Question

calcule sem o uso da calculadora a tangente de 105

Answer 1

Olá!

Use a soma de ângulos:

$tg\:105°=tg\:(\frac{\pi}{4}+\frac{\pi}{3})$

Sabendo que $tg(\alpha+\beta)=\frac{sen(\alpha+\beta)}{cos(\alpha+\beta)}$

$\frac{sen(\alpha)cos(\beta)}{cos(\alpha)cos(\beta)-sen(\alpha)sen(\beta)}+\frac{cos(\alpha)sen(\beta)}{cos(\alpha)cos(\beta)-sen(\alpha)sen(\beta)}$

$\frac{sen(\frac{\pi}{4})cos(\frac{\pi}{3})}{cos(\frac{\pi}{4})cos(\frac{\pi}{3})-sen(\frac{\pi}{4})sen(\frac{\pi}{3})}+\frac{cos(\frac{\pi}{4})sen(\frac{\pi}{3})}{cos(\frac{\pi}{4})cos(\frac{\pi}{3})-sen(\frac{\pi}{4})sen(\frac{\pi}{3})}$

$\frac{\frac{\sqrt{2}}{2}\frac{1}{2}}{\frac{\sqrt{2}}{2}\frac{1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}}+\frac{\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}\frac{1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}}$

$tg(\frac{\pi}{4}+\frac{\pi}{3})=\fbox{-2-\sqrt{3}}$

Espero ter ajudado.

Bons estudos!