I'm reviewing differential equations, and came across this problem.
In the MIT OCW lecture, the professor utilizes the trig formula
$A\cos t + B\sin t = C\cos(t - \phi)$
where $C$ is the amplitude and $\pi$ is $\arctan(\frac{B}{A})$.
But if you would look at this video and how the TA solves it, he gets the particular solution $X_p$ to be composed of a sine function.
(Video: http://www.youtube.com/watch?v=-0_vZ4t-q0I&list=PL64BDFBDA2AF24F7E&index=44 )
Little confused how he got there, considering the trig formula ends up being a function of cosine. Please advise.
Note: the way I solved it gives: $X_n = \frac{1}{(4n^2 + (4-n^2)^2) }cos(nt - \varphi )$
where $\varphi$ is $\arctan(\frac{-(4-n)^{2}}{2n})$
Sorry if the formatting is not working!!
All you need is
$$\sin(A)=\cos\left(A-\frac{\pi}{2}\right).$$
$\cos (f(t))$ is not a function of cosine by the way but a function of $t$.