Laws of rational trigonometry Rational trigonometry
1 laws of rational trigonometry
1.1 triple quad formula
1.2 pythagoras s theorem
1.3 spread law
1.4 cross law
1.5 triple spread formula
laws of rational trigonometry
wildberger states there 5 basic laws in rational trigonometry. states these laws can verified using high-school level mathematics. equivalent standard trigonometrical formulae variables expressed quadrance , spread.
in following 5 formulae, have triangle made of 3 points a1, a2, a3. spreads of angles @ points s1, s2, s3, , q1, q2, q3, quadrances of triangle sides opposite a1, a2, a3, respectively. in classical trigonometry, if know 3 of 6 elements s1, s2, s3, q1, q2, q3, , these 3 not 3 s, can compute other three.
triple quad formula
the 3 points a1, a2, a3 collinear if , if:
(
q
1
+
q
2
+
q
3
)
2
=
2
(
q
1
2
+
q
2
2
+
q
3
2
)
{\displaystyle (q_{1}+q_{2}+q_{3})^{2}=2\left(q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\right)}
where q1, q2, q3 represent quadrances between a1, a2, a3 respectively. can either proved analytic geometry (the preferred means within rational trigonometry) or derived heron s formula, using condition collinearity triangle formed 3 points has 0 area.
illustration of nomenclature used in proof.
the line ab has general form:
a
x
+
b
y
+
c
=
0
{\displaystyle ax+by+c=0}
where (non-unique) parameters a, b, c can expressed in terms of coordinates of points , b as:
a
=
a
y
−
b
y
b
=
b
x
−
a
x
c
=
a
x
b
y
−
a
y
b
x
{\displaystyle {\begin{aligned}a&=a_{y}-b_{y}\\b&=b_{x}-a_{x}\\c&=a_{x}b_{y}-a_{y}b_{x}\end{aligned}}}
so that, everywhere on line:
(
a
y
−
b
y
)
x
+
(
b
x
−
a
x
)
y
+
(
a
x
b
y
−
a
y
b
x
)
=
0.
{\displaystyle \left(a_{y}-b_{y}\right)x+\left(b_{x}-a_{x}\right)y+\left(a_{x}b_{y}-a_{y}b_{x}\right)=0.}
but line can specified 2 simultaneous equations in parameter t, t = 0 @ point , t = 1 @ point b:
x
=
(
b
x
−
a
x
)
t
+
a
x
,
y
=
(
b
y
−
a
y
)
t
+
a
y
,
{\displaystyle {\begin{aligned}x&=(b_{x}-a_{x})t+a_{x},\\y&=(b_{y}-a_{y})t+a_{y},\end{aligned}}}
or, in terms of original parameters:
x
=
b
t
+
a
x
,
y
=
−
a
t
+
a
y
.
{\displaystyle {\begin{aligned}x&=bt+a_{x},\\y&=-at+a_{y}.\end{aligned}}}
if point c collinear points , b, there exists value of t (for distinct points, not equal 0 or 1), call λ, these 2 equations simultaneously satisfied @ coordinates of point c, such that:
c
x
=
b
λ
+
a
x
.
c
y
=
−
a
λ
+
a
y
.
