Trigonometria

  • Materia: Trigonometria
  • Visto: 1919
  • Data: 09/01/2010
  • Di: Redazione StudentVille.it

$sin(x+30^\\circ)cosy-cosxcos(y+60^\\circ)-cos(30^\\circ)sin(x+y)$

esercizio svolto o teoria

A cura di: Francesco Speciale

Semplificare la seguente espressione
$sin(x+30^\circ)cosy-cosxcos(y+60^\circ)-cos(30^\circ)sin(x+y)$


$sin(x+30^\circ)cosy-cosxcos(y+60^\circ)-cos(30^\circ)sin(x+y)=$
Tenendo presente le formule di somma del seno e del coseno:

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

Riferendoci alla nosta espressione avremo:
$sin(x+30^\circ)cosy-cosxcos(y+60^\circ)-cos(30^\circ)sin(x+y)=$
$(sinxcos(30^\circ)+sin(30^\circ)cosx)cosy-cosx(cosycos(60^\circ)-sinysin(60^\circ))-(sqrt3)/2(sinxcosy+cosxsiny)=$
$((sqrt3)/2sinx+1/2cosx)cosy-cosx(1/2cosy-(sqrt3)/2siny)-(sqrt3)/2sinxcosy-(sqrt3)/2sinycosx=$
$(sqrt3)/2sinxcosy+1/2cosxcosy-1/2cosycosx+(sqrt3)/2sinycosx-(sqrt3)/2sinxcosy-(sqrt3)/2sinycosx=$
Semplificando si ha che l'espressione
$sin(x+30^\circ)cosy-cosxcos(y+60^\circ)-cos(30^\circ)sin(x+y)=0$.