Archive

Monthly Archives: October 2014

Rational Trigonometry:

deriving the TSF from a standard trigonometric identity

The Triple Spread Formula (TSF)

(SA + SB + SC)² = 2 (SA² + SB² + SC²) + 4 SASBSC

features prominently in rational trigonometry, but its derivation (via the Cross Law) may not satisfy everyone coming across it from a standard trigonometrical background, steeped in angles. In particular, how is the TSF really ‘equivalent’ to the familiar sum of angles condition?

Well, the TSF is derived from the sine-ratios of vertices, not the angular measures of those vertices, so the derivation must use a known relationship on the sines in a triangle to stand in for the simpler (but less tractable) angles themselves:

Since the angles in a triangle sum to 180º and the sine of an angle equals the sine of its supplement, it follows that:

Sin(A) = Sin(B+C) = Sin(B)Cos(C) + Sin(C)Cos(B).

Square both sides, noting Cos²(B) = 1 – Sin²(B) = 1 – SB etc

⇒ S = SB(1 – SC) + 2 Sin(B)Sin(C)Cos(B)Cos(C) + SC(1 – SB)

Distribute and collect like terms (i.e. spreads) on one side:

⇒ S – (SB + SC) + 2 SBSC  = 2 Sin(B)Sin(C)Cos(B)Cos(C)

Square both sides a second time to ‘clear’ the ordinary trigonometric ratios:

⇒ (S – (SB + SC))² + 4 (S – (SB + SC))SBSC + 4 (SBSC)² = 4 SBSC(1 – SB)(1 – SC)

 But we want a symmetric form to appear in the final formula, hence we substitute:

     (S + (SB + SC))² – 4 SA(SB + SC)  ≡  (S – (SB + SC))²

⇒ (S +  SB + SC)²  –  4 SA(SB + SC) + 4(S – (SB + SC))SBSC + 4 (SBSC

= 4 SBSC(1 – SB)(1 – SC)

= 4 SBSC(1 – (S+ SC) + SBSC)

Distribute both sides and cancel any like terms we find:

⇒ (S +  SB + SC)² – 4 SASB – 4 SASC + 4 SASBSC – 4 SBSC(SB + SC) + 4 (SBSC

= 4 SBSC                                                                            – 4 SBSC(S+ SC) + 4 (SBSC

⇒ (S +  SB + SC)² – 4 SASB –  4 SASC + 4 SASBSC

= 4 SBSC

We’re almost there! Take everything except the perfect square to the right hand side:

⇒ (S +  SB + SC)²  = 4 SASB + 4 SASC + 4 SBSC – 4 SASBSC

Note the three cross terms appearing together and recall the identity

(S +  SB + SC)²  ≡ (SA² +  SB² + SC²) + 2  (SASB + SASC+ SBSC)

⇒ 4 (SASB + SASC + SBSC)  ≡  2 (S +  SB + SC)²  –  2 (SA² +  SB² + SC²)

Therefore our equation becomes:

(S +  SB + SC)²  = 2 (S +  SB + SC)²  –  2 (SA² +  SB² + SC²) – 4 SASBSC

and simple rearrangement (subtract twice the perfect square and switch sign) gives

(S +  SB + SC)²  =   2 (SA² +  SB² + SC²) + 4 SASBSC

Which is what we required to show.

Hopefully this derivation is of some use to the reader since (after obtaining it one time to satisfy myself) I couldn’t easily recreate it, so it can sit here as a reference to what we ‘know’ to be the case but may not have at our ‘fingertips’

This gallery contains 1 photo.

[6] . Rational Trigonometry:   Spread in a circle   Part 6 –  Part 1   –    Part 2   –    Part 3   –    Part 4   –    Part 5   – putting everything together: three spreads in a single construction. Click for a larger image THE figure above has a pair of lines through B (broken) at position ‘1’ on …

Read More

[5]

.

Rational Trigonometry:   Spread in a circle   Part 5

 Part 1   –    Part 2   –    Part 3   –    Part 4   –    Part 6   –

constructions for measuring green spread

Due to the triadic nature of perpendicular relations noted, and since a green circle is a rotated form of red circle, we essentially have two ways to use our method of measuring spread to construct a measure of green spread for a pair of oblique lines, as illustrated in the following figure:

Figure 9Click for a larger image

Here, a green circle has been formed by rotating a red circle about an ‘orgin’ (the point of contact with the blue circle at A) instead of their common centres.

