Question

alguém ajuda aqui nao consigo fazer !! é a 18

Answer 1

$\tan(15\°)=\frac{\sin(15\°)}{\cos(15\°)}=\frac{\sin(60\°-45\°)}{\cos(60\°-45\°)}\\ \\ \sin(a-b)=\sin a\times\cos b-\sin b\times\cos a\\ \cos(a-b)=\sin a\times\sin b+\cos a\times \cos b\\ \\ \frac{\sin(60\°-45\°)}{\cos(60\°-45\°)}=\frac{\sin (60\°)\times\cos (45\°)-\sin(45\°)\times\cos(60\°)}{\sin(60\°)\times\sin(45\°)+\cos (60\°)\times \cos(45\°)}$ $\\ \\ \frac{\frac{\sqrt3}{2}\times\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\times\frac{1}{2}}{\frac{\sqrt3}{2}\times\frac{\sqrt2}{2}+\frac{1}{2}\times\frac{\sqrt2}{2}}=\frac{\frac{\sqrt6}{4}-\frac{\sqrt2}{4}}{\frac{\sqrt6}{4}+\frac{\sqrt2}{4}}=\frac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}=2-\sqrt3$