Triple spread formula

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’

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