How do I find the period of the graph, #f(x)=sqrt(sin^2(x/4)#, without graphing?

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#f(x)=sqrt(sin^2(x/4)#

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Answer 1 · 2018-06-22T22:23:43

The period of #f# is #rho=4npi# for any integer #n#.

Let #rho# be the general period of #f#. The principal period, #rho_0# is defined the smallest positive period of #f#.

#rho=nrho_0#, #n in ZZ#.

We ask what is the period of:

#f(x)=sqrt(sin^2(x/4))=|sin(x/4)|#

Basically, this means solving the equality below for #rho_0#:

#f(x+rho_0)=f(x)#

#|sin((x+rho_0)/4)| = |sin(x/4)|#

If #|a|=|b|#, then #a=b or a = -b#.

1. #sin((x+rho_0)/4) = sin(x/4)#

#sin(x/4+rho_0/4) = sin(x/4)#

#sin(x/4)cos(rho_0/4)+cos(x/4)sin(rho_0/4)=sin(x/4)#

#=> {(cos(rho_0/4) =1),(sin(rho_0/4)=0) :}#

With lowest positive value #rho_0 = 8pi#.

2. #sin((x+rho_0)/4) = -sin(x/4)#

#sin(x/4)cos(rho_0/4)+cos(x/4)sin(rho_0/4)=-sin(x/4)#

#=> {(cos(rho_0/4)=-1),(sin(rho_0/4)=0) :}#

This time, the lowest positive value is #rho_0=4pi#.

As such, the principal period is actually

#rho_0=4pi#

With the set of periods

#cc P={4npi | n in ZZ}#