Question

Aratati ca numarul A=1 supra sin 10 grade - radical din 3 supra cos 10 grade este intreg
Va rooog ajutor!

Answer 1

Salut,

$\dfrac{A}2=\dfrac{1}{2sin10^{\circ}}-\dfrac{\sqrt3}{2cos10^{\circ}}=\dfrac{\dfrac{1}2}{sin10^{\circ}}-\dfrac{\dfrac{\sqrt3}2}{cos10^{\circ}}=\\\\=\dfrac{sin30^{\circ}}{sin10^{\circ}}-\dfrac{cos30^{\circ}}{cos10^{\circ}}=\dfrac{sin30^{\circ}\cdot cos10^{\circ}-cos30^{\circ}\cdot sin10^{\circ}}{sin10^{\circ}\cdot cos10^{\circ}}=\\\\=\dfrac{sin(30^{\circ}-10^{\circ})}{sin10^{\circ}\cdot cos10^{\circ}}\Rightarrow\dfrac{A}4=\dfrac{sin20^{\circ}}{2\cdot sin10^{\circ}\cdot cos10^{\circ}}=\dfrac{sin20^{\circ}}{sin(2\cdot 10^{\circ})}=1.$

Deci A = 4, care este un număr întreg.

Green eyes.