How do you determine the amplitude, period, and phase shift for #y=sin(x-pi/2)#?

1 Answer
Alan P.
Apr 25, 2015

#sin(theta)# has
an amplitude of #1#
a period of #2pi#
and a phase shift of #0#

The only effect of replacing #theta " with " (x-pi/2)#
is to shift the variable to the left by #pi/2#
(e.g.. #sin(x - pi/2) " with " x=pi/2#
is equivalent to #sin(theta) " with " theta=0#)

So
the amplitude remains #1#
the period remains #2pi#
and the phase shift is #-pi/2#

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