# Thread: 12cos(ot) + 5sin(ot) + 3cos(ot − pi/3) in the form Acos(ot + phi)

1. ## 12cos(ot) + 5sin(ot) + 3cos(ot − pi/3) in the form Acos(ot + phi)

Write $\displaystyle f(t) = 12\cos(\omega t) + 5\sin(\omega t) + 3\cos(\omega t - \pi/3)$ in the form $\displaystyle f(t) = A\cos(\omega t + \phi)$.

My attempt:

$\displaystyle \\ f(t) = 12\cos(\omega t) + 5\sin(\omega t) + 3\cos(\omega t - \pi/3)\\ = 12\cos(\omega t) + 5\sin(\omega t) + 3\cos(\omega t)(cos(-\frac{\pi}{3}) - 3\sin(\omega t)\sin(-\frac{\pi}{3}) \\ = 12\cos(\omega t) + 5\sin(\omega t) + \frac{3}{2} \cos(\omega t) + \frac{3\sqrt{3}}{2}\sin(\omega t) \\ = \frac{27}{2}\cos(\omega t) + \frac{(10 + 3\sqrt{3})}{2}\sin(\omega t)$.

I don't know if there is anywhere that I can go from here!

Any help would be awesome.

2. ## Re: 12cos(ot) + 5sin(ot) + 3cos(ot − pi/3) in the form Acos(ot + phi)

$12\cos(\omega t)+5\sin(\omega t) +3\bigg[\cos(\omega t)\cos\left(\dfrac{\pi}{3}\right)+\sin(\omega t)\sin\left(\dfrac{\pi}{3}\right) \bigg]$

$\dfrac{27}{2}\cos(\omega t) + \left(\dfrac{10+3\sqrt{3}}{2}\right) \sin(\omega t)$

note ...

$A\cos{x}+B\sin{x} = R\cos(x-\alpha)$ where $R= \sqrt{A^2+B^2}$ and $\tan{\alpha} = \dfrac{B}{A}$

see what you can do from here ...

3. ## Re: 12cos(ot) + 5sin(ot) + 3cos(ot − pi/3) in the form Acos(ot + phi)

Originally Posted by skeeter

$A\cos{x}+B\sin{x} = R\cos(x-\alpha)$ where $R= \sqrt{A^2+B^2}$ and $\tan{\alpha} = \dfrac{B}{A}$

see what you can do from here ...
Thanks a lot for the response.

I did actually use that method and got $\displaystyle 15.49 \cos(\omega t - 0.51)$ which seems to be about right.

The context of the question seems to suggest that there is a way to get an exact expression without decimals. Is there a way to do this (perhaps involving exponentials?)?

Thanks again!

4. ## Re: 12cos(ot) + 5sin(ot) + 3cos(ot − pi/3) in the form Acos(ot + phi)

best I can get for an "exact" expression ...

$R \cos(\omega t - \alpha)$ where $R = \sqrt{15\sqrt{3}+214}$ and $\alpha = \arctan\left(\dfrac{10+3\sqrt{3}}{27}\right)$

... maybe someone else will come along and offer an improved version.