Long Journey of CosSin

First, note the factorization $f(x) = (\sin^2x+\cos^2x)(\sin^4x+\cos^4x) =\sin^4x+\cos^4x.$ Then, this becomes $(\sin^2x+\cos^2x)^2 - 2\sin^2 x\cos^2 x = 1-\frac{1}{2}\sin^2(2x).$ We can graph $\sin^2(2x)$ to see that it has a period of $\frac{\pi}{2}.$

Alternatively, we can think of the fact that $\sin(2x)$ has a period of $\pi$ (one trip around the unit circle), but since we are squaring it, the half of the unit circle where sine is negative is the same as the positive half, so the period is half of $\pi.$

We can find that f(x)is a symmetric polynomial （If we change the position of the sin(x) and the position of cos(x)， the equation we get is as same as the old one).And all terms are even power,which means we dont need to care about whether the term is positive or not.Since sin(x+2/π)=cos(x) and cos(x+2/π)=sin x ,the fundamental period of f(x) is 2/π.