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Why don't sin and arcsin cancel?

  • Thread starter jumbogala
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1. Homework Statement
This isn't really a problem, just a question...

If you had sin(x) / sin^2(x), you are left with 1/sin(x), right?

So how come when you have sin(x)*arcsin(x) you aren't left with 1? Isn't arcsin(x) the same as writing sin^-1 (x)?

Therefore writing sin(x)arcsin(x) would be like writing sin(x)/sin(x) = 1

But it isn't.... why not?


2. Homework Equations



3. The Attempt at a Solution
 

Tom Mattson

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Because [itex]\arcsin(x)[/itex] is not [itex]1/\sin(x)[/itex]. The reciprocal of the sine function is the cosecant function.

The following is a valid identity for all x in the domain of the cosecant function.

[tex]\sin(x)\csc(x)=\sin(x)\frac{1}{\sin(x)}=1[/tex]

The arcsine function is the inverse function for the sine function on the interval [itex][-\pi/2,\pi/2][/itex]. So they "cancel" each other under composition of functions, as follows.

[tex]\sin(\arcsin(x))=\arcsin(\sin(x))=x[/tex]
 
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The notation for inverse functions, f-1(x) is just that: notation, a shorthand way of writing the inverse of a function f. Even though the -1 looks like an exponent, it is meant to be understood in a different way. In function composition, it makes a bit more sense: [itex](f \circ f^{-1})(x) = x[/itex]. Keep in mind the operation indicated by the circle is not multiplication.
 
Because [itex]\arcsin(x)[/itex] is not [itex]1/\sin(x)[/itex]. The reciprocal of the sine function is the cosecant function.

The following is a valid identity for all x in the domain of the cosecant function.

[tex]\sin(x)\csc(x)=\sin(x)\frac{1}{\sin(x)}=1[/tex]

The arcsine function is the inverse function for the sine function on the interval [itex][-\pi/2,\pi/2][/itex]. So they "cancel" each other under composition of functions, as follows.

[tex]\sin(\arcsin(x))=\arcsin(\sin(x))=x[/tex]

I know this is an old thread, but it's exactly what I've been looking for. :smile:

If [tex]\sin(\arcsin(x))=x[/tex], would [tex]\sin(3\arcsin(x))=3x[/tex]?

3 multiplied by arcsin(x), the arcsin is cancelled and we're left with 3 multiplied by x?
 

Borek

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No, sin(arcsin(3x)) = 3x.
 
Thanks for the reply. :smile:

Hmmm...

I've already worked out [tex]y=arcsin(x)[/tex] in another question, so how about this? :

http://img231.imageshack.us/img231/1707/65841090.jpg [Broken]
 
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What about that? What is strange there? It is derivative of y= arcisin(x) .
 
I was referring to my previous question, whether [tex]\sin(3\arcsin(x))=3x[/tex]. :smile:
 
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I know this is an old thread, but it's exactly what I've been looking for. :smile:

If [tex]\sin(\arcsin(x))=x[/tex], would [tex]\sin(3\arcsin(x))=3x[/tex]?

3 multiplied by arcsin(x), the arcsin is cancelled and we're left with 3 multiplied by x?
No. You seem to be thinking in terms of multiplication instead of function composition, which is what sin(arcsin(x)) represents.

For a simpler example, if f(x) = x2, is f(2x) = 2f(x)? No. Why then should sin(3x) be equal to 3sin(x)? This is essentially what you're asking.
 
No. You seem to be thinking in terms of multiplication instead of function composition, which is what sin(arcsin(x)) represents.

For a simpler example, if f(x) = x2, is f(2x) = 2f(x)? No. Why then should sin(3x) be equal to 3sin(x)? This is essentially what you're asking.
It makes sense now. :smile:

I haven't done this before, so I'm just trying everything I can think of.

Did I get it right in that image I posted? :smile:
 

drizzle

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It makes sense now. :smile:

I haven't done this before, so I'm just trying everything I can think of.

Did I get it right in that image I posted? :smile:

yes, 3 is only a constant
 
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Yes. One of the properties of differentiation is that d/dx(k*f(x)) = k* d/dx(f(x)), where k is a constant. Differentiation and integration are linear in that respect, while most functions, including the trig functions, aren't.
 
Oh brilliant. Thank you very much. :smile:

I'll make a note of "d/dx(k*f(x)) = k* d/dx(f(x))". I actually understand it. :biggrin:
 
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This confusion is a good argument for always writing [tex]\arcsin[/tex] and never writing [tex]\sin^{-1}[/tex] . A lot of nineteenth-century notation like that can nowadays be left behind. All we have to do is convince everyone of this!

Second point. Strictly speaking [tex]\arcsin(\sin(x)) = x[/tex] is not always true. For example:
[tex]0 = \arcsin(0) = \arcsin(\sin(\pi)) \ne \pi[/tex]
 
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This confusion is a good argument for always writing [tex]\arcsin[/tex] and never writing [tex]\sin^{-1}[/tex] . A lot of nineteenth-century notation like that can nowadays be left behind. All we have to do is convince everyone of this!
I for one am not convinced. How would you write the composition of an arbitrary function and its inverse without this notation?
 
