}-\pi\right)\cup\left(\frac{\pi}{2}-\pi,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\frac{\pi}{2}+\pi\right)\cup\left(\frac{\pi}{2}+\pi,\frac{\pi}{2}+2\pi\right)\cup\cdots\\
=\cdots\cup\left(-\frac{3\pi}{2},-\frac{\pi}{2}\right)\cup\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\frac{3\pi}{2}\right)\cup\left(\frac{3\pi}{2},\frac{5\pi}{2}\right)\cup\cdots\\
={\large\bigcup_{n=-\infty}^\infty}\left(\frac{\pi}{2}+n\pi,\frac{\pi}{2}+(n+1)\pi\right)\]
You can actually forget all this, because you only need to know the range of secant, which is \((-\infty,-1]\cup[1,\infty)\).

As for the inverse sine function... recall that \(x=\sin y\) is bounded between -1 and 1; that is, the range of \(y=\sin x\) is \(y\in[-1,1]\). If a function has an inverse, then the function's range is equivalent to the function's inverse's domain. This means \(\sin^{-1}x=y\) has domain \(x\in[-1,1]\).

So, since the domain of the inverse sine is \([-1,1]\), we can only allow values of \(\sec x\) that belong in this interval. The only values that we can then allow are those that make \(\sec x=1\), because those are the only numbers we can allow as input for the inverse sine function.

Which values of \(x\) make \(\sec x=1\)?
\[\sec x=\frac{1}{\cos x}=1~~\iff~~x=n\pi,~\text{for }n=0,\pm1,\pm2,...\]
So the domain of \(\sin^{-1}(\sec x)\) is
\[\{x~|~ \sec x=1\}~~~~\text{(set builder notation)}\\
\{...,-2\pi,\pi,0,\pi,2\pi,...\}~~~~\text{(roster notation)}\\
\{x~|~x=n\pi,~n\in\mathbb{Z}\}~~~~\text{(another variant of set builder)}\]

Answer 5

@SithsAndGiggles i can't seem to get why values that make secx=1
couldn't we have also secx=-1?
can you show this graphically^_^

Answer 6

why you choose*

Answer 7

@xapproachesinfinity

A function is defined (more or less) to be a rule that takes elements from a set (the domain) and matches them up with elements from another set (the range).

If \(f\) is our function, we can denote this by \(f:X\to Y\), so \(X\) is the domain and \(Y\) is the range.

For example, if \(f:\mathbb{R}\to\mathbb{R}\) is defined by \(f(x)=x^2\), then it's easy to see that plugging in real numbers gives real numbers for your output.

Now consider a composite function \(f\circ g\), which we could write in the nested notation, \(f(g)\). Since \(g\) is its own function, it has its own domain and range set, \(g:X\to Y\). Whatever the output is of \(g\) is used as the input of \(f\), so you can write \(f:Y\to Z\), or \(f:(X\to Y)\to Z\).

As you can see, the only set that matters regarding the domain of \(f\) is the range set of \(g\).

Answer 8

And I agree, graphs are very helpful. Here's \(y=\sec x\):

Like we've agreed before, odd multiples of \(\dfrac{\pi}{2}\) produce asymptotes. The graph has local minima at \(y=1\) and local maxima at \(y=-1\). You can see that the domains and ranges are as I described earlier.