#### Rational Trigonometry:

#### deriving the TSF from a standard trigonometric identity

The Triple Spread Formula (TSF)

##### (S_{A} + S_{B} + S_{C})² = 2* *(S_{A}² + S_{B}² + S_{C}²) + 4* *S_{A}S_{B}S_{C}

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 – S_{B} etc

⇒ S_{A } = S_{B}(1 – S_{C}) + 2 Sin(B)Sin(C)Cos(B)Cos(C) + S_{C}(1 – S_{B})

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

⇒ S_{A } – (S_{B} + S_{C}) + 2 S_{B}S_{C} = 2 Sin(B)Sin(C)Cos(B)Cos(C)

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

⇒ (S_{A } – (S_{B} + S_{C}))² + 4 (S_{A } – (S_{B} + S_{C}))S_{B}S_{C} + 4 (S_{B}S_{C})² = 4 S_{B}S_{C}(1 – S_{B})(1 – S_{C})

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

(S_{A } + (S_{B} + S_{C}))² – 4 S_{A}(S_{B} + S_{C}) ≡ (S_{A } – (S_{B} + S_{C}))²

⇒ (S_{A } + S_{B} + S_{C})² – 4 S_{A}(S_{B} + S_{C}) + 4(S_{A } – (S_{B} + S_{C}))S_{B}S_{C} + 4 (S_{B}S_{C})²

= 4 S_{B}S_{C}(1 – S_{B})(1 – S_{C})

= 4 S_{B}S_{C}(1 – (S_{B }+ S_{C}) + S_{B}S_{C})

Distribute both sides and cancel any like terms we find:

⇒ (S_{A } + S_{B} + S_{C})² – 4 S_{A}S_{B} – 4 S_{A}S_{C} + 4 S_{A}S_{B}S_{C} – 4 S_{B}S_{C}(S_{B} + S_{C}) + 4 (S_{B}S_{C})²

= 4 S_{B}S_{C} – 4 S_{B}S_{C}(S_{B }+ S_{C}) + 4 (S_{B}S_{C})²

⇒ (S_{A } + S_{B} + S_{C})² – 4 S_{A}S_{B} – 4 S_{A}S_{C} + 4 S_{A}S_{B}S_{C}

= 4 S_{B}S_{C}

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

⇒ (S_{A } + S_{B} + S_{C})² = 4 S_{A}S_{B} + 4 S_{A}S_{C} + 4 S_{B}S_{C} – 4 S_{A}S_{B}S_{C}

Note the three cross terms appearing together and recall the identity

(S_{A } + S_{B} + S_{C})² ≡ (S_{A}² + S_{B}² + S_{C}²) + 2 (S_{A}S_{B} + S_{A}S_{C}+ S_{B}S_{C})

⇒ 4 (S_{A}S_{B} + S_{A}S_{C} + S_{B}S_{C}) ≡ 2 (S_{A } + S_{B} + S_{C})² – 2 (S_{A}² + S_{B}² + S_{C}²)

Therefore our equation becomes:

(S_{A } + S_{B} + S_{C})² = 2 (S_{A } + S_{B} + S_{C})² – 2 (S_{A}² + S_{B}² + S_{C}²) – 4 S_{A}S_{B}S_{C}

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

##### (S_{A } + S_{B} + S_{C})² = 2 (S_{A}² + S_{B}² + S_{C}²) + 4 S_{A}S_{B}S_{C}

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’