# Homework Help: Proof of sqrt(x^2)/x=sin (x)

1. Oct 18, 2012

### Albeaver89

This isn't a homework question, but I felt it was appropriate....

Proof that √(x^2)/x=sin(x)

2. Oct 18, 2012

### Staff: Mentor

You're going to have a very difficult time proving this - it isn't true.

3. Oct 18, 2012

### SammyS

Staff Emeritus
I think that should be:

$\displaystyle \frac{\sqrt{x^2}}{x}=\text{sign}(x)\ .$

4. Oct 18, 2012

### Albeaver89

5. Oct 18, 2012

### lendav_rott

Oh nevermind, so basically 1 = sinx .
So x = arcsin 1
What is really amasing is how they waste the ink to write X as sqrt(x²)
Unless there's something hidden here, I cannot see the point.
Well sin X = 1 if X is Pi and the way that the sine's sinusoidal graph repeats itself you can get the other possibilities for X.

Last edited: Oct 18, 2012
6. Oct 18, 2012

### SammyS

Staff Emeritus
Edit your original post, & make a note there that you've edited it.

7. Oct 18, 2012

### Staff: Mentor

No for a couple of reasons. First, the OP meant sign(x) not sin(x). Second,
$\frac{\sqrt{x^2}}{x} \neq 1$
No. x ≠ $\sqrt{x^2}$

8. Oct 18, 2012

### lendav_rott

Wait, x =/= sqrt(x²)??

I cannot see what you are trying to say - do you mean that sqrt(x²) = +/- X?
This is actually something I was arguing over with my math's lector and he said that the square root of X or X² for that matter, is defined as sqrt(X²) = |X|

A^x = B
x lnA = lnB
if B were negative then this wouldn't hold true and the only explanation is that Sqrt(A²) = |A|

9. Oct 18, 2012

### Staff: Mentor

Yes, that's exactly what I mean. As an example, do you think that $\sqrt{(-2)^2} = -2$?
No, I don't mean that either. The square root of a nonnegative expression produces a single value, not two of them, as ± x implies.
Also, your second step isn't valid if A ≤ 0, because ln(A) wouldn't be defined.

10. Oct 20, 2012

### Akshay_Anti

the reason why x is not equal to root of x squared is this-

If a, b are positive numbers, and you take root of that (in complex numbers, of course), then the relation √-a × √-b = √(-×-)ab = √(+ab) does not hold valid

instead, √-a × √-b = i√a × i√b = -√ab

that's the reason.... and also, √x2 / x should be equal to signum function of x which equals modulus of x divided by x