Question

Please explain the sine rule for a triangle? (Its derivation and explanation in short)

Answer 1

The Law of Sines is one of the laws of Trigonometry. It equates three ratios of a triangle, each the quotient of the length of a side to the sine of its opposite vertex.

The Law of Sines and the Law of Cosines are the two major tools for solving triangles. The Law of Sines involves two sides and two angles; when we're missing one we can use the Law of Sines to get it. The Law of Cosines similarly involves three sides and one angle.

When the goal is to determine an angle, it's generally preferable to use the Law of Cosines if possible. The cosine uniquely determines a triangle angle. By contrast, there will in general be two triangle angles, supplementary angles, one acute, one obtuse, with the same sine.

It's straightforward to derive all the Laws of Trigonometry, including the Law of Sines, from the Law of Cosines.

`c^2 = a^2 + b^2 - 2 a b cos C`

`(c^2-a^2-b^2)^2= 4 a^2 b^2 cos^2 C = 4 a^2 b^2 (1 - sin ^2 C)`

`4 a^2 b^2 sin ^2 C = 4a^2 b^2 - (c^2-a^2-b^2)^2`

By the way, both sides are 16 times the squared area of the triangle here. Focus, Dean.

`4 a^2 b^2 sin ^2 C = 4a^2b^2 - ( c^4+a^4+b^4 - 2(a^2c^2+b^2c^2- a^2b^2 ) )`

`4 a^2 b^2 c^2 (sin^2 C/c^2) =2(a^2b^2 + a^2c^2+b^2c^2) - ( a^4+b^4+c^ 4)`

`sin^2 C/c^2 = { 2(a^2b^2 + a^2c^2+b^2c^2) - ( a^4+b^4+c^ 4) } / { 4 a^2 b^2 c^2 }`

The right hand side is completely symmetric with respect to interchange of the sides. That means we'd get the same right side if we'd started the Law of Cosines `a^2 =...` or `b^2 = ...`. So we conclude

`sin^2 A/a^2 = sin^2 B/b^2 = sin^2 C/c^2`

For triangle angles the sines are always positive, so we can safely equate the square roots and the proof is done.

It doesn't really matter that we have the sines in the numerator here; both ways work.