Converting the final result of a trigonometric identity back into its original form

[PainterGuy]

TL;DR Summary

I was working on a trigonometric identity and out of curiosity wanted to see how one could go back to the original expression.

Hi,

K₁cos(θt+φ)=K₁cos(θt)cos(φ)-K₁sin(θt)sin(φ)=K₁K₂cos(θt)-K₁K₃sin(θt)

Let's assume φ=30° , K₁=5
5cos(θt+30°) = 5cos(θt)cos(30°)-5sin(θt)sin(30°) = (5)0.866cos(θt)-(5)0.5sin(θt) = 4.33cos(θt)-2.5sin(θt)

If only the final result, 4.33cos(θt)-2.5sin(θt), is given, how do I find the original φ=30° and K₁=5? Or, how do I convert the final result back into 5cos(θt+30°) without knowing any intermediate steps?

Could you please help me with it? Thank you.

Note to self:
sin(A+90°)=sinAcos(90°)+cosAsin90°=cosA therefore cosA=sin(A+90°)
or, cos(A+90°)=cosAcos90°-sinAsin90°=-sinAsin90°

 

[mfb]

From your derivation you can see that $4.33 = K_1 \cos(\phi)$ and $-2.5=-K_1 \sin(\phi)$. This is a system with two equations and two unknowns that you can solve for φ and K₁. The equations are nonlinear but you can get a result by dividing them (for phi) and by looking at the sum of squares (for K₁).

 

[PainterGuy]

Thank you.

I hope it's correct. Could you please give it a look?

 

[mfb]

Looks good, and the result is correct of course.

One caveat: There is a bit of ambiguity. An angle that is larger/smaller by 180 degree and the opposite sign for K1 at the same time lead to the same result, and changing the angle by 360 degree doesn't impact the result either, so you can't notice that. Usually this isn't a problem in applications, but it is something to keep in mind.