confusion about negative trigonometric identities Given that $\cos A=1/2$ and $\cos A$ and $\sin A$ have the same sign, find the value of $\sin(-A)$. If the question is referring to the first quadrant, where all trigonometric identities are positive, why is the value of $\sin (-A)$: $-\sqrt{3}/2$? Is it because this rule only applies to positive angles? Also is the rule $\sin(-A)= -\sin A $ applicable for every quadrant? 

Just do it.
$\cos A = \frac 12$
So what is $\sin A$?  Well $\cos^2 A + \sin^2 A = 1$ and so $(\frac 12)^2 + \sin^2 A = 1$ so $\sin^2 A= 1 - \frac 14$ so $\sin A =\pm \frac{\sqrt {3}}2$.  
But we are told $\sin A$ and $\cos A$ are the same sign so $\sin A =\frac {\sqrt 3}2$.
So what is $\sin -A$?  Well, we have a rule that $\sin (-A) = -\sin A$ (and ,yes, that is true for all quadrants) so $\sin (-A) = -\sin A = -\frac {\sqrt 3}2$.
I'm not entirely sure where you trouble lies.
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Given that $\cos A=1/2$

Okay, that means that $\cos A > 0$ so $A$ is in the first or fourth quadrant.

and $\cos A$ and $\sin A$ have the same sign

Okay, that means that $A$ is in first or third.
In combo with $\cos A > 0$ that means $\sin A > 0$ and $A$ is in the first.

Is it because this rule only applies to positive angles?

What rule?  There's nothing wrong and off here.  We have $\sin A > 0$ and so $\sin (-A) = - \sin A < 0$.  This is exactly what we'd expect.
Somehow I suspect you are reading on of those $-A$ as though it were positive $A$ but I don't know where.

Also is the rule $\sin(−A)=−\sin A$ applicable for every quadrant? 

Yes, it is.
