2 sin (θ/2) has a geometric meaning: it's the length of the base of an isosceles triangle whose unit-length legs are separated by θ, or equivalently the length of a chord of the unit circle whose endpoints are separated by θ, which to me feels like a more natural function than one based on right triangles. That makes the results be √0̅, √1̅, √2̅, √3̅, √4̅, and the angles in question become 0, π/3, π/2, π/1½, and π/1.
The relationships between the π/6, π/4 pair and the π/3, π/2 pair are of course a consequence of the half-angle or double-angle identity — sin 2θ = 2 sin θ cos θ — and the relationship between sin and cos, namely cos θ = √(̅1̅ ̅-̅ ̅s̅i̅n̅²̅ ̅θ̅)̅. Substituting, sin 2θ = 2 sin θ √(̅1̅ ̅-̅ ̅s̅i̅n̅²̅ ̅θ̅)̅. In the case where θ = π/4 and sin θ = √2̄/2, we have sin 2θ = 2 √2̄/2 √(̅1̅ ̅-̅ ̅(̄√̄2̄̄/̄2̄)̅²̅)̅ = 2 √2 √(̅1̅-̅2̅/̅4̅)̅ = √2̅ √½̅ = √2̅ √2̅/2 = 2/2 = 1. I don't know that this sheds any real light on the topic, though...
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no subject
The relationships between the π/6, π/4 pair and the π/3, π/2 pair are of course a consequence of the half-angle or double-angle identity — sin 2θ = 2 sin θ cos θ — and the relationship between sin and cos, namely cos θ = √(̅1̅ ̅-̅ ̅s̅i̅n̅²̅ ̅θ̅)̅. Substituting, sin 2θ = 2 sin θ √(̅1̅ ̅-̅ ̅s̅i̅n̅²̅ ̅θ̅)̅. In the case where θ = π/4 and sin θ = √2̄/2, we have sin 2θ = 2 √2̄/2 √(̅1̅ ̅-̅ ̅(̄√̄2̄̄/̄2̄)̅²̅)̅ = 2 √2 √(̅1̅-̅2̅/̅4̅)̅ = √2̅ √½̅ = √2̅ √2̅/2 = 2/2 = 1. I don't know that this sheds any real light on the topic, though...
This Unicode abuse is brought to you by my compose key (http://canonical.org/~kragen/setting-up-keyboard.html) and my keyboard map (https://github.com/kragen/xcompose).