simple square root factoring question

[Odyssey]

Hi,

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

If not, what should it be? :confused:

Appreciate the help! :smile:

[Sirus]

That is correct.

[Odyssey]

Thank you for the help Sirus!

[Tide]

It is correct within a sign!

[Odyssey]

It is correct within a sign!

What do you mean?

[maverick280857]

That is correct.

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\|

In your case

\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

[Sirus]

I stand corrected. I forgot about that.