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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.

Thanks

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The sine function satisfies $\sin(x+\pi)=\sin(-x)$. Therefore,

$$\sin(4-3t)=\sin(3t-4+\pi)$$

And thus, the relative phase to $\sin(3t)$ is $\pi-4\approx -0.85$.

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You're running into trouble with the negative sign. Remember that $-\theta=\theta + \pi$, then reduce it to a coterminal angle. $$\sin(4 - 3t)$$ $$\sin(-3t + 4)$$ $$\sin(3t + \pi + 4)$$ $$\sin(3t + (\pi + 4)\mod 2\pi)$$ $$\sin(3t + 0.858)$$

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