# How to think about/what is the justification for $\pm \sqrt{x^2} = x$?

It seems really strange that $$\pm$$ can be eliminated, as an algebraic manipulation... it doesn't seem like an algebraic rule.

I think it's as simple as, if $$x\ge0$$, then $$x = \sqrt{x^2}$$, and if $$x\le0$$, then $$x = -\sqrt{x^2}$$, so we don't need the gosh-I-don't-know $$\pm$$, but this doesn't really make sense to me.

Just for context, this came up in trig. I worked out the derivation, but am puzzled at the step in the title.

\begin{align} \tan \frac{\theta}{2} &= \pm \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\\ &= \frac{1-\cos\theta}{\sin\theta} \end{align}

I'm missing something more basic than trig, but I don't recall this ever coming up before. Thanks for any clarity!

• √x >0 always. The plus/minus sign arises here as tan(x/2) can be positive/negative according to which quadrant it lies. – user600016 May 16 at 8:00

In general the $$\pm$$ can't be eliminated, as you've already noted. What the derivation should have said is $$\frac{1-\cos\theta}{\sin\theta}=\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\tan\frac{\theta}{2}.$$