The Real Law of Sines - A Proof in Two Parts


 , where r is the radius of the circumscribed circle.

Part I - Any inscribed angle in a circle is half the arc inscribed.

Given Circle O, and any three points A, B, and C.

Prove:


First, some constructions. Since AO, BO, and CO are radii, triangles AOB, AOC, and BOC are isosceles. Add sides AC and OB, and label angles 1, 2, and 3. The proof is easy now, based simply on each triangle having a total of 180o.

From Triangle ABC:

    

    (1)   Algebra

From Angle B:

    

      Substitution into (1)

    (2)   Algebra

Now, from Triangle AOC:

    

    (3)

And therefore,   Transitive with (2) and (3)

Part II - The Complete Law of Sines

Given any triangle ABC and the circle defined.

Prove:


First, some constructions, as usual. Since AO, BO, and CO are radii, triangles AOB, AOC, and BOC are (again) isosceles. Since the sides are arbitrary, just drop the one altitude h to side b at point D.

From Angle B:

          From Part I

         Isosceles triangle

    

From Side b:

          Isosceles triangle

    So,        Definition of Sine

         Algebra,

         and more alg.

Since the same steps hold for sides a and c,


The Complete Law of Sines has some interesting visual features.

One is the connection between sine and the distance to the circumscribed center from the midpoint of each side. As a side of the triangle approaches the center, its length approaches 2r, of course, so of course the value of sine approaches 1 as the side gets closer to the center and becomes a diameter.

Another feature is the meaning in the so-called "two-unit" circle, a circle with a diameter of 1. In a circle of this size, the actual lengths of the sides of an inscribed triangle are equal to the sine of the angle opposite that side. That is, a = sin A, b = sin B, and c = sin C. Very cool.