# Trigonometry in the DE

I have a very simple 2nd order homogeneous DE of the form:

$$y''=-a^2y$$

So, a solution will be of the form $$y(t)=c_1 e^{+iat}+c_2 e^{-iat} \quad (1)$$

which you can rewrite to $$y(t)=d_1\cos(at)+d_2 \sin(at) \quad (2)$$

The final answer of the problem has the form of $C \sin(at+d)$, where $d$ is a phase term. No initial conditions are given.

My question is, can eq. (2) be rewritten in some way that only a sine term emerges, with a constant $d$ appearing in there? I looked at some trig. relations and couldn't see a quick way. I think the initial conditions have to be used. What do you think?

(A teaching assistent told me it could be done, but I distrust her advice a little since she has given me wrong answers before)

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Expand on the trig expression you get from the answer:

$$\sin(at+d) = \sin(at) \cos(d) + \cos(at) \sin(d)$$

So you can take eqn.(2) and factor out the constant $C = \sqrt{d_1^2 + d_2^2}$ and then pick $d$ so that $\cos(d) = d_1/C$ and $\sin(d) = d_2/C$.

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WE did it at the same time! – Babak S. Feb 27 '13 at 13:43
@BabakS. :) yes, indeed – gt6989b Feb 27 '13 at 13:44

As far as I know for a suitable $\eta$, we can have $a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\eta)$

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Yay for $\eta$ and Babak +1 – amWhy Feb 28 '13 at 0:15