# Simple square root factoring question

1. Oct 4, 2004

### Odyssey

Hi,

is $$\sqrt{a^2-a^2\sin^2{x}} = a\cos{x}?$$

If not, what should it be?

Appreciate the help!

Last edited: Oct 4, 2004
2. Oct 4, 2004

### Sirus

That is correct.

3. Oct 4, 2004

### Odyssey

Thank you for the help Sirus!

4. Oct 4, 2004

### Tide

It is correct within a sign!

5. Oct 4, 2004

### Odyssey

What do you mean?

6. Oct 4, 2004

### maverick280857

No, that is not quite correct if you think of how a secondary definition of the modulus (absolute value) follows from the square root of a square,

$$\sqrt{x^2} = \|x\|$$

$$\sqrt{a^2-a^2\sin^2{x}} = a\cos{x}$$

can be written if and only if a and cos(x) are both positive or both negative; before you brought them out of the square root sign, you had the intermediate step,

$$\sqrt{a^2-a^2\sin^2{x}} = \sqrt{a^2\cos^2{x}}$$

so

$$\sqrt{a^2-a^2\sin^2{x}} = \sqrt{a^2\cos^2{x}} = \|a\cos x\|$$

which should hold true anyway since the left hand side is the positive square root.

In general however, you can write it as a cos x if you have no major problems with the signs (you won't have any if theta lies between 0 and pi/2 and a > 0 for instance). But if you're proving something which involves this substitution, you had rather take it into account.

Cheers
Vivek

7. Oct 5, 2004

### Sirus

I stand corrected. I forgot about that.