How Does Trigonometric Identity Simplify Forced Vibration Equations?

[jellicorse]

I am trying to work though an example on this topic in my book and have reached a point that I am not sure about. I was wondering if anyone could help me clear this up.

The equation of motion for a spring-mass system with no damping and a periodic external force is

mu'' + ku = Fcos\omega t

The general solution to this is:

u = Acos\omega_0t + B sin\omega_0 t +\frac{F}{m(\omega_0^2+\omega^2)}cos\omega t

If the mass is initially at rest, so that u(0)=0 and u'(0)=0, then the solution to this equation is

u=\frac{F}{m(\omega_0^2-\omega^2)}(cos\omega t -cos\omega_0 t)


I have managed to follow it to here but I can not see how they have completed the next step:

"making use of the trigonometric identities for cos(A\pmB) with A=(\omega_0+\omega)t/2 and B=(\omega_0-\omega)t/2 we can write the equation in the form":

u=\left[\frac{2F}{m(\omega_0^2-\omega^2)}sin\frac{(\omega_0-\omega)t}{2}\right]sin\frac{(\omega_0+\omega)t}{2}

I can not see how the penultimate equation becomes the final equation here; can anyone tell me how this works?

 

[olivermsun]

Write out A+B and A-B.
Then write out the angle addition formula for cos(A±B).
You'll see how it works.

 

[jellicorse]

Thanks a lot for the tip, Oliver. I've got it now.