The spread at B can be translated, unchanged, to C (the rotated position of point B) on the green circle. Thus, measuring the green spread at B using the red circle as reference (along diameter AB) becomes geometrically equivalent  to measuring the red spread at C using the green circle as reference (along AC) instead.

In this position, with respect to axis AC, each of the lines forming the spread at C meet the green circle at points ‘red’ perpendicular from A. Due to rotation of axes about A however, these directions are also ‘green’ perpendicular to those lines with respect to the axes of the red circle as noted.

Starting from A, when we follow the same sequence of directions between the untranslated lines (measured along AB) or between the translated lines (measured along AC), whilst moving differing amounts in each direction we return to either axis of measurement at the same relative displacement – the measured green spread at B (or C)

Start

______________________________________________________________________________________________________________

[4]

.

Rational Trigonometry:   Spread in a circle   Part 4

 Part 1   –    Part 2   –    Part 3   –    Part 5   –    Part 6   –

further (non-Euclidean) perpendicularity & parallelism

By considering the dot products which are the basis of blue and red quadrance, namely:

BLUE:    (x– x1)(x– x1) + (y– y1)(y– y1)  = (x2x1)² + (y2y1)² = QB(1,2) [⇒  k1(a,b) k2,(-b’,a’)]

RED:     (x– x1)(x– x1) –  (y– y1)(y– y1)  = (x2x1)²  – (y2y1)² = QR(1,2) [⇒  k1(a,b) k2,(b’,a’)]

(with a = a’ and b = b’) we note there is a third dot product available. If a line or vector passes through a point in direction (a,b) from the origin, and another lines passes a point lying in direction   (-a’,b’)  from that origin, the lines are ‘green’ perpendicular since, for any scalars k1 and k2, the inner product

k1(a, b) dotG k2(-a’, b’) =  k1k2(ab’ + b(-a’)) ≡ 0 ⇔ k1(a,b) k2,(-a’,b’)

We therefore introduce a third form of quadrance (and associated perpendicularity) derived from this green dot product:

GREEN:    (x– x1)(y– y1) + (y– y1)(x– x1)  = 2 (x2x1)(y– y1) = QG(1,2) [⇒  k1(a,b) k2,(-a’,b’)]

Green quadrance from a point to the origin is ‘1’ on the rectangular hyperbola y = 1/(2x), and is 0 along the usual coordinate axes (asymptotic directions). Such a hyperbola is also obtained by rotating a standard curve,  –  = 0, 45 degrees about a central origin, with its axes along these lines. Green perpendicular lines through (a,b) and (-a’,b’) are symmetric with respect to the usual (red/blue) axes, while a green perpendicular to a line is blue perpendicular to that line’s red perpendicular direction. More succinctly, with respect to an origin of measurement:

k1(a,b)  k2,(b’,a’) k3,(-a’,b’) k4(a’,b’)

This is shown also by the following figure:

Figure 8Click for a larger image

A green circle is therefore a rotated red circle.

A fourth variation occurs when we consider vectors from the origin through points (a,b) and (-a’,-b’) (with a = a’ and b = b’ again) that describe a parallelism of lines which necessarily have 0, null or ‘vanishing’ quadrance (and spread) ‘between’ them

PARALLEL: Q(1,2) = (x2x1)(y1y2) + (x1x2)(y2y1) ≡ 0 [⇒.  k1(a,b) k2,(-a’,-b’)]

Each variation of coloured geometry: blue, red and green thus uses it own perpendicular measure (spread =1) together with a common parallel measure (spread =0) and can be related by composition.

Start

______________________________________________________________________________________________________________

[3]

.

Rational Trigonometry:   Spread in a circle   Part 3

 Part 1   –    Part 2   –    Part 4   –    Part 5   –    Part 6   –

Spread measurement outside a circle involving rotation

.                                                                                                                                             IN PART 2 we saw how a single line will make both a red and blue spread measured by the x coordinates of their meeting point on each circle. [As a side note, the x positions giving each spread are related by circle inversion].                                                   IN PART 1 Where a spread was made by two oblique lines, meeting at B, a rigid rotation was initially used to bring one into alignment with the x-axis, and the blue spread was measured once more by dropping a perpendicular from the point on the circle to the diameter.

With red geometry, the idea of a rotation must be adapted as shown in the following figure.