{\displaystyle {\begin{aligned}c_{x}&=b\lambda +a_{x}.\\c_{y}&=-a\lambda +a_{y}.\end{aligned}}}
now, quadrances of 3 line segments given squared differences of coordinates, can expressed in terms of λ:
q
(
a
b
)
≡
(
b
x
−
a
x
)
2
+
(
b
y
−
a
y
)
2
=
b
2
+
(
−
a
)
2
=
a
2
+
b
2
q
(
b
c
)
≡
(
c
x
−
b
x
)
2
+
(
c
y
−
b
y
)
2
=
(
(
b
λ
+
a
x
)
−
b
x
)
2
+
(
(
−
a
λ
+
a
y
)
−
b
y
)
2
=
(
b
λ
+
(
a
x
−
b
x
)
)
2
+
(
−
a
λ
+
(
a
y
−
b
y
)
)
2
=
(
b
λ
+
(
−
b
)
)
2
+
(
−
a
λ
+
a
)
2
=
b
2
(
λ
−
1
)
2
+
a
2
(
−
λ
+
1
)
2
=
b
2
(
λ
−
1
)
2
+
a
2
(
λ
−
1
)
2
=
(
a
2
+
b
2
)
(
λ
−
1
)
2
q
(
a
c
)
≡
(
c
x
−
a
x
)
2
+
(
c
y
−
a
y
)
2
=
(
(
b
λ
+
a
x
)
−
a
x
)
2
+
(
(
−
a
λ
+
a
y
)
−
a
y
)
2
=
(
b
λ
+
a
x
−
a
x
)
2
+
(
−
a
λ
+
a
y
−
a
y
)
2
=
(
b
λ
)
2
+
(
−
a
λ
)
2
=
b
2
λ
2
+
(
−
a
)
2
λ
2
=
b
2
λ
2
+
a
2
λ
2
=
(
a
2
+
b
2
)
λ
2
{\displaystyle {\begin{aligned}q(ab)&\equiv (b_{x}-a_{x})^{2}+(b_{y}-a_{y})^{2}\\&=b^{2}+(-a)^{2}\\&=a^{2}+b^{2}\\[10pt]q(bc)&\equiv (c_{x}-b_{x})^{2}+(c_{y}-b_{y})^{2}\\&={\bigl (}(b\lambda +a_{x})-b_{x}{\bigr )}^{2}+{\bigl (}(-a\lambda +a_{y})-b_{y}{\bigr )}^{2}\\&={\bigl (}b\lambda +(a_{x}-b_{x}){\bigr )}^{2}+{\bigl (}-a\lambda +(a_{y}-b_{y}){\bigr )}^{2}\\&={\bigl (}b\lambda +(-b){\bigr )}^{2}+(-a\lambda +a)^{2}\\&=b^{2}(\lambda -1)^{2}+a^{2}(-\lambda +1)^{2}\\&=b^{2}(\lambda -1)^{2}+a^{2}(\lambda -1)^{2}\\&=\left(a^{2}+b^{2}\right)(\lambda -1)^{2}\\[10pt]q(ac)&\equiv (c_{x}-a_{x})^{2}+(c_{y}-a_{y})^{2}\\&={\bigl (}(b\lambda +a_{x})-a_{x}{\bigr )}^{2}+{\bigl (}(-a\lambda +a_{y})-a_{y}{\bigr )}^{2}\\&=(b\lambda +a_{x}-a_{x})^{2}+(-a\lambda +a_{y}-a_{y})^{2}\\&=(b\lambda )^{2}+(-a\lambda )^{2}\\&=b^{2}\lambda ^{2}+(-a)^{2}\lambda ^{2}\\&=b^{2}\lambda ^{2}+a^{2}\lambda ^{2}\\&=\left(a^{2}+b^{2}\right)\lambda ^{2}\end{aligned}}}
where use made of fact (−λ + 1) = (λ − 1).
substituting these quadrances equation proved:
(
q
(
a
b
)
+
q
(
b
c
)
+
q
(
a
c
)
)
2
=
2
(
q
(
a
b
)
2
+
q
(
b
c
)
2
+
q
(
a
c
)
2
)
(
(
a
2
+
b
2
)
+
(
a
2
+
b
2
)
(
λ
−
1
)
2
+
(
a
2
+
b
2
)
λ
2
)
2
=
2
(
(
a
2
+
b
2
)
2
+
(
(
a
2
+
b
2
)
(
λ
−
1
)
2
)
2
+
(
(
a
2
+
b
2
)
λ
2
)
2
)
(
a
2
+
b
2
)
2
(
1
+
(
λ
−
1
)
2
+
λ
2
)
2
=
2
(
a
2
+
b
2
)
2
(
1
+
(
(
λ
−
1
)
2
)
2
+
(
λ
2
)
2
)
{\displaystyle {\begin{aligned}{\bigl (}q(ab)+q(bc)+q(ac){\bigr )}^{2}&=2\left(q(ab)^{2}+q(bc)^{2}+q(ac)^{2}\right)\\\left(\left(a^{2}+b^{2}\right)+\left(a^{2}+b^{2}\right)(\lambda -1)^{2}+\left(a^{2}+b^{2}\right)\lambda ^{2}\right)^{2}&=2\left(\left(a^{2}+b^{2}\right)^{2}+\left(\left(a^{2}+b^{2}\right)(\lambda -1)^{2}\right)^{2}+\left(\left(a^{2}+b^{2}\right)\lambda ^{2}\right)^{2}\right)\\\left(a^{2}+b^{2}\right)^{2}\left(1+(\lambda -1)^{2}+\lambda ^{2}\right)^{2}&=2\left(a^{2}+b^{2}\right)^{2}\left(1+\left((\lambda -1)^{2}\right)^{2}+\left(\lambda ^{2}\right)^{2}\right)\end{aligned}}}
