If si$n^4$θ + co$s^4$θ = 5/8 and θ is acute, then sinθ·cosθ equals:

$si$$n^4$θ + $co$$s^4$θ = 1 − 2$(sin\theta \cdot cos\theta {)}^2$, so $1-2p^2=5/8$ where $p$ = sinθ·cosθ. Then $p^2=3/16$ and, since θ is acute, $p$ = √3/4 (about 0.433, which is legal because sinθ·cosθ never exceeds 1/2).