How do you find the period and amplitude of #y=1/4sin(2pix)#?

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Answer 1 · 2018-07-15T08:27:46

The period is #T=1#. The amplitude is #=1/4#

The period #T# of a periodic function #f(x)# is defined by

#f(x)=f(x+T)#

Here,

#f(x)=1/4sin(2pix)#............................#(1)#

Therefore,

#f(x+T)=1/4sin2pi(x+T)#

#=1/2sin(2pix+2piT)#

#=1/4sin(2pix)cos(2piT)+1/4cos(2pix)sin(2piT)#...........................#(2)#

Comparing equations #(1)# and #(2)#

#{(cos2piT=1),(sin2piT=0):}#

#=>#, #2piT=2pi#

The period is #T=1#

For the sine function

#-1<=sinx<=1#

#-1/4<=1/4sinx<=1/4#

#-1/4<=1/4sin2pix<=1/4#

The amplitude is #=1/4#

graph{1/4sin(2pix) [-0.587, 2.113, -0.696, 0.654]}