now, if , b represent distinct points, such + b ≠ 0, may divide both sides q(ab) = (a + b):
(
1
+
λ
2
−
2
λ
+
1
+
λ
2
)
2
=
2
(
1
+
(
λ
2
−
2
λ
+
1
)
2
+
λ
4
)
(
2
λ
2
−
2
λ
+
2
)
2
=
2
(
1
+
λ
4
−
2
λ
3
+
λ
2
−
2
λ
3
+
4
λ
2
−
2
λ
+
λ
2
−
2
λ
+
1
+
λ
4
)
4
(
λ
2
−
λ
+
1
)
2
=
2
(
2
λ
4
−
4
λ
3
+
6
λ
2
−
4
λ
+
2
)
4
(
λ
4
−
λ
3
+
λ
2
−
λ
3
+
λ
2
−
λ
+
λ
2
−
λ
+
1
)
=
4
(
λ
4
−
2
λ
3
+
3
λ
2
−
2
λ
+
1
)
λ
4
−
2
λ
3
+
3
λ
2
−
2
λ
+
1
=
λ
4
−
2
λ
3
+
3
λ
2
−
2
λ
+
1
{\displaystyle {\begin{aligned}\left(1+\lambda ^{2}-2\lambda +1+\lambda ^{2}\right)^{2}&=2\left(1+\left(\lambda ^{2}-2\lambda +1\right)^{2}+\lambda ^{4}\right)\\\left(2\lambda ^{2}-2\lambda +2\right)^{2}&=2\left(1+\lambda ^{4}-2\lambda ^{3}+\lambda ^{2}-2\lambda ^{3}+4\lambda ^{2}-2\lambda +\lambda ^{2}-2\lambda +1+\lambda ^{4}\right)\\4\left(\lambda ^{2}-\lambda +1\right)^{2}&=2\left(2\lambda ^{4}-4\lambda ^{3}+6\lambda ^{2}-4\lambda +2\right)\\4\left(\lambda ^{4}-\lambda ^{3}+\lambda ^{2}-\lambda ^{3}+\lambda ^{2}-\lambda +\lambda ^{2}-\lambda +1\right)&=4\left(\lambda ^{4}-2\lambda ^{3}+3\lambda ^{2}-2\lambda +1\right)\\\lambda ^{4}-2\lambda ^{3}+3\lambda ^{2}-2\lambda +1&=\lambda ^{4}-2\lambda ^{3}+3\lambda ^{2}-2\lambda +1\end{aligned}}}
pythagoras s theorem
the lines a1a3 (of quadrance q1) , a2a3 (of quadrance q2) perpendicular (their spread 1) if , if:
q
1
+
q
2
=
q
3
.
{\displaystyle q_{1}+q_{2}=q_{3}.}
where q3 quadrance between a1 , a2.
this equivalent pythagorean theorem (and converse).
there many classical proofs of pythagoras s theorem; 1 framed in terms of rational trigonometry.
the spread of angle square of sine. given triangle △abc spread of 1 between sides ab , ac,
q
(
a
b
)
+
q
(
a
c
)
=
q
(
b
c
)
{\displaystyle q(ab)+q(ac)=q(bc)}
where q quadrance , i.e. square of distance.
illustration of nomenclature used in proof.
construct line ad dividing spread of 1, point d on line bc, , making spread of 1 db , dc. triangles △abc, △dba , △dac similar (have same spreads not same quadrances).
this leads 2 equations in ratios, based on spreads of sides of triangle:
s
c
=
q
(
a
b
)
q
(
b
c
)
=
q
(
b
d
)
q
(
a
b
)
=
q
(
a
d
)
q
(
a
c
)
.
s
b
=
q
(
a
c
)
q
(
b
c
)
=
q
(
d
c
)
q
(
a
c
)
=
q
(
a
d
)
q
(
a
b
)
.
