The result is entirely correct. Perhaps you're assuming that sin works in degrees? If that is the case, use sind instead of sin.
By the way, there is no reason to use a cell array for your example, a vector would make the code a lot simpler:
StartingValue(1).Next(1).x = [0.0025, 0.0075, 0.0125, 0.0175, 0.0225, 0.0275, 0.0325, 0.0375, 0.0425, 0.0475]
f = @(x) sin(5.1*pi.*x+0.5).^6;
StartingValue(1).Next(1).fn = f(StartingValue(1).Next(1).x)
Also note that the fn field name is misleading since its content is not a function but a vector.
