I guess this is a really dumb question, but i've been trying quite a lot and I can't figure out how to determine the solutinons.
I need to determine the values of two angles and they don't seem to be ok. But I do know for sure that the values are right. The equation is as it follows:
$$\begin{align*} &\sin(\alpha)=1/3\;;\\ &\cos(\alpha)\cos(\beta)=0\;;\text{ and}\\ &\cos(\alpha)\sin(\beta)=0.94280899908173\;. \end{align*}$$
Because if $\sin(\alpha)$ is not $1$, neither will $\cos(\alpha)$ be $0$. Also means that $\cos(\beta)$ is $0$, resulting that $\sin(\beta)$ is $1$. Determined by calculus that $\cos(\alpha)$ is not $0.94280899908173$, I seem to be in a real trouble..
Your reasoning to the effect that $\cos\beta=0$ is correct, and it follows that $\sin\beta=\pm1$. Since $\sin\alpha=\frac13$, you know that $\cos^2\alpha=1-\sin^2\alpha=\frac89$, and therefore $$\cos\alpha=\pm\frac{2\sqrt2}3\approx\pm0.9428090415821$$ and $$\cos\alpha\sin\beta\approx0.9428090415821$$ as well. Your figure of $0.94280899908173$ simply isn’t quite correct, though it’s close: the three values that you give aren’t quite consistent.