How do I calculate the phase $\phi$ of $\sin(4-3t)$ relative to $\sin(3t)$? Also what would the angular frequency $\omega$ be?
With something like $\sin(2t + 2)$, I can see that the phase relative to $\sin(2t)$ is clearly $+2$.
For $\cos(2-t)$ relative to $\cos(t)$, I thought of it as $\cos(-t+2)$ and instinctively rearranged it as $\cos(t-2)$ to make $\phi = -2$, however this "logic" doesn't translate to the problem above, as the phase is apparently $0.858$ and not $-4$.
I'd appreciate any help.