# Trigonometric equation

1. Apr 13, 2014

### alejandrito29

Is correct the following procediment?

$A \sin (\omega t) = A \sin (\phi) \to \phi= \sin^{-1} \sin (\omega t )$

Is correct to say that $\phi = \omega t$ is oscillatory in this case ??

2. Apr 13, 2014

### Staff: Mentor

You don't need inverse trig functions here. We can ignore A, assuming that it is a nonzero constant.
If sin(ωt) = sin(θ), then θ = ωt + k($2\pi$), where k is an integer.

3. Apr 13, 2014

### HallsofIvy

Staff Emeritus
But you certainly cannot "say that $\phi= \omega t$ is oscillatory"! It is the function $sin(\phi)$ that is oscillatory, not just the argument, [itex]\phi[itex].