# Sinusoid question

## 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

berkeman
Mentor

## 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

berkeman
Mentor
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?

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.

The Electrician
Gold Member
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.

Redbelly98
Staff Emeritus
Homework Helper
See the link berkeman gave in post #5. You eventually want to get things in the form

A cos(2t) + B sin(2t)​

To do that you'll need to use the addition identity for cosine,

cos(x+y) = ???​

Ok I am getting 5.1 which sounds right. Thanks everyone.