Find angle in rad from sin I can't solve this and I have a test tomorrow, can you help me please:
$\alpha - \beta = \frac{3}{5} + \pi$
$\cos \alpha = \frac{1}{3}$
Determine $\sin\alpha - \beta$
That's the problem. Here's what I know:
$\sin \alpha = (\sqrt 24) / 5$
How do I progress? Thanks for the help.

$$\alpha - \beta = \frac{3}{5} + \pi$$
$$\cos \alpha = \frac{1}{3}$$
so
$$\sin\alpha=\pm\frac{2\sqrt{2}}{3}$$
So the answer is
$$\pm\frac{2\sqrt{2}}{3}-\beta$$
and because of
$$\alpha - \beta = \frac{3}{5} + \pi$$
it is
$$\beta=\alpha-\pi-\frac{3}{5}$$
substituting to 
$$\pm\frac{2\sqrt{2}}{3}-\beta$$
;
$$\pm\frac{2\sqrt{2}}{3}-\alpha+\pi+\frac{3}{5}$$
where
$$\cos \alpha = \frac{1}{3}$$
and 
$$\alpha =2n\pi  \mp\cos^{-1}\frac{1}{3}$$
Fixing the plus-minus and minus-plus, the answer is
$$2n\pi\pm\frac{2\sqrt{2}}{3}\mp\cos^{-1}\frac{1}{3}+\pi+\frac{3}{5}$$