{\displaystyle {\begin{aligned}s_{c}&={\frac {q(ab)}{q(bc)}}&&={\frac {q(bd)}{q(ab)}}&&={\frac {q(ad)}{q(ac)}}.\\s_{b}&={\frac {q(ac)}{q(bc)}}&&={\frac {q(dc)}{q(ac)}}&&={\frac {q(ad)}{q(ab)}}.\end{aligned}}}
now in general, 2 spreads resulting dividing spread 2 parts, line ad spread cab, not add original spread since spread non-linear function. first prove dividing spread of 1, results in 2 spreads add original spread of 1.
for convenience, no loss of generality, orient lines intersecting spread of 1 coordinate axes, , label dividing line coordinates (x1, y1) , (x2, y2). 2 spreads given by:
s
1
=
(
x
2
−
x
2
)
2
+
(
y
2
−
y
1
)
2
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
=
(
y
2
−
y
1
)
2
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
,
s
2
=
(
x
2
−
x
1
)
2
+
(
y
2
−
y
2
)
2
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
=
(
x
2
−
x
1
)
2
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
.
{\displaystyle {\begin{aligned}s_{1}&={\frac {(x_{2}-x_{2})^{2}+(y_{2}-y_{1})^{2}}{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}&&={\frac {(y_{2}-y_{1})^{2}}{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}},\\s_{2}&={\frac {(x_{2}-x_{1})^{2}+(y_{2}-y_{2})^{2}}{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}&&={\frac {(x_{2}-x_{1})^{2}}{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}.\end{aligned}}}
hence
s
1
+
s
2
=
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
=
1
,
{\displaystyle s_{1}+s_{2}={\frac {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}=1,}
so that
s
c
+
s
b
=
1.
{\displaystyle s_{c}+s_{b}=1.}
using first 2 ratios first set of equations, can rewritten:
q
(
a
b
)
q
(
b
c
)
+
q
(
a
c
)
q
(
b
c
)
=
1.
{\displaystyle {\frac {q(ab)}{q(bc)}}+{\frac {q(ac)}{q(bc)}}=1.}
multiplying both sides q(bc):
q
(
a
b
)
+
q
(
a
c
)
=
q
(
b
c
)
.
{\displaystyle q(ab)+q(ac)=q(bc).}
q.e.d.
spread law
for triangle △a1a2a3 nonzero quadrances:
s
1
q
1
=
s
2
q
2
=
s
3
q
3
.
{\displaystyle {\frac {s_{1}}{q_{1}}}={\frac {s_{2}}{q_{2}}}={\frac {s_{3}}{q_{3}}}.}
this law of sines, squared.
cross law
for triangle △a1a2a3,
(
q
1
+
q
2
−
q
3
)
2
=
4
q
1
q
2
(
1
−
s
3
)
.
{\displaystyle (q_{1}+q_{2}-q_{3})^{2}=4q_{1}q_{2}(1-s_{3}).}
this analogous law of cosines. called cross law because (1 − s3), square of cosine of angle, called cross .
triple spread formula
for triangle △a1a2a3,
(
s
1
+
s
2
+
s
3
)
2
=
2
(
s
1
2
+
s
2
2
+
s
3
2
)
+
4
s
1
s
2
s
3
.
{\displaystyle (s_{1}+s_{2}+s_{3})^{2}=2\left(s_{1}^{2}+s_{2}^{2}+s_{3}^{2}\right)+4s_{1}s_{2}s_{3}.}
this relation can derived formula sine of compound angle: in triangle (whose 3 angles sum 180°) have,
sin
(
a
)
=
sin
(
b
+
c
)
=
sin
(
b
)
cos
(
c
)
+
sin
(
c
)
cos
(
b
)
{\displaystyle \sin(a)=\sin(b+c)=\sin(b)\cos(c)+\sin(c)\cos(b)}
.
equivalently, describes relationship between spreads of 3 concurrent lines, spread (like angle) unaffected when sides of triangle moved parallel meet in common point.
knowing 2 spreads allows third calculated solving associated quadratic formula but, since 2 solutions possible, further triangle spread rules must used select appropriate one. (the relative complexity of process contrasts simpler method of obtaining supplementary angle of 2 others.)
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