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? 

Remember the negative angle identities: 
$$\sin(-A)=-\sin(A)$$
$$\cos(-A)=\cos(A)$$
To find the value of A, we have to solve the equation $\cos^{-1}(A)=\frac{1}{2}$ (Remember your 30-60-90 triangle)
Assuming that this is over the interval $[0, 360]$, you can get either $60^{\circ}$ or $300^{\circ}$ (Remember your reference angle formulas)
Because the question stated that the value of $\cos(A)$ and $\sin(-A)$ have the same sign, we will have to evaluate the $\sin(-300^{\circ})$ Why? Because only the cosine is positive in the fourth quadrant and because $\sin(-300) = -\sin(300) = --\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$