Figure 6

Click for a larger image

In the case of a blue rotation, the lower (brown) figure enclosing the required arc AA’ is reflected in the (dotted) radial line so that AA’ now lies on C’C. This is also equivalent to either a reflection in the parallel line through A and a translation from A to C’ or a translation from A to C’ followed by reflection in the parallel line through C’. The two arcs are copied in the limit, considering smaller and smaller rotations from A’ to A, because each quadrance, and each spread made with parallel lines, is preserved by isometries of reflection and translation. (A rigid rotation captures this idea, geometrically, but does not define it.)

For a red rotation, the transformations use the same radial line but ‘reflection’ corresponds to an opposite, ‘equi-quadrant’, displacement of a point, across the line, measured in the red perpendicular sense. In the limit, this also results in arc A’RA being ‘copied’ to arc C’RCR. Although appearing unequal the arcs, if measured by their chords alone, have equal red quadrance and spread with the parallel lines through A and C respectively.

Hence, under a red rotation, arc A’RAC’R corresponds with AC’RCR.

With bases of the red triangles being met by red perpendicular bisectors, through the third vertices at A and C’R, as in the case of blue reflections, red congruent triangles are formed. However, it now becomes apparent that while a red rotation is a useful description of the process for measuring spread for oblique lines geometrically, it is not the rotation which matters but the use of red isometry to construct a chord of equal red quadrance to the initial separation A’RC’R (not drawn) measured from the desired endpoint A – namely chord ACR. This also shows how spread can be measured without an auxiliary red circle as in the blue case.

Start

______________________________________________________________________________________________________________

[2]

.

Rational Trigonometry:   Spread in a circle   Part 2

 Part 1   –    Part 3   –    Part 4   –    Part 5   –    Part 6   –

Spread measurement outside a circle

Figure 5

Click for a larger image

By symmetry, the spread at vertex B can be seen to equal the spread at several other vertices in the above figure, for example the spread at vertex C:

S(C)   = x²/h²   =  x² / (x² + y²)   = x²/(x² +  x (1 – x))   =  x

And, even more directly:

S(B)   =  h²/1²   =    (x² + y²)      =   (x² +  x (1 – x))     =  x

(In a unit circle therefore, the quadrance of a chord equals the opposite subtended spread, and is given by the chord’s projection on a diameter on which it sits)


We now introduce a new form of quadrance, based on a difference rather than a sum of squares – and measure its associated spread value, so-called red quadrance (and spread):

QR(1,2) = (x2 – x1)² – (y2 – y1)².

From similarity in the figure above we have:

   ED/EB (in DEB)  = y’/ (1 + x’)

= EA/ED (in AED)  = x’/y’,

therefore

y’/ (1 + x’)   =  x’/y’      ⇒      y’² = (x’)(1 + x’)

in terms of the definition of red quadrance therefore, 

QR(AD)    =   x’² –  y’²     =    x’² – (x’)(1 + x’)     =    -x’ 

Note that the vertex at D with spread S(DA,DB) is formed by reflecting AC (perpendicular to DB) through the vertical line at A, making lines DA and DB symmetrical with respect to lines of slope +1 and -1 through D as may be verified from the figure above (AD is rotated as far from the vertical – and in an opposite sense – as BD is rotated from the horizontal). This condition of being reflected in lines at ’45 degrees’ to the usual coordinate axes defines perpendicularity in the geometry of red quadrances and spreads. 

Specifically, if one line passes through a point in direction (a,b) from the origin, and another passes through a point lying in direction (b’,a’) from that origin, the lines are red perpendicular since their red dot product is 0. That is, for any scalars k1 and k2 (with a=a’ and b=b’)

k1(a, b) dotR k2(b’, a’) =  k1k2(ab’  (ba’)) =  0 ⇔ k1(a,b)  k2,(b’,a’) 

This contrasts with the usual Euclidean dot product (which henceforth we identify by the prefix blue) that vectors in directions (a,b) and (b’,-a’) through the origin are blue perpendicular since, for any scalars k1 and k2

k1(a, b) dotB k2(b’, -a’) =  k1k2(ab’ + b(-a’)) =  0 ⇔ k1(a,b)  k2,(b’,-a’) 

Thus in triangle DAB vertex D is a red perpendicular, so QR(AD)/QR(AB) defines an equivalent spread ratio at (opposite) vertex B. Being parallel to the x direction, the (red) hypotenuse quadrance of diameter AB remains a unit and therefore 

SR(B)    =   QR(AD)/1   =   x’ 

Note that the sign of this red spread and quadrance corresponds to the displacement of segment AE to the left of the ‘0’ position used in the measurement of blue spread.  This is not accidental. Allowing for sign, the red spread of vertex B corresponds to the x coordinate of the exterior similar triangle formed.

