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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!!

Matthew Cassell
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naiu
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  • Please consult this formatting tutorial (and do ask if you can't format something): http://meta.math.stackexchange.com/q/5020/166535 – Joonas Ilmavirta Dec 03 '14 at 13:12
  • Just as you can write $A\cos t + B\sin t = C\cos(t - \phi)$, you can also write $$A\cos t + B\sin t = D\sin(t - \psi).$$ Can you find the formula for $D$ and $\psi$? – Simon S Dec 03 '14 at 13:57
  • is D supposed to equal C? can't figure out how you got there... and also the way the TA solved it in the video. I must be not seeing something, and I do get that if you shift it by \frac{pi}{2} then sine becomes cosine. Still lost. – naiu Dec 03 '14 at 17:35

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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 $t$.

JP McCarthy
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