What is $\sqrt{\sin^2(-60^o)}$ equal to?

Question

I begin with the assumptions that:

1. $\sin(-60^o)= -\frac{\sqrt {3}}{2}$
2. $\sin^2(-60^o)=(\sin(-60^o))^2$
3. $(\sin(-60^o))^2= \sin(-60^o)*sin(-60^o) =-\frac{\sqrt {3}}{2}*(-\frac{\sqrt {3}}{2})=\frac {3}{4} $
4. $\sqrt[a]{x^a}=x$

So in my mind,

I. $\sqrt{\sin^2(-60^o)}=\sqrt{\sin(−60^o)∗\sin(−60^o)}=\sqrt{-\frac{\sqrt {3}}{2}∗(-\frac{\sqrt {3}}{2})}=\sqrt{\frac {3}{4}}=\frac{\sqrt{3}}{2}$

which is is what I would normally assume. But what happens in the case where I use 4. $\sqrt[a]{x^a}=x$?

Now when I try to find $\sqrt{\sin^2(-60^o)}$ , I get:

II. $\sqrt{\sin^2(-60^o)}=\sin(-60^o)=-\frac{\sqrt {3}}{2}$

But I also know, that $\sqrt{x}$ is defined only for $x\ge0$, so is II. even possible, or does $\sqrt{\sin^2(-60^o)}$ have an imaginary part, or am I making a mistake somewhere?

Answer 1

Inside -out:

$-60$ is "fourth quandrant" so $\sin -60 < 0$ and $\cos -60 > 0$.

$\sin -60 = - \sin 60 = -\frac {\sqrt 3}2$

$\sin^2 -60 = (\frac{\sqrt 3}2)^2 = \frac 34$

$\sqrt{\sin^2 -60} = \sqrt {\frac 34} = \frac {\sqrt 3}2$.

Outside in:

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

so $\sqrt{\sin^2 (-60)} = |\sin (-60)|>0$ so equivalent to the angle that gives a positive value. As $\sin(x) = -\sin(-x) = \sin(\pi - x) = -\sin(\pi + x)$ etc. so the positive value is $\sin 60 = \frac {\sqrt 3} 2$.

Without handwaving and keeping the internal monolog internal (where is belongs):

I) $\sqrt {\sin^2 (-60)} = \sqrt {(-\frac {\sqrt{3}}2)^2} = \sqrt{(\frac {\sqrt{3}}2)^2} = \frac {\sqrt 3}2$

II) $\sqrt{\sin^2(-60)} = |\sin(-60)| = |-\frac {\sqrt 3}2| = \frac {\sqrt 3}2$.

.... Basically, keep the basic understanding of signs basic: Squaring something has a positive result; what's under a radical sign must be positive; the square root is always the non-negative square root; $\sin (-x) = - \sin (x)$ etc. Keeping those in mind and know that $\sin {60} = \frac {\sqrt 3}2$, then..... you just do it.

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In other words:

I begin with the assumptions that:

4) $\sqrt[a]{x^a} = a$.

That is a wrong assumption.

If $a=2n; n\in \mathbb N$ is even then $\sqrt[a]{x^a} = |x|$.

If $a=2n-1; n \in \mathbb N$ is odd then $\sqrt[a]{x^a} = x$.

If $a \not \in \mathbb N$ then we have to define $\sqrt[a]{x^a}$ in terms of logarithms and that requires $x > 0$ and we will get $\sqrt[a]{x^a} = x$ provided $a \ne 0$>

Answer 2

$$\sqrt{\sin^2(-60^{\circ})}=\sqrt{\frac{3}{4}}=\frac{\sqrt3}{2}.$$ Another way: $$\sqrt{\sin^2(-60^{\circ})}=|\sin(-60^{\circ})|=\left|-\frac{\sqrt3}{2}\right|=\frac{\sqrt3}{2}.$$