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\begin{align*} x(t) &= e^{-t/2}\left(\cos(\sqrt{11}t/2)+\frac1{\sqrt{11}}\sin(\sqrt{11}t/2)\right)\\ &= \frac{\sqrt{12}}{\sqrt{11}} e^{-t/2} \cos(\sqrt{11}t/2)-\phi \end{align*} where $\phi=\tan^{-1}(1/\sqrt{11})$.

Original picture: https://i.stack.imgur.com/QCiFW.png

How is arctan derived in this example? Probably some kind of identity has been used, like $\frac{\cos(x)}{\sin(x)}=\frac{1}{\tan(x)}$

Martin Sleziak
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user42141
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  • Some older posts with similar identities: http://math.stackexchange.com/questions/643835/solve-equation-cos-x-sin-x-0/643836#643836, http://math.stackexchange.com/questions/181676/how-to-see-sin-x-cos-x, http://math.stackexchange.com/questions/444887/solving-the-trigonometric-equation-a-cos-x-b-sin-x-c, http://math.stackexchange.com/questions/201399/solving-sin-x-sqrt-3-cos-x-1-is-my-solution-correct, http://math.stackexchange.com/questions/397984/identity-for-a-weighted-sum-of-sines-sines-with-different-amplitudes – Martin Sleziak Nov 16 '14 at 09:30

1 Answers1

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Just consider the case of $$A=\cos(x)+a\sin(x)$$ and define $a=\tan(\phi)$. So $$A=\cos(x)+\tan(\phi)\sin(x)=\frac{\cos(x)\cos(\phi)+\sin(\phi)\sin(x)}{\cos(\phi)}=\frac{\cos(x-\phi)}{\cos(\phi)}$$ Now, using $\sin^2(\phi)+\cos^2(\phi)=1$, so $\frac{\sin^2(\phi)}{\cos^2(\phi)}+1=\frac{1}{\cos^2(\phi)}=1+a^2$ and then $\cdots$

I am sure that you can take from here.

Claude Leibovici
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