I for one am not convinced. How would you write the composition of an arbitrary function and its inverse without this notation?
Neither am I. If anything, sticking to such notation saves you the trouble of having to memorize for stuff, which in turn will probably confuse you further.
 

Borek

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Well, it is a matter of finding some balance between things that can be confusing. What do I mean by:

[tex]x^{-1} = sin^{-1}(x)[/tex]

is it

[tex]\frac 1 x = arcsin(x)[/tex]

or

[tex]\frac {1} {x} = \frac {1} {sin(x)}[/tex]
 

cristo

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Well, it is a matter of finding some balance between things that can be confusing. What do I mean by:

[tex]x^{-1} = sin^{-1}(x)[/tex]

is it

[tex]\frac 1 x = arcsin(x)[/tex]
I would hope you mean this.

One simply needs to remember that "^(-1)" means "inverse." So, when applied to a variable as [itex]x^{-1}[/itex], it means multiplicative inverse, but when applied to a function as [itex]f^{-1}(x)[/itex], it means the inverse function.
 

Borek

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So, how do we understand [tex]f^{-2}(x)[/tex]?
 

cristo

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So, how do we understand [tex]f^{-2}(x)[/tex]?
I would never write that. I would write [itex]f(x)^{-2}[/itex].
 
This confusion is a good argument for always writing [tex]\arcsin[/tex] and never writing [tex]\sin^{-1}[/tex] . A lot of nineteenth-century notation like that can nowadays be left behind. All we have to do is convince everyone of this!

Second point. Strictly speaking [tex]\arcsin(\sin(x)) = x[/tex] is not always true. For example:
[tex]0 = \arcsin(0) = \arcsin(\sin(\pi)) \ne \pi[/tex]
[tex]Arcsin(x)[/tex] can only return values of y [tex]\frac{-\pi}{2}\leq y\leq \frac{\pi}{2}[/tex]
If you're going to define a function [tex]Sin(x)[/tex], then its domain would have to conform to the range of [tex]Arcsin(x)[/tex] , wouldn't it?

Once you're using non-functions though, that's a whole story altogether.
 
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Mentallic

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I agree with borek here. Rather than having two completely different representations of the inverse for trigs, such as [itex]sin^{-1}x[/itex] and [itex]arcsinx[/itex], the first of which can be confused for [itex]cosecx[/itex] and the second seems to have no initial relationship with the inverse function [itex]f^{-1}(x)[/itex], there should instead be a change for the representation of inverse functions.
Something that is completely different, and cannot be confused for something else.

e.g. the inverse of [itex]f(x)[/itex] could be something like [itex]f^*(x)[/itex] so the inverse of a trig function wouldn't be confused for anything else since there is nothing else that represents the *

Just a bad call of judgement for choice of notation I believe.
 

Borek

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To make things more confusing, [tex]f^*(x)[/tex] means to me conjugate function :devil:
 

cristo

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I agree with borek here. Rather than having two completely different representations of the inverse for trigs, such as [itex]sin^{-1}x[/itex] and [itex]arcsinx[/itex], the first of which can be confused for [itex]cosecx[/itex] and the second seems to have no initial relationship with the inverse function [itex]f^{-1}(x)[/itex], there should instead be a change for the representation of inverse functions.
Something that is completely different, and cannot be confused for something else.
There's no confusion if one is consistent! Positive powers of trigonometric (and all) functions can be written [itex]f^n(x)[/itex] since there is no ambiguity. Negative powers are written [itex]f(x)^{-n}[/itex] to avoid confusion with inverses.

e.g. the inverse of [itex]f(x)[/itex] could be something like [itex]f^*(x)[/itex] so the inverse of a trig function wouldn't be confused for anything else since there is nothing else that represents the *

Just a bad call of judgement for choice of notation I believe.
Asterisks are used for lots of things, so it's probably best not to choose that!
 

Mentallic

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To make things more confusing, [tex]f^*(x)[/tex] means to me conjugate function :devil:
I was afraid this was going to happen! :yuck:
It's not even possible to think of a new, original, notation idea. Maybe thats the reason for expressing inverses in that such way.

[tex]f^{-1}(x)[/tex]

[tex]f(x^{-1})[/tex]

[tex]f(x)^{-1}[/tex]

All meaning completely different things. I like it :biggrin:

There's no confusion if one is consistent! Positive powers of trigonometric (and all) functions can be written [itex]f^n(x)[/itex] since there is no ambiguity.
It's just this ambiguity that you need to watch out for. Remembering to apply the inverse rather than reciprocal is quite easy to remember and not get confused with after a bit of practice, but the whole idea is very unoriginal and confusing for the newbie mathematician.

By the way, were the reciprocal trigo functions created to avoid confusing it with the inverse?

[tex]sin(x)^{-1}=cosec(x)\neq sin^{-1}(x)[/tex]

[tex]cos(x)^{-1}=sec(x)\neq cos^{-1}(x)[/tex]

[tex]tan(x)^{-1}=cot(x)\neq tan^{-1}(x)[/tex]

If so, how much nicer would it have been if the inverse was represented by something else that didn't cause this ambiguity in the last column of trigo functions.
 
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