Simplification of Trigonometic Expression

[bomerman218]

\frac{1}{7} e^{-2t} \cos(4 \sqrt 6 t)+\frac{\sqrt 6}{21} e^{-2t} \sin(4 \sqrt 6 t) =\frac{\sqrt 15}{21} e^{-2t} \cos(4 \sqrt 6 t+\arctan \sqrt 6/2)

I am having trouble figuring out how to prove this relation. Any help would be greatly appreciated. My initial thought was to use this formula:

\cos(u+v)=\cos(u) \cos(v)+\sin(u) \sin(v)

 

[adriank]

It looks a bit daunting at first, but we can clean that up a little, by dividing everything by (\sqrt{15}/21) e^{-2t} and substituting \theta = 4\sqrt{6} t:

\frac{3}{\sqrt{15}} \cos \theta + \frac{\sqrt{6}}{\sqrt{15}} \sin \theta = \cos \left( \theta + \arctan \frac{\sqrt{6}}{2} \right).

Now can you do it?

 

[bomerman218]

Yeah I can. Unfortunately when I was solving the governing differential equation I was not given the solution so I'm wondering how I would approach this problem not knowing the solution.

 

[adriank]

What was the differential equation?

 

[bomerman218]

\frac{1}{8}u''+\frac{1}{2}u'+\frac{25}{2}=0

u(0)=\frac{1}{7} u'(0)=\frac{6}{7}