# Simplification of Trigonometic Expression

1. Dec 5, 2008

### 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:

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

2. Dec 5, 2008

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?

3. Dec 5, 2008

### 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.

4. Dec 5, 2008

$$\frac{1}{8}u''+\frac{1}{2}u'+\frac{25}{2}=0$$
$$u(0)=\frac{1}{7}$$ $$u'(0)=\frac{6}{7}$$