How to conclude that $r^t(E\cos(\theta t)+Fsin(\theta t))$ ossilates with increasing magnitude? Given: 
$r^t(E\cos(\theta t)+Fsin(\theta t))$
Assume $r>1$, $E>0$, $t\ge0$ and $F$ is not known.
How do we conclude that the given expression oscillates with increasing magnitude?
My attempt:


*

*Since $r>1$, $r^t$ will tend to infinity as $t$ tends to infinity.

*Therefore, if the expression is supposed to oscillate with increasing magnitude, then $(E\cos(\theta t)+Fsin(\theta t))$ will need to change regularly change sign. I know that the $\cos$ and $sin$ function will alternate from -1 to 1 but I don't know how to apply that property to this question.


Please help.
 A: No matter what $E$ and $F$ are, we can write
$r^t(E\cos(\theta t)+Fsin(\theta t)) = r^t \sqrt{E^2 + F^2}(\frac{E}{\sqrt{E^2 + F^2}}\cos(\theta t) + \frac{F}{\sqrt{E^2 + F^2}}\sin(\theta t)), \tag{1}$
and since
$(\frac{E}{\sqrt{E^2 + F^2}})^2 + (\frac{F}{\sqrt{E^2 + F^2}})^2 = 1 \tag{2}$
there exists a $\phi \in \Bbb R$ with
$\sin \phi = \frac{E}{\sqrt{E^2 + F^2}}; \; \cos \phi = \frac{F}{\sqrt{E^2 + F^2}}. \tag{3}$
Using (3) in (1) we have
$r^t(E\cos(\theta t)+Fsin(\theta t)) = r^t \sqrt{E^2 + F^2}(\sin \phi \cos(\theta t) + \cos \phi\sin(\theta t)), \tag{4}$
and using the usual trigonometric identity $\sin (\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha$ we have
$r^t(E\cos(\theta t)+Fsin(\theta t)) = r^t \sqrt{E^2 + F^2}\sin(\theta t + \phi). \tag{5}$
Next, note that $r > 1$ allows us to write $r^t = e^{t \ln r}$; since $\ln r > 0$, this is an increasing exponential as $t \to \infty$.  So
$r^t(E\cos(\theta t)+Fsin(\theta t)) = e^{t\ln r} \sqrt{E^2 + F^2}\sin(\theta t + \phi) \tag{6}$
is now easily seen to undergo oscillations of ever-increasing amplitude as $t \to \infty$, since the $\sin$ factor is simply phase-shifted by $\phi$.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
