Half angle identity

[Albert Chan]

cos(105°) = sin(90-105°) = -sin(15°) = -√((1-cos(30°)/2) = -√((1-√3/2)/2)

We can remove nested square roots with identity (easily confirmed by squaring both side)
If x,y square root free, RHS (assumed ≥ 0) have no nested square roots.

\(\sqrt{2\;(x ± \sqrt{x^2-y^2})} = \sqrt{x+y}\;± \sqrt{x-y} \)

-√((1-√3/2)/2)
= -1/2 * √(2*(1 - √(1-1/4)))
= -1/2 * (√(1+1/2) - √(1-1/2))
= (-√6 + √2)/4
= -1/(√6 + √2)