# Why don't sin and arcsin cancel?

1. Mar 10, 2009

### jumbogala

1. The problem statement, all variables and given/known data
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. Relevant equations

3. The attempt at a solution

2. Mar 10, 2009

### Tom Mattson

Staff Emeritus
Because $\arcsin(x)$ is not $1/\sin(x)$. 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.

$$\sin(x)\csc(x)=\sin(x)\frac{1}{\sin(x)}=1$$

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

$$\sin(\arcsin(x))=\arcsin(\sin(x))=x$$

3. Mar 10, 2009

### Staff: Mentor

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: $(f \circ f^{-1})(x) = x$. Keep in mind the operation indicated by the circle is not multiplication.

4. Jun 30, 2009

### Matty R

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

If $$\sin(\arcsin(x))=x$$, would $$\sin(3\arcsin(x))=3x$$?

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

5. Jun 30, 2009

### Staff: Mentor

No, sin(arcsin(3x)) = 3x.

6. Jun 30, 2009

### Matty R

Hmmm...

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

http://img231.imageshack.us/img231/1707/65841090.jpg [Broken]

Last edited by a moderator: May 4, 2017
7. Jun 30, 2009

### Дьявол

What about that? What is strange there? It is derivative of y= arcisin(x) .

8. Jun 30, 2009

### Matty R

I was referring to my previous question, whether $$\sin(3\arcsin(x))=3x$$.

9. Jun 30, 2009

### Staff: Mentor

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.

10. Jun 30, 2009

### Matty R

It makes sense now.

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?

11. Jun 30, 2009

### drizzle

yes, 3 is only a constant

12. Jun 30, 2009

### Staff: Mentor

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.

13. Jun 30, 2009

### Matty R

Oh brilliant. Thank you very much.

I'll make a note of "d/dx(k*f(x)) = k* d/dx(f(x))". I actually understand it.

14. Jun 30, 2009

### g_edgar

This confusion is a good argument for always writing $$\arcsin$$ and never writing $$\sin^{-1}$$ . 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 $$\arcsin(\sin(x)) = x$$ is not always true. For example:
$$0 = \arcsin(0) = \arcsin(\sin(\pi)) \ne \pi$$

15. Jun 30, 2009

### Staff: Mentor

I for one am not convinced. How would you write the composition of an arbitrary function and its inverse without this notation?

16. Jul 1, 2009

### queenofbabes

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.

17. Jul 1, 2009

### Staff: Mentor

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

$$x^{-1} = sin^{-1}(x)$$

is it

$$\frac 1 x = arcsin(x)$$

or

$$\frac {1} {x} = \frac {1} {sin(x)}$$

18. Jul 1, 2009

### cristo

Staff Emeritus
I would hope you mean this.

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

19. Jul 1, 2009

### Staff: Mentor

So, how do we understand $$f^{-2}(x)$$?

20. Jul 1, 2009

### cristo

Staff Emeritus
I would never write that. I would write $f(x)^{-2}$.