(In reply to

proof found by Steven Lord)

Well, lets just skip the grunt work - deriving the formula for sin(5x) = poly in sin(x).

We can look that up; one place is https://mathworld.wolfram.com/Multiple-AngleFormulas.html . {For brevity let S=sin(x)} sin(5x) = 5S-20S^3+16S^5.

Now onto the actual problem: evaluate this at x=18 degrees.

Then 1 = 5S-20S^3+16S^5.

Then subtract 1 from each side and factor: (S-1)*(4S^2+2S-1)^2 = 0.

We know sin(18) = S != 1, so S must be one of the roots of 4S^2+2S-1 = 0.

The roots are S=(sqrt(5)-1)/4 and S=(-sqrt(5)-1)/4. The second one is negative so it is rejected since we know that sin(18) is between 0 and 1. Therefore sin(18) = (sqrt(5)-1)/4.

Now we can do what K Sengupta did and just directly evaluate the expression S^3+S^2 at (sqrt(5)-1)/4 to get 1/8.