For every angle θ, the value of 3(si$n^4$θ + co$s^4$θ) − 2(si$n^6$θ + co$s^6$θ) is:

With $p$ = sinθ·cosθ the two standard reductions are si$n^4$θ + co$s^4$θ = $1-2p^2$ and si$n^6$θ + co$s^6$θ = $1-3p^2$. The expression is therefore $3(1-2p^2)-2(1-3p^2)=3-6p^2-2+6p^2=1$: the $p^2$ terms cancel, so the value never depends on θ at all. Check at θ = 45°: $3(1/2)-2(1/4)=1$. Adding the two outside coefficients suggests 5, and using $1-3p^2$ for the fourth powers as well would give 0.