# What steps are implemented to derrive $\cos(xt)+\sin(xt)=…=\cos(xt-\phi)$, when $\phi=\arctan(x)$

\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: http://i.stack.imgur.com/QCiFW.png

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

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 $\cdots$
• What is $\frac{1}{\cos(\phi)}$ as defined before the dots ? – Claude Leibovici Nov 16 '14 at 9:36