ECE 300 Help! Need to transform equations using trigonometric identities, (AC curcuit analysis)

Transform the following equations into the form x(t) = Xmcos(wt+theta) using the sum and difference trigonometric formulas

1: v1(t) = 1.5 cos(2πt) − 0.75 sin(2πt)

if someone could just help me get started, i looked up the sum and difference formulas but this problem doesn't seem to fit the format of the first question. There are a couple more after this one, but once i figure this one out i think the rest will just fall into place.

Note that in general cos(wt + Θ) = cos(wt)cosΘ - sin(wt)sinΘ. So the object is to use the coefficients that you have---1.5 and 0.75---and somehow get a cosΘ and a sinΘ out of them.

You can not just say cosΘ = 1.5 because that makes no sense (cosΘ can't be greater than 1.) Plus, whatever cosΘ and sinΘ are, they have to obey the "law" cos²Θ + sin²Θ = 1. Here is a technique that will work in these situations:

Set X_m = √(1.5² + 0.75²). Multiply and divide your expression by this factor.

v1(t) = X_m (1.5/X_m cos(2πt) - 0.75/X_m sin(2πt)).

Define Θ by the following equalities:

cosΘ = 1.5/X_m = 1.5/√(1.5² + 0.75²), and

sinΘ = 0.75/X_m = 0.75/√(1.5² + 0.75²).

Note that both of these numbers are less than 1, and if you square them and add them together you get 1. So these define a legitimate "angle" Θ. Since sinΘ > and cosΘ > 0, Θ is an acute angle. You can use the inverse sine function to approximate it.

Θ = arcsin(0.75/√(1.5² + 0.75²)) = arcsin(1/√(5)) ≈ 0.464

That is roughly 26.6°. Also

X_m = 3√(5)/4 ≈ 1.677.

This gives

v1(t) = 3√(5)/4 cos(2πt + Θ) where Θ = arcsin(1/√(5)).

This general approach will work. You can even write a function as sin(wt + Θ) in a similar way.