What is the period of #f(t)=sin( ( 3t)/4 )#?

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Answer 1 · 2018-05-26T08:04:00

The period is #=8/3pi#

We need

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

The period of a periodic function is #T# iif

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

Here,

#f(t)=sin(3/4t)#

Therefore,

#f(t+T)=sin(3/4(t+T))#

where the period is #=T#

So,

#sin(3/4t)=sin(3/4(t+T))#

#sin(3/4t)=sin(3/4t+3/4T)#

#sin(3/4t)=sin(3/4t)cos(3/4T)+cos(3/4t)sin(3/4T)#

Then,

#{(cos(3/4T)=1),(sin(3/4T)=0):}#

#<=>#, #3/4T=2pi#

#<=>#, #T=8/3pi#