If si$n^4$θ + co$s^4$θ = 7/9, then si$n^8$θ + co$s^8$θ equals:

Write $p$ = sinθ·cosθ. Since $si$$n^4$θ + $co$$s^4$θ = $1-2p^2$, the given value 7/9 forces $p^2=1/9$. Squaring the fourth-power sum, $si$$n^8$θ + $co$$s^8$θ = $(si{n}^4\theta +co{s}^4\theta {)}^2-2(sin\theta \cdot cos\theta {)}^4=49/81-2p^4=49/81-2/81=47/81$ ≈ 0.580. Squaring and stopping there gives 49/81; the angle is θ ≈ 20.9°, where $0.3568^8+0.9342^8=0.5803$, confirming the value.