How do you find the amplitude, period, and shift for #y=2cos(3x+pi/2)#?

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Answer 1 · 2016-02-15T10:24:11

Amplitude#=abs(2)=2#
Period #=(2pi)/3=2.0943951023932#
Shift #=-pi/6=-0.5235987755983#

Let the form be #y=a cos(bx-c)#

Amplitude #= abs(a)#

Period #=(2pi)/abs(b)#

Shift #=c/b#

Computations:

Arrange the equation first

#y = 2 cos (3x+pi/2)#

#y=2* cos (3x- -pi/2)#

Amplitude #= abs(a)=abs(2)=2#

Period #=(2pi)/abs(b)=(2pi)/abs(3)=(2pi)/3=2.0943951023932#

Shift #=c/b=(-pi/2)/3=-pi/6=-0.5235987755983#

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