What is the amplitude and period of #y=3cos (1/2 x)#?

1 Answer
Apr 29, 2015

Consider the basic form:
#y = cos(theta)#
which has an amplitude of #1#
(#y# has a range of #[-1,+1]#)
and
a period of #2pi#
(#cos(0) " to " cos(2pi)# forms one recurring cycle)

In #3 cos(1/2 x)#
the coefficient #3# stretches the range of #y# to #[-3,+3]#
so its amplitude is #3#

The coefficient #1/2 " of "x# forces #x# to have to extend from
#0 " to " 4pi# to cover the set of argument values from #0 " to " 2pi#
so the period of #y=3cos(1/2x)# is #4pi#