Problem simplifying the solution of an ODE.

[-Dragoon-]

So, I was following the derivation in my physics book of:
x(t) = c_1e^-(\frac{\gamma t}{2})\cos(\omega_d t)+c_2e^-(\frac{\gamma t}{2})\sin(\omega_d t)

Until they simply get to this in one step:
Ae^-(\frac{\gamma t}{2})\cos(\omega_d t + \phi)

I've tried reading many other sources for this derivation of the underdamped oscillator, and I follow up until this last critical step and they don't hint at the omitted steps.

 

[HallsofIvy]

That's a pretty standard step. Use the identity cos(A+ B)= cos(A)cos(B)- sin(A)sin(B).

Comparing that to \alpha cos(\theta)+ \beta sin(\theta)
(\alpha= c_1e^{-\gamma t/2}, \beta= c_2e^{-\gamma t/2} and \theta= \omega_d t.)
We need cos(A)= \alpha and sin(A)= \beta. Of course, that is not possible unless \alpha^2+ \beta^2= 1. If that is not true, then we multiply and divide by \alpha^2+ \beta^2:
(\alpha^2+ \beta^2)\left(\frac{\alpha}{\alpha^2+ \beta^2}cos(\theta)+ \frac{\beta}{\alpha^2+ \beta^2}sin(\theta)\right)

 

[LCKurtz]

I think Halls meant$$
\sqrt{\alpha^2+ \beta^2}\left(\frac{\alpha}{\sqrt{\alpha^2+ \beta^2}}cos(\theta)+ \frac{\beta}{\sqrt{\alpha^2+ \beta^2}}sin(\theta)\right)$$

 

[-Dragoon-]

Got it, thanks everyone.