sines law proof

The goal is to prove the sine law:

\\frac{\\sin A}{a}=\\frac{\\sin B}{b}=\\frac{\\sin C}{c}=\\frac{1}{2R}

where the variables are defined by the triangle

\\xy,(0,0);(40,0)**@{-};(60,30)**@{-};(0,0)**@{-},(50,12)*{a},(30,18)*{b},(20,-%3)*{c},(9,2)*{A},(39,3)*{B},(54,25)*{C}

and where $R$ is the radius of the circumcircle that encloses our triangle.

Let’s add a couple of lines and define more variables.

\\xy,(0,0);(40,0)**@{-};(60,30)**@{-};(0,0)**@{-},(50,12)*{a},(30,18)*{b},(20,-%3)*{c},(9,2)*{A},(39,3)*{B},(54,25)*{C},(40,0);(60,0)**@{--};(60,30)**@{--},(5%0,-3)*{x},(63,15)*{y}

So, we now know that

\\sin A=\\frac{y}{b}

and, therefore, we need to prove

\\frac{\\sin B}{b}=\\frac{y}{ba}

\\sin B=\\frac{y}{a}

From geometry, we can see that

\\sin\\left(\\pi-B\\right)=\\frac{y}{a}

So the proof is reduced to proving that

\\sin\\left(\\pi-B\\right)=\\sin B

This is easily seen astrue after examining the top half of the unit circle. So, putting allof our results together, we get

$\\displaystyle\\frac{\\sin A}{a}$	$\\displaystyle=$	$\\displaystyle\\frac{y}{ba}$
$\\displaystyle\\frac{\\sin A}{a}$	$\\displaystyle=$	$\\displaystyle\\frac{\\sin\\left(\\pi-B\\right)}{b}$
$\\displaystyle\\frac{\\sin A}{a}$	$\\displaystyle=$	$\\displaystyle\\frac{\\sin B}{b}$	(1)

The same logic may be followed to show that each of these fractions isalso equal to $\\frac{\\sin C}{c}$ .

For the final step of the proof, we must show that

2R=\\frac{a}{\\sin A}

We begin by defining our coordinate system. For this, it isconvenient to find one side that is not shorter than the others andlabel it with length $b$ . (The concept of a “longest” side is notwell defined in equilateral and some isoceles triangles, but there isalways at least one side that is not shorter than the others.) Wethen define our coordinate system such that the corners of thetriangle that mark the ends of side $b$ are at the coordinates $\\left(0,0\\right)$ and $\\left(b,0\\right)$ . Our third corner (withsides labelled alphbetically clockwise) is at the point $\\left(c\\cos A,c\\sin A\\right)$ . Let the center of our circumcircle be at $\\left(x_{0},y_{0}\\right)$ . We now have

$\\displaystyle x_{0}^{2}+y_{0}^{2}$	$\\displaystyle=$	$\\displaystyle R^{2}$	(2)
$\\displaystyle\\left(b-x_{0}\\right)^{2}+y_{0}^{2}$	$\\displaystyle=$	$\\displaystyle R^{2}$	(3)
$\\displaystyle\\left(c\\cos A-x_{0}\\right)^{2}+\\left(c\\sin A-y_{0}\\right)^{2}$	$\\displaystyle=$	$\\displaystyle R^{2}$	(4)

as each corner of our triangle is, by definition of the circumcircle,a distance $R$ from the circle’s center.

Combining equations (3) and (2), we find

$\\displaystyle\\left(b-x_{0}\\right)^{2}+y_{0}^{2}$	$\\displaystyle=$	$\\displaystyle x_{0}^{2}+y_{0}^{2}$
$\\displaystyle b^{2}-2bx_{0}$	$\\displaystyle=$	$\\displaystyle 0$
$\\displaystyle\\frac{b}{2}$	$\\displaystyle=$	$\\displaystyle x_{0}$

Substituting this into equation (2) we find that

y_{0}^{2}=R^{2}-\\frac{b^{2}}{4}

(5)

Combining equations (4) and (5) leaves us with

$\\displaystyle\\left(c\\cos A-x_{0}\\right)^{2}+\\left(c\\sin A-y_{0}\\right)^{2}$	$\\displaystyle=$	$\\displaystyle x_{0}^{2}+y_{0}^{2}$
$\\displaystyle c^{2}\\cos^{2}A-2x_{0}c\\cos A+c^{2}\\sin^{2}A-2y_{0}c\\sin A$	$\\displaystyle=$	$\\displaystyle 0$
$\\displaystyle c-2x_{0}\\cos A-2y_{0}\\sin A$	$\\displaystyle=$	$\\displaystyle 0$
$\\displaystyle\\frac{c-b\\cos A}{2\\sin A}$	$\\displaystyle=$	$\\displaystyle y_{0}$
$\\displaystyle\\frac{\\left(c-b\\cos A\\right)^{2}}{4\\sin^{2}A}$	$\\displaystyle=$	$\\displaystyle R^{2}-\\frac{b^{2}}{4}$
$\\displaystyle\\left(c-b\\cos A\\right)^{2}+b^{2}\\sin^{2}A$	$\\displaystyle=$	$\\displaystyle 4R^{2}\\sin^{2}A$
$\\displaystyle c^{2}-2bc\\cos A+b^{2}$	$\\displaystyle=$	$\\displaystyle 4R^{2}\\sin^{2}A$
$\\displaystyle a^{2}$	$\\displaystyle=$	$\\displaystyle 4R^{2}\\sin^{2}A$
$\\displaystyle\\frac{a}{\\sin A}$	$\\displaystyle=$	$\\displaystyle 2R$

where we have applied the cosines law in the second to last step.