formule goniometriche

SIN(α + β + γ) =

SIN((α + β) + γ) = SIN(α + β)·COS(γ) + SIN(γ)·COS(α + β)

Dati i seni:

SIN(α) = 1/2----> COS(α) = √(1 - (1/2)^2)

COS(α) = √3/2

SIN(β) = 4/5---> COS(β) = √(1 - (4/5)^2)

COS(β) = 3/5

SIN(γ) = 1/3----> COS(γ) = √(1 - (1/3)^2)

COS(γ) = 2·√2/3

Quindi:

SIN(α + β) = SIN(α)·COS(β) + SIN(β)·COS(α)

SIN(α + β) = 1/2·(3/5) + 4/5·(√3/2) = 2·√3/5 + 3/10

COS(α + β) = √3/2·(3/5) - 1/2·(4/5) = 3·√3/10 - 2/5

SIN((α + β) + γ) = (2·√3/5 + 3/10)·2·√2/3 + 1/3·(3·√3/10 - 2/5) =

=(4·√6/15 + √2/5) + (√3/10 - 2/15)= (8·√6 + 3·√3 + 6·√2 - 4)/30

Se {α, β, γ} sono nel primo quadrante allora
* α = arcsin(1/2) = π/6
* β = arcsin(4/5) ~= 0.93
* γ = arcsin(1/3) ~= 0.34
e con
* s = β + γ
* sin(s) = sin(β + γ) = cos(γ)*sin(β) + cos(β)*sin(γ) =
= cos(arcsin(1/3))*sin(arcsin(4/5)) + cos(arcsin(4/5))*sin(arcsin(1/3)) =
= (2*√2/3)*(4/5) + (3/5)*(1/3) =
= (8*√2 + 3)/15
* cos(s) = √(1 - sin^2(s)) = √(1 - ((8*√2 + 3)/15)^2) =
= 2*(3*√2 - 2)/15
si ha
* sin(α + β + γ) = sin(α + s) = cos(α)*sin(s) + cos(s)*sin(α) =
= cos(π/6)*sin(s) + cos(s)*sin(π/6) =
= ((√3)*sin(s) + cos(s))/2 =
= ((√3)*(8*√2 + 3)/15 + 2*(3*√2 - 2)/15)/2 =
= (6*√2 + 3*√3 + 8*√6 - 4)/30 ~= 0.9759
Verifica
http://www.wolframalpha.com/input?i=simplify+sin%28arcsin%281%2F2%29--arcsin%284%2F5%29--arcsin%281%2F3%29%29