text: # Thread: find a different form for trigonometric expression
  
  1. ## find a different form for trigonometric expression
  
  Hi, I need to prove that:
  x[t]=cos(ω0*t)+ cos( ωo*t + Δω*t)
  
  can be transform into the form:
  x[t]=A(t)*cos[ωo*t + θ(t)]
  
  where A(t) and θ(t) are function of Δω.
  
  I have the solution but I cannot find out the way to solve it
  A(t)=2|cos(Δω*t)|
  
  and
  θ(t)= ArcTan[sin(Δω*t)/(1+cos(Δω*t))]
  
  I have started by using the trigon identity cos(a+b).
  I first factor cos[ωo*t] to have 1+cos(Δω*t) then I factor 1+cos(Δω*t) to have the expression under the Arctan.
  
  At one point I get:
  
  x[t]= (1+cosΔω*t)*[cos[ωo*t]-sin(ωo*t)*{sin(Δω*t)/(1+cosΔω*t)]}]
  
  from here I can not figure out how to fin A(t) and θ(t).
  
  please can someone help me to finsh the computation?
  thank you
  B
  
  Hi, I need to prove that:
  x[t]=cos(ω0*t)+ cos( ωo*t + Δω*t)
  If you have $\displaystyle A\cos x + B\sin y$ where the coefficients are non-zero you can write,
  $\displaystyle \sqrt{A^2+B^2} \left( \frac{A}{\sqrt{A^2+B^2}}\cos x + \frac{B}{\sqrt{A^2+B^2}}\sin x \right)$
  If $\displaystyle \theta = \tan^{-1} \frac{B}{A}$ then,
  $\displaystyle \sqrt{A^2+B^2} \left( \cos x \cos \theta + \sin x \sin \theta \right) = \sqrt{A^2+B^2} \cos (x - \theta)$
  
  Here expand,
  $\displaystyle \cos (\omega_0 t) + \cos (\omega_0 t) \cos (\Delta \omega t) - \sin (\omega_0 t) \sin (\Delta \omega t)$
  Thus,
  [$\displaystyle [1+\cos (\Delta \omega t) ] \cos (\omega_0 t) - \sin (\omega_0 t)$
  Now apply the trick I used above.
  
  3. Originally Posted by ThePerfectHacker
  If you have $\displaystyle A\cos x + B\sin y$ where the coefficients are non-zero you can write,
  $\displaystyle \sqrt{A^2+B^2} \left( \frac{A}{\sqrt{A^2+B^2}}\cos x + \frac{B}{\sqrt{A^2+B^2}}\sin x \right)$
  If $\displaystyle \theta = \tan^{-1} \frac{B}{A}$ then,
  $\displaystyle \sqrt{A^2+B^2} \left( \cos x \cos \theta + \sin x \sin \theta \right) = \sqrt{A^2+B^2} \cos (x - \theta)$
  
  Here expand,
  $\displaystyle \cos (\omega_0 t) + \cos (\omega_0 t) \cos (\Delta \omega t) - \sin (\omega_0 t) \sin (\Delta \omega t)$
  Thus,
  $\displaystyle [1+\cos (\Delta \omega t) ] \cos (\omega_0 t) - \sin (\omega_0 t)$
  Now apply the trick I used above.
  Hi perfectHacker,
  I tried but I am still stuck. First
  [$\displaystyle [1+\cos (\Delta \omega t) ] \cos (\omega_0 t) - \sin (\omega_0 t)$(You forgot the last term sin( delta omega*t))
  
  OK I have:
  $\displaystyle [1+\cos (\Delta \omega t) ] [\cos (\omega_0 t) - \sin (\omega_0 t)\frac{\sin (\Delta \omega t)}{1+\cos (\Delta \omega t)}]$
  
  Now let $\displaystyle \theta(t)=\arctan(\frac{\sin (\Delta \omega t)}{1+\cos (\Delta \omega t)})$
  
  Now the all square root factor confuses me since I have not exactly the same form.
  Please can push just a little bit so I can finish it?
  
  4. (Yes I forgot that term).
  $\displaystyle [1+\cos (\Delta \omega t) ] \cos (\omega_0 t) - \sin (\Delta \omega t)\sin (\omega_0 t)$
  So $\displaystyle A = 1+\cos (\Delta \omega t)$ and $\displaystyle B=\sin (\Delta \omega t)$.
  Thus,
  $\displaystyle \sqrt{A^2+B^2} = \sqrt{ 1 + 2 \cos (\Delta \omega t) + \cos^2 (\Delta \omega t) + \sin (\Delta \omega t) }$
  Simplify,
  $\displaystyle \sqrt{2+2 \cos (\Delta \omega t)} = \sqrt{ 2(1+\cos (\Delta \omega t)}$
  Half-angle idenity,
  $\displaystyle \sqrt{4 \cos^2 \frac{\Delta \omega t}{2} } = 2 \left| \cos \frac{\cos (\Delta \omega t)}{2} \right|$
  
  (You can probably ignore absolute values because in this problem whatever you are modelling the angle is small enough so it remains positive).
  
  5. Thank you. Sorry I could not come earlier!!