How do you find the period, amplitude and sketch #y=1/100sin120pit#?

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Answer 1 · 2018-05-26T08:31:06

The period is #=1/60# and the amplitude is #=1/100#

We need

#sin(a+b)=sinacosb+sinbcosa#

The period of a periodic function is #T# iif

#f(t)=f(t+T)#

Here,

#f(t)=1/100sin(120pit)#

Therefore,

#f(t+T)=1/100sin(120pi(t+T))#

where the period is #=T#

So,

#1/100sin(120pit)=sin(120pi(t+T))#

#sin(120pit)=sin(120pit+120piT)#

#sin(120pit)=sin(120pit)cos(120piT)+cos(120pit)sin(120piT)#

Then,

#{(cos(120piT)=1),(sin(120piT)=0):}#

#<=>#, #120piT=2pi#

#<=>#, #T=1/60#

As

#-1<=sinx<=1#

Therefore,

#-1<=sin(120pit)<=1#

#-1/100<=1/100sin(120pit)<=1/100#

The amplitude is #=1/100=0.01#

See the graph below

graph{1/100sin(120pix) [-0.0374, 0.04145, -0.02057, 0.01883]}