Find $\sin(A)$ and $\cos(A)$ given $\cos^4(A) - \sin^4(A) = \frac{1}{2}$ and $A$ is located in the second quadrant. 
Question: Find $\sin(A)$ and $\cos(A)$, given $\cos^4(A)-\sin^4(A)=\frac{1}{2}$ and $A$ is located in the second quadrant.

Using the fundamental trigonometric identity, I was able to find that:
•　$\cos^2(A) - \sin^2(A) = \frac{1}{2}$
and 
• $$ \cos(A) \cdot \sin(A) = -\frac{1}{4} $$
However, I am unsure about how to find $\sin(A)$ and $\cos(A)$ individually after this.
Edit: I solved the problem through using the Fundamental Trignometric Identity with the difference of $\cos^2(A)$ and $\sin^2(A)$.

We can use the double angle identities for cosine to solve the problem.  They are
\begin{align*}
\cos(2x) & = \cos^2x - \sin^2x\\
         & = 2\cos^2x - 1\\
         & = 1 - 2\sin^2x
\end{align*}
You correctly found that 
$$\cos^2A - \sin^2A = \frac{1}{2}$$
where $A$ is a second-quadrant angle.  Substituting $2\cos^2A - 1$ for $\cos^2A - \sin^2A$ yields
$$2\cos^2A - 1 = \frac{1}{2}$$
Solve for $\cos A$, keeping in mind that $\cos A < 0$ in the second quadrant.  
Substituting $1 - 2\sin^2A$ for $\cos^2A - \sin^2A$ yields
$$1 - 2\sin^2A = \frac{1}{2}$$
Solve for $\sin A$, keeping in mind that $\sin A > 0$ in the second quadrant.
Alternatively, you could determine $\cos A$, then take its inverse to find $A$, then evaluate $\sin A$.
