A cura di: Francesco Speciale

Calcolare il valore esatto delle funzioni goniometriche del seguente arco: $48^circ$.


Svolgimento
Osserviamo che $(48^circ)=(30^circ)+(18^circ)$; per le formule di addizione del seno, coseno e tangente, si ha:

$sin(alpha+beta)=sin(alpha)cos(beta)+cos(alpha)sin(beta)$
$cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)$
$tg(alpha+beta)=(tg(alpha)+tg(beta))/(1-tg(beta)tg(alpha))$
Nel nostro caso $alpha=30^circ , beta=18^circ$, sostituendo otteniamo

$sin(48^circ)=sin(30^circ+18^circ)=sin(30^circ)cos(18^circ)+cos(30^circ)sin(18^circ)=$
$=1/2(1/4(sqrt(10+2sqrt5)))+(sqrt3)/2*1/4(sqrt5-1)=(sqrt(10+2sqrt5))/8+(sqrt(15)-sqrt3)/8=1/8(sqrt(10+2sqrt5)+sqrt(15)-sqrt3)$.
$cos(48^circ)=cos(30^circ+18^circ)=cos(30^circ)cos(18^circ)-sin(30^circ)sin(18^circ)=$
$=(sqrt3)/2*(1/4(sqrt(10+2sqrt5)))-1/2*1/4(sqrt5-1)=(sqrt3)/8(sqrt(10+2sqrt5))-1/8(sqrt5-1)=1/8(sqrt(3(10+2sqrt5))-sqrt5+1)$.
$tg(48^circ)=tg(30^circ+18^circ)=(tg(30^circ)+tg(18^circ))/(1-tg(18^circ)tg(30^circ))=$
$=((sqrt3)/3+sqrt(1-2/5sqrt5))/(1-(sqrt3)/3*(sqrt(1-2/5sqrt5)))=((sqrt3+3sqrt(1-2/5sqrt5))/3)/((3-sqrt3-6/5sqrt5)/3)=$
$=(sqrt3+3sqrt(1-2/5sqrt5))/(3-sqrt3-6/5sqrt5)$.