How do I find the period of the graph, f(x)=sqrt(sin^2(x/4)f(x)=sin2(x4), without graphing?

f(x)=sqrt(sin^2(x/4)f(x)=sin2(x4)

1 Answer
Jun 22, 2018

The period of ff is rho=4npiρ=4nπ for any integer nn.

Explanation:

Let rhoρ be the general period of ff. The principal period, rho_0ρ0 is defined the smallest positive period of ff.

rho=nrho_0ρ=nρ0, 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}