Here’s a simple calculation that I’ve done often enough that I’d like to save the result for my future reference and for the benefit of anyone searching on this.
A linear combination of sines and cosines
a sin(x) + b cos(x)
can be written as a sine with a phase shift
A sin(x + φ).
Going between {a, b} and {A, φ} is the calculation I’d like to save. For completeness I also include the case
A cos(x + ψ).
Derivation
Define
f(x) = a sin(x) + b cos(x)
and
g(x) = A sin(x + φ).
Both functions satisfy the differential equation
y″ + y = 0
and so f = g if and only if f(0) = g(0) and f′(0) = g′(0).
Setting the values at 0 equal implies
b = A sin(φ)
and setting the derivatives at 0 equal implies
a = A cos(φ).
Taking the ratio of these two equations shows
b/a = tan(φ)
and adding the squares of both equations shows
a² + b² = A².
Equations
First we consider the case
a sin(x) + b cos(x) = A sin(x + φ).
Sine with phase shift
If a and b are given,
A = √(a² + b²)
and
φ = tan−1(b / a).
If A and φ are given,
a = A cos(φ)
and
b = A sin(φ)
from the previous section.
Cosine with phase shift
Now suppose we want
a sin(x) + b cos(x) = A cos(x + ψ)
If a and b are given, then
A = √(a² + b²)
as before and
ψ = − tan−1(a / b).
If A and ψ are given then
a = − A sin(ψ)
and
b = A cos(ψ).
Interesting to see your thinking on this one. I think the usual way is via a trig identity rather than the differential equation.