sine of angle of triangle


The cosines law allows to express the cosine of an angle of triangle through the sides:

cosα=b2+c2-a22bc. (1)

Substituting this to the “fundamental formula of trigonometryMathworldPlanetmath”,

sin2α+cos2α= 1,

we can calculate as follows:

sinα =+1-(b2+c2-a22bc)2
=(2bc)2-(b2+c2-a2)2(2bc)2
=(2bc+b2+c2-a2)(2bc-b2-c2+a2)2bc
=[(b+c)2-a2][a2-(b-c)2]2bc
=(b+c+a)(b+c-a)(a+b-c)(a-b+c)2bc

Thus we have the beautiful formula

sinα=(-a+b+c)(a-b+c)(a+b-c)(a+b+c)2bc.

Substituting (1) similarly to the general formula for the sine of half-angle (http://planetmath.org/GoniometricFormulae)

sinα2=±1-cosα2,

one can obtain the formula

sinα2=(a-b+c)(a+b-c)4bc.
Title sine of angle of triangle
Canonical name SineOfAngleOfTriangle
Date of creation 2013-03-22 18:27:16
Last modified on 2013-03-22 18:27:16
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 5
Author pahio (2872)
Entry type Derivation
Classification msc 51M04
Related topic DifferenceOfSquares