Formules de duplication, cos(a+b), sin(a+b)….

 
Ci-dessous quelques démonstrations d’inspirations très diverses des formules clés de la trigonométrie .

CALCUL DE \cos(\alpha+\beta) :

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\cos(\alpha+\beta)=OK=OP\times \cos(\alpha)

PH=\tan(\alpha)\times \sin(\beta)

OP+PH=\cos(\beta)

et donc \cos(\alpha+\beta)=(\cos(\beta)-\tan(\alpha)\times \sin(\beta))\times \cos(\alpha)

=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)

CALCUL DE \sin(\alpha+\beta) :

On se souvient de la formule des sinus: \dfrac{BC}{\sin\hat{A}}=\dfrac{AC}{\sin\hat{B}}=\dfrac{AB}{\sin\hat{C}} (*)
valable dans tout triangle.
Soit (AA') la hauteur issue de A

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BC=BA'+A'C=AB\cos\hat{B}+AC \cos \hat{C}
=BC\dfrac{\sin\hat{C}}{\sin\hat{A}}\cos\hat{B}+BC\dfrac{\sin\hat{B}}{\sin\hat{A}}\cos \hat{C} d’après (*)

Puis \sin(\hat{B}+\hat{C})=\sin(\pi-\hat{A})=\sin \hat{A}

=\sin\hat{C}\cos\hat{B}+\sin\hat{B}\cos \hat{C} !

……une tout autre approche, avec le birapport de quatre droites:

Si d_{1},d_{2},d_{3},d_{4} sont quatre droites passant par m, soit \Delta une droite ne passant pas par m et soient A,B,C,D les traces de d_{1},d_{2},d_{3},d_{4} sur \Delta.
Alors on a [[d_{1},d_{2},d_{3},d_{4}]]= [[A,B,C,D]].

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On a bien sur : [[A,B,C,D]] = [[A',B',C',D']]

En particulier: Si \Delta ' parallèle à d_{4} (D’ en l’infini) alors:

\dfrac{\overline{AC}}{\overline{AB}} \div \dfrac{\overline{DC}}{\overline{DB}}=\dfrac{\overline{A'C'}}{\overline{A'B'}}
et \dfrac{\overline{AD}}{\overline{AB}} \div \dfrac{\overline{CD}}{\overline{CB}}=\dfrac{\overline{C'B'}}{\overline{A'B'}}

…et d’après la relation de Chasles:\overline{A'C'}+\overline{C'B'}=\overline{A'B'}
donc \dfrac{\overline{A'C'}}{\overline{A'B'}}+\dfrac{\overline{C'B'}}{\overline{A'B'}}=1

donc \dfrac{\overline{AC}}{\overline{AB}} \div \dfrac{\overline{DC}}{\overline{DB}}+ \dfrac{\overline{AD}}{\overline{AB}} \div \dfrac{\overline{CD}}{\overline{CB}}=1

ce qui se traduit par la relation fondamentale :

    \[\overline{AB}.\overline{CD}+\overline{AC}.\overline{DB}+\overline{AD}.\overline{BC}=0\]

Comme [[d_{1},d_{2},d_{3},d_{4}]]=\dfrac{\sin(d_1,d_3)}{\sin(d_1,d_4)} \div \dfrac{\sin(d_2,d_3)}{\sin(d_2,d_4)},

La relation ci-dessus s’écrit :

\sin(d_1,d_2)\sin(d_3,d_4)+\sin(d_1,d_3)\sin(d_4,d_2)+\sin(d_1,d_4)\sin(d_2,d_3)=0

Si les quatre droites sont issues du centre du cercle unité, les arcs ab, cd, etc., mesurent alors les angles de ces droites.

Dans le cas particulier où bd=\frac{\pi}{2}, \sin(db)=-1, \sin(ad)=\cos(ab) et \sin(cd)=\cos(cb)=\cos(bc) et l’équation ci-dessus devient:

    \[\sin(ab+bc)=\sin(ab).\cos(bc)+\sin(bc).\cos(ab)\]

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