Trigonometria

  • Materia: Trigonometria
  • Visto: 1808
  • Data: 09/01/2010
  • Di: Redazione StudentVille.it

$(sin(\\alpha+60^\\circ)+sin(\\alpha-60^\\circ)-2sin(\\alpha))/(cos(\\alpha+60^\\circ)+cos(\\alpha-60^\\circ)

esercizio svolto o teoria

A cura di: Francesco Speciale

Semplificare la seguente espressione
$(sin(\alpha+60^\circ)+sin(\alpha-60^\circ)-2sin(\alpha))/(cos(\alpha+60^\circ)+cos(\alpha-60^\circ)-2cos(\alpha))$



$(sin(\alpha+60^\circ)+sin(\alpha-60^\circ)-2sin(\alpha))/(cos(\alpha+60^\circ)+cos(\alpha-60^\circ)-2cos(\alpha))=$
Tenendo presente le formule di somma e differenza del seno e del coseno:

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

Riferendoci alla nosta espressione avremo:
$(sin(\alpha+60^\circ)+sin(\alpha-60^\circ)-2sin(\alpha))/(cos(\alpha+60^\circ)+cos(\alpha-60^\circ)-2cos(\alpha))=$
$(sin(\alpha)cos(60^\circ)+cos(\alpha)sin(60^\circ)+sin(\alpha)cos(60^\circ)-cos(\alpha)sin(60^\circ)-2sin(\alpha))/(cos(\alpha)cos(60^\circ)-sin(\alpha)sin(60^\circ)+cos(\alpha)cos(60^\circ)+sin(\alpha)sin(60^\circ)-2cos(\alpha))=$
Semplificando
$(2sin(\alpha)cos(60^\circ)-2sin(\alpha))/(2cos(\alpha)cos(60^\circ)-2cos(\alpha))=$
Essendo $cos(60^\circ)=1/2$, sostituendo nell'espressione otteniamo:
$(2*1/2sin(\alpha)-2sin(\alpha))/(2*1/2cos(\alpha)-2cos(\alpha))=$
$=(sin(\alpha)-2sin(\alpha))/(cos(\alpha)-2cos(\alpha))=(-sin(\alpha))/(-cos(\alpha))=tg(\alpha)$.