If f(x)=sin2(pi/8+x/2) -sin2(pi/8-x/2),then the period of f is

We are given the function:

f(x) = sin(2(π/8 + x/2)) - sin(2(π/8 - x/2))

Step 1: Simplify the Angles
Expanding the angles:

f(x) = sin(π/4 + x) - sin(π/4 - x)

Using the identity:

sin(A) - sin(B) = 2 cos((A+B)/2) sin((A-B)/2)

where A = (π/4 + x) and B = (π/4 - x), we get:

(A + B)/2 = [(π/4 + x) + (π/4 - x)] / 2 = (π/2) / 2 = π/4

(A - B)/2 = [(π/4 + x) - (π/4 - x)] / 2 = (2x) / 2 = x

Thus,

f(x) = 2 cos(π/4) sin(x)

Since cos(π/4) = 1/√2, we get:

f(x) = (2/√2) sin(x) = √2 sin(x)

Step 2: Find the Period
The function reduces to f(x) = √2 sin(x), which is a simple sine function with a period of 2π.

Thus, the period of f(x) is 2π.