where x(t) = cos(2/3)t-(3/4)sin(2/3)t express x(t) in the form Acos((2/3)t+y)

Question

Hint says to use the formula 

cos(x+y)=cos(x) cos(y) - sin(x) sin(y)

Answer 1

Multiply and divide by the square root of the sum of the squares of the coefficients, and distribute the square root in the denominator through the sum:

 

√[1 + (-3/4)²] = √(25/16) = 5/4

 

x(t) = (5/4) {[1/(5/4)]cos(2/3)t - [(3/4)/(5/4)]sin(2/3)t}

 

Now consider a right triangle whose hypotenuse is 5/4, base is 1, and height is 3/4.  Let θ be the angle between the base and the hypotenuse.  Note that

 

cosθ = 1/(5/4) = 4/5 

sinθ = (3/4)/(5/4) = 3/5

 

Then

 

x(t) = (5/4) [cosθ cos(2/3)t - sinθ sin(2/3)t]

 

where 

 

θ = tan^-1[(3/5)/(4/5)] = tan^-1(3/4)

 

Now use the given identity and see that

 

x(t) = (5/4)cos[θ + (2/3)t] = (5/4)cos[(2/3)t + θ]

 

Then

 

A = 5/4

θ = tan^-1(3/4)