Question

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

Answer 1

Let's start with the more basic form:
`y = sin(theta)`
which has:
an amplitude of `1`
a period of `2pi`
and a phase shift of `0`

`y=a*sin(theta)`
The constant `a` causes the values of `sin(theta)` to be stretched by a factor of `a`.

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

`y= 1/2sin(x+pi)`
has an amplitude of `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 `2pi`

(A change in period would be the result of a value `c` in an equation of the form
`y = sin(c*theta)`
which would cause `theta epsilon [0,(2pi)/c]" to cover the normal single period range of "[0,2pi] " )"`