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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

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

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:

Forum graph - svgphp-0123-1-2-312345-1xy

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:`

Forum graph - svgphp-1123-1-2-3123-1-2-3xy

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.

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