A simple rule for the simple sines

[kpreid]

It is an often-used fact (in examples of elementary trigonometry and problems that involve it) that the sines of certain simple angles have simple expressions themselves. Specifically,

sin(0)			=	0
sin(π/6)	=	sin(30°)	=	1/2
sin(π/4)	=	sin(45°)	=	√(2)/2
sin(π/3)	=	sin(60°)	=	√(3)/2
sin(π/2)	=	sin(90°)	=	1

Furthermore, these statements about angles in the first quadrant can be reflected to handle similar common angles in the other three quadrants, and cosines. There is a common diagram illustrating this, considering sine and cosine as the coordinates of points on the unit circle.

When attempting to memorize these facts as one is expected to, I observed a pattern: each of the values may be expressed in the form √(x)/2.

sin(0)			=	√(0)/2
sin(π/6)	=	sin(30°)	=	√(1)/2
sin(π/4)	=	sin(45°)	=	√(2)/2
sin(π/3)	=	sin(60°)	=	√(3)/2
sin(π/2)	=	sin(90°)	=	√(4)/2

I found this pattern to be quite useful. It does not, however, explain why those particular angles (quarters, sixths, eighths, and twelfths — but not tenths! — of circles) form this pattern of sines.

Comments

[(Anonymous)]

You should ask at http://math.stackexchange.com/ ; someone might know :)

[byorgey.wordpress.com (from livejournal.com)]

I showed this to my students when I used to teach high school pre-calculus, it's a cute way to remember those facts! As to why those particular angles show up, I don't have a great explanation for you, but there is definitely some deep structure surfacing here, related to finite reflection groups. It is related to the fact that the "special triangles" with the angles in question (45-45-90 and 30-60-90) are precisely those that constitute a fundamental domain for one of the three regular tilings of the plane (45-45-90 corresponds to the square tiling, and 30-60-90 to the equilateral triangle and regular hexagonal tilings).

[kragen.livejournal.com]

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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