Adding Sinusoids of differing Magnitute & Phases

[jeff1evesque]

Homework Statement

Find A and \theta given that:
Acos(\omega t + \theta) = 4sin(\omega t) + 3 cos(\omega t)
Could someone elaborate on how to solve this. I mean it looks to me that one simply takes the magnitude of the coefficient and the inverse tangent of the same coefficients. But I feel there needs to be justification as to why we're allowed to do this.

Homework Equations

not sure.

The Attempt at a Solution

But the solution (according to my notes) is: A = \sqrt{4^2 + 3^2} = 5 and \theta^{-1} = \frac{4}{3} = 53.1

Thanks,

JL

 

[jeff1evesque]

What if instead of

<br /> Acos(\omega t + \theta) = 4sin(\omega t) + 3 cos(\omega t)<br />,

I wanted to write it as,

<br /> Asin(\omega t + \theta) = 4sin(\omega t) + 3 cos(\omega t)<br />,

would the values of A and \theta change at all?

 

[jeff1evesque]

What if instead of

<br /> Acos(\omega t + \theta) = 4sin(\omega t) + 3 cos(\omega t)<br />,

I wanted to write it as,

<br /> Asin(\omega t + \theta) = 4sin(\omega t) + 3 cos(\omega t)<br />,

would the values of A and \theta change at all?

Oh, actually I think I know. The x coordinate axis is Acos(\omega t) (negative for the negative x-axis), and the y coordinate axis is Asin(\omega t) (negative for the negative y-axis). So this means the left side value in the equality is the actual vector, and the terms on the right are the x and y components.

Thannks,JL

 

[turin]

You need addition and subtraction formulas for sine and cosine (or you can use Euler's identity, but that is more advanced).