If sinθ + cosθ = $a$ and si$n^3$θ + co$s^3$θ = $b$, then a and $b$ always satisfy:

Squaring the first equation gives 1 + 2·sinθ·cosθ = $a^2$, so sinθ·cosθ = $(a^2-1)/2$. The sum of cubes factors as $b$ = (sinθ + cosθ)(1 − sinθ·cosθ) = $a[1-(a^2-1)/2]=a(3-a^2)/2$. Multiplying out, $2b=3a-a^3$, i.e. $a^3-3a+2b=0$. Check at θ = 37°: $a=1.400451$ and $b=0.727351$, and 2.7466 − 4.2014 + 1.4547 = 0.000. Sign slips in moving the terms across produce the other three relations, none of which survives that numerical test.