Question

calcular tg 7pi sobre 12

Answer 1

Tg 7pi/12                       lembrando que pi na geometria equivale a 180°
Tg 7 . 180 /12
Tg 105°= 1,83 ...

Espero ter ajudado.

Answer 2

Podemos reescrever o arco de $\frac{7\pi}{12}$ como

$\frac{7\pi}{12}=\frac{3\pi+4\pi}{12}\\ \\ \frac{7\pi}{12}=\frac{3\pi}{12}+\frac{4\pi}{12}\\ \\ \frac{7\pi}{12}=\frac{\pi}{4}+\frac{\pi}{3}$


Aplicando tangente aos dois lados, temos

$\mathrm{tg}(\frac{7\pi}{12})=\mathrm{tg}(\frac{\pi}{4}+\frac{\pi}{3})$


A fórmula da tangente da soma de dois arcos:

$\mathrm{tg}(\alpha+\beta)=\dfrac{\mathrm{tg\,}\alpha+\mathrm{tg\,}\beta}{1-\mathrm{tg\,}\alpha\cdot \mathrm{tg\,}\beta}$


Então, temos que

$\mathrm{tg}(\frac{7\pi}{12})=\dfrac{\mathrm{tg}(\frac{\pi}{4})+\mathrm{tg}(\frac{\pi}{3})}{1-\mathrm{tg}(\frac{\pi}{4})\cdot \mathrm{tg}(\frac{\pi}{3})}\\ \\ \\ \mathrm{tg}(\frac{7\pi}{12})=\dfrac{1+\sqrt{3}}{1-1\cdot\sqrt{3}}\\ \\ \\ \mathrm{tg}(\frac{7\pi}{12})=\dfrac{1+\sqrt{3}}{1-\sqrt{3}}\\ \\ \\ \mathrm{tg}(\frac{7\pi}{12})=\dfrac{(1+\sqrt{3})\cdot (1+\sqrt{3})}{(1-\sqrt{3})\cdot (1+\sqrt{3})}\\ \\ \\ \mathrm{tg}(\frac{7\pi}{12})=\dfrac{(1+\sqrt{3})^{2}}{1^{2}-(\sqrt{3})^{2}}\\ \\ \\ \mathrm{tg}(\frac{7\pi}{12})=\dfrac{1^{2}+2\sqrt{3}+(\sqrt{3})^{2}}{1^{2}-(\sqrt{3})^{2}}\\ \\ \\ \mathrm{tg}(\frac{7\pi}{12})=\dfrac{1+2\sqrt{3}+3}{1-3}\\ \\ \\ \mathrm{tg}(\frac{7\pi}{12})=\dfrac{4+2\sqrt{3}}{-2}\\ \\ \\ \mathrm{tg}(\frac{7\pi}{12})=\dfrac{2\cdot (2+\sqrt{3})}{-2}\\ \\ \\ \mathrm{tg}(\frac{7\pi}{12})=-(2+\sqrt{3})\\ \\ \boxed{\mathrm{tg}(\frac{7\pi}{12})=-2-\sqrt{3}}$