$sqrt(sin^2(105^\circ)+cos^2(15^\circ)-2sin(60^\circ)cos(60^\circ))$ - Studentville

$sqrt(sin^2(105^\circ)+cos^2(15^\circ)-2sin(60^\circ)cos(60^\circ))$

esercizio svolto o teoria

A cura di: Francesco Speciale

Calcolare il valore della seguente espressione:
$sqrt(sin^2(105^circ)+cos^2(15^circ)-2sin(60^circ)cos(60^circ))$


$sqrt(sin^2(105^circ)+cos^2(15^circ)-2sin(60^circ)cos(60^circ))$
Noi sappiamo che $sin(105^circ)=sin(60^circ+45^circ)=sin(60^circ)cos(45^circ)+cos(60^circ)sin(45^circ)=$
$=(sqrt3)/2((sqrt2)/2)+1/2(sqrt2)/2=(sqrt6)/4+(sqrt2)/4=1/4(sqrt6+sqrt2)$,
$sin(60^circ)=(sqrt3)/2 , cos(60^circ)=1/2$,
$cos(15^circ)=cos(45^circ-30^circ)=cos(45^circ)cos(30^circ)+sin(45^circ)sin(30^circ)=$
$=(sqrt2)/2(sqrt3)/2+(sqrt2)/2*1/2=(sqrt6)/(4)+(sqrt2)/(4)=1/4(sqrt6+sqrt2)$,

sostituiamo i valori noti nell’espressione e risolviamola:
$=sqrt((1/4(sqrt6+sqrt2))^2+(1/4(sqrt6+sqrt2))^2+2*((sqrt3)/2)*1/2)=$
$=sqrt(2(1/4(sqrt6+sqrt2))^2-(sqrt3)/2)=sqrt(2(1/(16)(6+2+2sqrt(12)))-(sqrt3)/2)=$
$=sqrt(1/8(8+4sqrt3)-(sqrt3)/2)=sqrt(1+1/2sqrt3-(sqrt3)/2)=sqrt1=1$.

  • Trigonometria

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