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

1 Answer
Apr 26, 2015

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] " )"#