Finally we note the locus of point D, as segment AC is moved on auxiliary circle, (x – ½)² + y² = (½)² is the rectangular hyperbola (x – ½)² – y² = (½)², with a common centre at (½,0), touching on common ‘diameter’ AB. Accordingly we henceforth identify the locus of red perpendicularity on this diameter as a red circle and our previous circle with the prefix blue, putting both objects on equal footing. The noted equal but opposite rotations of lines passing through fixed points A and B always meet (red) perpendicularly on this curve, just as equal rotations of lines passing through fixed points will meet (blue) perpendicularly on a classic circle.

Start

______________________________________________________________________________________________________________

[1]

.

Rational Trigonometry:   Spread in a circle   Part 1

 Part 2   –    Part 3   –    Part 4   –    Part 5   –    Part 6   –

Spread measurement in a circle 

Following the definition of spread as the ratio of an opposite quadrance to the hypotenuse quadrance in a right-angled triangle, consider (following Euclid’s theorem on intersecting chords) forming two similar right-angled triangles in the semicircles ACB and AC’B. Let the diameter AB equal 1.

Figure 1

Then, comparing corresponding sides (noting OC = OC’ by symmetry) 

x/y (in AOC’) = y/(1 – x) (in COB) and therefore y² = x (1 – x).

But since (in COB) h² = y² + (1 – x)² = x (1 – x) + (1 – x) (1 – x) = (1) (1 – x)

the spread at vertex B, s(B) = y² / h² = x (1 – x) / (1 – x) = x (*)

This only measures spread when one of the lines is a diameter of the circle however, so what about a spread between two oblique lines?


Figure 2

Beginning with chords A’B and C’B, if we rotate A’B onto the diameter AB this takes C’B onto chord CB. We can see that chords/arcs A’A and C’C need to be equal.

The perpendicular to C’B through A’, which meets the circle also at C, is parallel to chord AC’, also perpendicular to C’B (a right angle subtended by a diameter). Parallel chords, with midpoints collinear with the circle centre, are bisected by a common radius which bisects the associated inner and outer arcs A’AC’C and AC’ leaving arcs A’A and C’C equal by symmetry, and making A’AC’ and AC’C equal, This allows spread to again be measured by dropping a perpendicular from C to diameter AB at O.

It thus appears a circle provides direct connection with the geometric measurement of spread. However, it is not the circle that determines this construction but the property of perpendicularity. A circle serves simply as a device for constructing these perpendiculars, suggesting that spread could instead be constructed without an auxiliary circle. Is this possible?


Let us re-examine our basic figure by adding further parallel segments:


Figure 3

Here a second perpendicular, this time to chord A’B through point C’, has been drawn. As expected by symmetry this meets the circle, at point D, on the same perpendicular line through C and O.

We observe that new segments, C’G and A’G, are parallel and equal to existing segments AA’ and AC’ respectively. Since AA’ was equal by symmetry to C’C, C’G equals C’C. Since A’C is perpendicular to C’B, triangles C’GE and C’CE are congruent, making segments GE and CE equal, and E (where perpendicular A’C meets C’B) the midpoint of GC. Likewise, point F (on A’B) is midpoint to GD. Finally, O is necessarily the midpoint of chord CD.

Putting this together, triangle OEF will be medial to triangle CDG, ensuring that perpendicular segments from A to lines C’B or A’B followed by parallel segments  C’F or A’E followed by parallel segments from  F or E  meet on the diameter at the same position as the previously dropped perpendicular from C or D on the circle.

This construction measures the spread of oblique lines without the need for an auxiliary circle.  We amend our basic figure accordingly:


Figure 4

Here we leave chord CD of the previous construction visible for reference. The only remaining objects necessary to obtain a measurement of spread are:

  • a pair of oblique lines through point B (whose spread is to be measured)
  • a reference point A (defining the unit segment AB)
  • a pair of perpendiculars through A to the given pair of lines through B
  • a segment parallel to the alternate line from A perpendicular to the alternate line through B
  • a second segment parallel to the original line from A meeting AB at position O.

The spread at B is then the ratio of segment AO to AB.

Start

______________________________________________________________________________________________________________