How do you find the amplitude and period of #y=cos2x#?

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Answer 1 · 2018-05-28T05:59:36

The period is #T=pi#.
The amplitude is #=1#

We need

#cos(a+b)=cosacosb-sinasinb#

The period of a periodic function is #T# iif

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

Here,

#f(x)=cos(2x)#

Therefore,

#f(x+T)=cos(2(x+T))#

where the period is #=T#

So,

#cos(2x)=cos(2(x+T))=cos(2x+2T)=cos2xcos2T-sin2xsin2T#

#=>#, #{(cos2T=1),(sin2T=0):}#

#=>#, #2T=2pi#

#=>#, #T=pi#

As

#-1<=cosx<=1#

Therefore,

#-1<=cos(2x)<=1#

The amplitude is #=1#

graph{cos(2x) [-1.322, 9.777, -2.333, 3.214]}