Solve for Amplitude and Phase of a Sinusoidal Function | Homework Help

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Homework Statement


Find the amplitude and phase of the sinusoid (4√3-3)cos(2t + 30°) + (3√3 – 4)cos(2t + 60°)


Homework Equations


Acos wt + Bsin wt


The Attempt at a Solution


Answer should be amplitude 5 and phase 36.9° but I get 4.11 and 5.7
 
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ecas said:

Homework Statement


Find the amplitude and phase of the sinusoid (4√3-3)cos(2t + 30°) + (3√3 – 4)cos(2t + 60°)


Homework Equations


Acos wt + Bsin wt


The Attempt at a Solution


Answer should be amplitude 5 and phase 36.9° but I get 4.11 and 5.7

Welcome to the PF. Please show us your work, so that we can try to help you. That's how it works here on the PF.
 
The square root of (4√3-3)^2 + (3√3 – 4)^ 2 = 4.11 and θ = tan^-1 (4√3-3)/(3√3 – 4) = 16.9 degrees
 
ecas said:
The square root of (4√3-3)^2 + (3√3 – 4)^ 2 = 4.11 and θ = tan^-1 (4√3-3)/(3√3 – 4) = 16.9 degrees

Not sure what that's meant to represent. You have two sinusoidal functions of time, which can be added to make a new sinusiodal function, with a combined amplitude, and a combined resultant phase. It should look something like this:

f(t) = A cos(2t + phase)

What would be a way that you can add two cosine functions that have different phases?
 
Ah, I see the way that you are wanting to solve this:

http://www.cs.sfu.ca/~tamaras/sinusoids318/Adding_two_sinusoids.html

Are you sure you have the original equations in the right format to apply the equations?
 
I tried changing the second function from cos to sin to match the identity but this does not seem to affect the magnitude in any way. I still do not know how to arrive at A=5.
 
Does this have anything to do with the study of electronics? If so, have you studied phasor arithmetic yet?
 
Actually the problem comes before the section on phasors. Supposedly it is supposed to be solved with the trig identities we are given.
 
Ok I am getting 5.1 which sounds right. Thanks everyone.