# Adding Sinusoids of differing Magnitute & Phases

1. Jul 7, 2009

### jeff1evesque

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
not sure.

3. 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

Last edited: Jul 7, 2009
2. Jul 7, 2009

### jeff1evesque

$$Acos(\omega t + \theta) = 4sin(\omega t) + 3 cos(\omega t)$$,

I wanted to write it as,

$$Asin(\omega t + \theta) = 4sin(\omega t) + 3 cos(\omega t)$$,

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

3. Jul 7, 2009

### jeff1evesque

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

4. Jul 7, 2009

### turin

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