# find a different form for trigonometric expression

• Oct 23rd 2007, 12:59 PM
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
• Oct 23rd 2007, 02:05 PM
ThePerfectHacker
Quote:

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.
• Oct 23rd 2007, 06:31 PM
Quote:

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?
• Oct 24th 2007, 05:26 PM
ThePerfectHacker
(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).
• Oct 29th 2007, 05:13 AM