How do you determine the amplitude and period for #y = -1/2 cos(pi/3)x#?

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Answer 1 · 2018-06-20T17:15:13

The amplitude is #=1/2# and the period is #=6#

The amplitude is #=1/2# as #-1<=sinx<=1#

This a periodic function such that

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

Therefore,

#-1/2cos(pi/3x)=-1/2cos(pi/3(x+T))#

#=-1/2cos(pi/3x+pi/3T)#

#=-1/2(cos(pi/3x)cos(pi/3T)-sin(pi/3x)sin(pi/3T))#

Therefore,

#{(cos(pi/3T)=1),(sin(pi/3T)=0):}#

#<=>#, #pi/3T=2pi#

#<=>#, #T=6#

The period is #T=6#

graph{-1/2cos(pi/3x) [-4.05, 11.75, -3.027, 4.875]}