# What is the amplitude, period and the phase shift of y=sin(θ - 45°)?

Dec 7, 2015

Given a generic trigonometric function like

$A \cos \left(\omega x + \phi\right) + k$, you have that:

1. $A$ affects the amplitude
2. $\omega$ affects the period via the relation $T = \frac{2 \setminus \pi}{\setminus} \omega$
3. $\phi$ is a phase shift (horizontal translation of the graph)
4. $k$ is a vertical translation of the graph.

In your case, $A = \omega = 1$, $\phi = - {45}^{\circ}$, and $k = 0$.

This means that the amplitude and the period remain untouched, while there is a shift phase of ${45}^{\circ}$, which means that your graph is shifted of ${45}^{\circ}$ to the right.