# Difference Identity problem

I have a homework problem that I don't know what to do with. We were just introduced to sum and difference identities. We've always been provided values in degrees both in class and in homework until this problem. I checked the book to see if a similar problem had been worked out; it hadn't. Any help would be appreciated.

Use the information below to find the exact value of sin (A-B): $\cos A = \dfrac1{3}, 0 < A < \dfrac{\pi}{2}, \sin B = -\dfrac1{2}, \dfrac{3\pi}{2} < B < 2\pi$

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Hint: You should have $\sin (A-B)= \sin A \cos B - \cos A \sin B$ and $\sin^2 \theta + \cos^2 \theta = 1$. To find $\sin A$, we use the second:
$\sin^2 A + \cos^2 A=1$
$\sin^2 A = \frac 89$
$\sin A = \pm \frac {2 \sqrt 2}3$
The restriction on $A$ should let you resolve the $\pm$ sign. Now do the same to find $\cos B$ and you have all you need for the first equation.