I've come across a circuit differential equation problem that I'm frustratingly close to solving:
"For a circuit described with the equation $I' + 10I = 30cos60t + 40sin60t$, write the steady periodic solution in the form $I(t) = Ccos(ωt-α)$"
Through solving a differential equation, I found values $A=-21/37$, $B=-22/37$. So we have:
$C = \sqrt{A^2+B^2} = 0.82$ (This is correct)
$ω = 60$ (This is correct)
$α = tan^{-1}(\frac{B}{A})$ (for $0<α<\pi$) $ = 0.8$
The last expression is NOT correct; according to my key the true answer is $α = \pi $ MINUS $0.8$. Why would this need to be done? $0.8$ is already between $0$ and $\pi$, and I don't know of any other condition that would require you to modify $α$'s value in this way. Can anyone point out what I'm missing with this?
$B$ and $A$ are really $x$ and $y$ coordinates. Since both are negative in your problem, you're in the 3rd Quadrant. That makes $\alpha = \pi+.08$. Then since you need an angle between $0$ and $\pi$, we just note that $\cos(\pi+0.8) = \cos(\pi -0.8)$ to move the angle to the 2nd Quadrant.