# inverse functions [Solved!]

### My question

I have the problem sec(arccos 5) and I understand that sec is the reciprocal of cos. But if the domain of arccos is [0,pi] then how can this be 1/5?

### Relevant page

Wolfram|Alpha: Computational Knowledge Engine

### What I've done so far

I have a good understanding of the inverse functions and their domains, so I just don't get why you can even do arccos of 5.

X

I have the problem sec(arccos 5) and I understand that sec is the reciprocal of cos. But if the domain of arccos is [0,pi] then how can this be 1/5?
Relevant page

<a href="https://www.wolframalpha.com/">Wolfram|Alpha: Computational Knowledge Engine</a>

What I've done so far

I have a good understanding of the inverse functions and their domains, so I just don't get why you can even do arccos of 5.

## Re: inverse functions

A corrections first. The domain of y=arccos(x) is actually -1 <=x <= 1, x ne 0. The range however, is 0 <= y <= pi.

Here's the graph:

But either way, we shouldn't be able to evaluate sec(arccos 5) since arccos 5 is not defined. (For those not sure how this works, see: Composition of functions.)

Here's the graph of y=sec(arccos(x)). It's actually the same as y=1/x for -1 <=x <= 1:

Wolfram|Alpha's domain solution correctly states the domain for sec(arccos x), but then goes on to plot it for -5.1 to 5.1.

They also give sec(arccos 5) = 1/5 as per your question. I've left feedback on this issue on the W|A site and we'll see what they say.

X

Hi ShayT: I share your view! There is something odd about this.

A corrections first. The domain of y=arccos(x) is actually -1 &lt;=x &lt;= 1, x ne 0. The <b>range</b> however, is 0 &lt;= y &lt;= pi.

Here's the graph:

[graph]310,250;-3,3;-1,5,1,1;arccos(x),[/graph]

But either way, we shouldn't be able to evaluate sec(arccos 5) since arccos 5 is not defined. (For those not sure how this works, see: <a href="http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-composite-2009-1.pdf">Composition of functions</a>.)

Here's the graph of y=sec(arccos(x)). It's actually the same as y=1/x for -1 &lt;=x &lt;= 1:

[graph]310,250;-3,3;-3,3,1,1;1/cos(arccos(x)),[/graph]

Wolfram|Alpha's domain solution correctly states the domain for sec(arccos x), but then goes on to plot it for -5.1 to 5.1.

They also give sec(arccos 5) = 1/5 as per your question. I've left feedback on this issue on the W|A site and we'll see what they say.