Prove that Sin(6°) = (1/8)(- √5 - 1 + √(30 - 6√5))?

Question

Show work! Hurry!!!!!

This was a fun sideline. I should have asked, "What's the smallest integer n degrees such that Sin(n) is expressible with an algebraic number? Ans: 3. This is the reason why one cannot construct a nonagon, for example, using a compass and a ruler, because were it possible, then Sin(1) and Sin(2) would also have algebraic values.

Answer 1

I would complete the solution though the person above has mentioned the steps
we know

sin(36°)={ √ [10-2√5] } /4

cos(36°) = { (√5) + 1 } / 4

sin(30°)= 1/2

cos(30°) = (√3) / 2

sin(6°) = sin(36° - 30°) = sin(36°) cos (30°) - cos(36°) sin (30°)
= { √ [10-2√5] } /4 * (√3) / 2 - { (√5) + 1 } / 4 / 2
= { √ [30-6√5] } /8 - ( (√5) + 1 ) / 8
= 1/8(( √ (30-6√5) ) /8 - ( (√5) + 1 ))
= 1/8(-√5 - 1 + √ (30-6√5) )

rearanging the terms

Answer 2

if you know the values
sin(36°)={ √ [10-2√5] } /4

cos(36°) = { (√5) + 1 } / 4

you could go for

sin(6°) = sin(36° - 30°)

use sin(A-B) formula

substitute and simplify !