# How do you determine the amplitude, period, and phase shift of the function y = 1/2 sin(x + pi)?

Apr 26, 2015

$y = \sin \left(\theta\right)$
which has:
an amplitude of $1$
a period of $2 \pi$
and a phase shift of $0$

$y = a \cdot \sin \left(\theta\right)$
The constant $a$ causes the values of $\sin \left(\theta\right)$ to be stretched by a factor of $a$.

$y = \sin \left(\theta + b\right)$
Shifts the values of $\theta$ needed to be equivalent to $\sin \left(0\right)$ to the left along the x-axis; that is $+ b$ causes a phase shift of $b$ to the left.

$y = \frac{1}{2} \sin \left(x + \pi\right)$
has an amplitude of $\frac{1}{2}$
and
a phase shift of $\pi$ (technically to the left, but a phase shift of $\pi$ is the same in both directions).

There is no change in the period; so the period remains $2 \pi$

(A change in period would be the result of a value $c$ in an equation of the form
$y = \sin \left(c \cdot \theta\right)$
which would cause $\theta \epsilon \left[0 , \frac{2 \pi}{c}\right] \text{ to cover the normal single period range of "[0,2pi] " )}$