Y= radical sign (| sinx |+| cosx |) 2 = radical sign (1+|sin2x|)

= radical sign {1+ radical sign [( 1-cos4x)/2]}

So the minimum positive period = 2pi/4 = pi/2.

Another definition can be used to prove that the minimum positive period =pi/2.

Because f (x+pi/2) = | sin (x+pi/2) | +| cos (x+pi/2) |

=|cosx|+|sinx|=f(x)

It shows that π/2 is the period of a function.

Suppose π/2-t(0

Then f (x+pi/2-t) = | sin (x+pi/2-t) | +| cos (x+pi/2-t) |.

= | cos(x-t)|+| sin(x-t)| = f(x)= | cosx |+| sinx |

So |sin(2x-2t)|=|sin2x|

Therefore | sin (2x-2t) | 2 = | sin2x | 2.

So [1-cos (4x-4t)]/2 = [1-cos4x]/2.

That is, cos(4x-4t)=cos4x, which is true for any X ..

So take x=pi/4, and get cos(pi-4t)=cospi=- 1, that is, cos4t= 1.

Another 0

Therefore, π/2 is a period of the function and the smallest positive period.